S.K o ze I, E R as I-i Ba , S.SlAVATiNskii
MiR PublishERS Moscow
COLLECTED
PDOELEMS
IN
PHYSICS
C.M. Koscji, 9.H. PamGa
€ .A. CnauaTHHCKiiii
CSopH H K
3a,a,a*i
no <|)H3HKe
HaA^TCjiiiCTno «IIayKa»
MocKBa
rjiamiaa pe/iai<u,nn
(J)H3nKO-MaTCMaTnqocKoii juiTcpaTypi.i
First published 1986
Revised from the 1979 Russian edition
Ha ansjiuucKOM nabixe
(6) TjiaBHafl peflai<i;HH $m3Hko-matoMat.viiiec ko ii jiHTepaTypu
H3AaTejitcTBa «HayKa», 1978.
© English translation, Mir Publishers, 1986.
S.KozeI, ER asI-iBa , S.SlAVATiNskii
COLLECTED
PPCDLEHS
IN
PHYSICS
Translated from the Russian
by VALERII ILYUSHCUENKO ,
Cand. Sc. (Phys.-Math.)
MiR PublishERs Moscow
CONTENTS
Preface
........................................................
Authors’N o l o .........................................................
0
9
P r o b le m s ................................................
I.
II.
Mechanics
......................................
Thermodynamics and Molecular
Physics
...........................................
III. Electricity and Magnetism . . . .
!
IV. O p t i c s .............................................
V. Atomic P h y s ic s ................. .
VI.
Radiation
......................................
VII. Solid S t a t e ...................... .
VIII.
Nuclear Physics
11
19
27
41
50
56
57
03
Answers and S o l u t i o n s ..........................................................
69
A p p e n d i c e s ........................................................................
173
173
175
I. Constants and T a b le s ...........
II.
Some Non-SystemUnits . . . . .
III. Units of Some Physical Quantities
in the SI and CGS Systems and
Their Relationships
. . . . ; .
IV. Elementary Particle Table . . . .
V. Prefixes and Powers for Decimal
Multiplo and Fractional Units . . .
Periodic Table of E lcm o n ts..............................
176
178
ISO
182
PREFACE
Just as fairy-tales assist children to learn about the adult
world, so problems serve the same purpose by introducing a
student to physics. Problems truly do have much in common
with Jolklore, giving the student an idea of the world of phys­
ics, how it is described, and how it can be understood. Elemen­
tary problems deal with an imaginary world of point masses,
weightless threads, ideal gases, and other perfect bodies, in the
same way as fairy-tale worlds arc populated with fierce ser­
pents and handsome princes who travel on magic flying carpets
in search for firebirds. In such a world the Powers of Good
and Evil are distinct and moral problems arc notable for their
clarity and single answers. Problems also allow us to consider
situations that are almost unreal or even fantastic, and hence
like fairy-tales stimulate our imagination. Step by step more
advanced problems bring us nearer to our complicated picture
of the reality of scientific research, where in many cases a great
deal of effort Is needed even to formulate the questions and where
in the end deeper investigation often results in an expansion
of our understanding, enabling us to reinterpret our initial
problem. The same is true with the problems in this collection.
In many cases a more profound analysis will require either
new calculations or more serious reflections.
Problems and fairy-tales are also alike in respect of their
authorship. It is rare to know exactly who originated a par­
ticular problem while established sections such as mechanics
contain a collection of traditional plots which are rearranged
by each generation in its own fashion. Most of these problems
have been extracted from the archives of the Moscow PhysicoTechnlcal Institute (MPTI). The compilers of this collection.
Professors S. M. Kozel, E. I. Ilashba, and S. A. Slavattnskii, have accomplished a great deal by systematically editing
the whole, of this voluminous archives for the first time: until
now problems were only published individually. The compilers
0
have succeeded In preserving the vivid and ing.nuous approach
which seems to be characteristic of modern actiiv research
physicists.
This problems book exemplifies the “
Phystech System", in
which active scientists take direct part in the education and
training of the next generation of researchers and engineers.
This direct link is especially Important when teaching the fun­
damental branches of the natural sciences of which physics and
mathematics are the principal ones. The gap between a stu­
dent's speciality and general education Is thus narrowed and
eliminates the delay between demands and supply which arises
when general courses are taught exclusively by full-time edu­
cators. The problems book thus continues the traditions of
teaching physics laid down by Academicians P. L. Kapitza
and L.D. Landau when they founded the MPTI. We are all
aware of the originality of P. L. Kapitza's own problems, as
they have been published in a number of books, and the posi­
tion given to problems in the theoretical physics curriculum de­
vised by L.D. lAindau and E. M. Lifshiti.
To a certain extent the problems themselves define the stand­
ards required of students on the general physics course at the
Moscow Physlco-Technical Institute. This is because problems
are a much better measure of the demands placed on a student
than curricula. Il is thus no coincidence that many problems
have been taken from the Final Examimilion, which is the sum­
mit of an MPTI course of general physics. The Final Examina­
tion is set by a commission of physicists from all the basic chairs
and departments at the MPTI. A student must present both
written solutions to the problems set and an essay on a subject
he or she has selected. The problems suggested for this examination
are designed so as to include several branches oj physics at once.
When solving a problem a student must have all the skills of
general physics, viz. the ability to idealize phenomena, to ap­
ply conservation laws, symmetry and invariance conditions,
and similarity principles, to isolate dimensionless parameters,
and to interpret results under limiting conditions. Another
demand consistently required of students ts that the solution
be carried through to a numerical answer. It is important that
the consequent numbers hare not only a real sense, hut are use­
ful and help the student remember necessary data.
New physical problems mirror the lime and circumstances
of their formulation much better than do the curricula of
courses in general physics. In fad, wc ran and must reflect every
significant advance in physics in the problems we set if we are
7
to leach students at the pace of the advance in scientific thought.
The two most active brunches vj fundamental research, i.e.
the physics of outer space and astrophysics, on one hand, and
elementary particle physics, on the other, are an inexhaustible
source of topics and inspiration for problems in general
physics. The many years of mankind's experience of learning
how to understand natural phenomena demonstrate that the
most important fundamental properties of matter become clear
when it is in some limiting form. The extremum conditions of
nature engender a vast scope for imagination and can be used
to test estimates and conclusions by direct and often instruc­
tive applications of the principal laws of physics. It is to be
hoped, therefore, that some problems will form the basis of
more thorough analyses in seminars, during which the solutions
given by the compilers can be developed.
Finally, / would ask both instructors and, especially, stu­
dents to search for and invent their own problems. With our
modern, somewhat pragmatic, predilection in education, or
rather in teaching methods, solving ready-made problems is
given much thought, but it is important that in the necessary
inculcation of the habits needed to solve problems, in which
are sown the seeds of scientific work, we should not inhibit
imagination or suppress creative intuition.
Of all tasks the most important is to correctly stale the prob­
lem. This proposition is valid both when a scientist is being
educated and during his later career. Yet to teach someone hou)
to do this is the hardest of all tasks.
S. P. Kapilza
AUTHORS’NOTE
This collodion includes problems that have boon set
over many years to physics students at the (Moscow PhysicoTechnical Institute. The initiator of this (project was Pro­
fessor S. P. Kapitza, the chairman of the Physics Chair.
The problems we chose from the department's voluminous
archives are those which scorned the most interesting from
the physical point of view. There is a rather large proportion
of advanced probloms, and so many of jthe problems arc
provided with solutions or detailed instructions. At tho
same time we felt in a number of cases that there is a noed
for some to remain without full solution^ and they only
have final answers. The SI system of units is preferred
though tho absolute C.iiiissiim system is u(lso used.
Note that most of the problems arc original and have been
suggested by the si aIT of ihn dopnrlinonl. In this sonse tho
collection is the joint work of the wholo department al­
though all responsibility for potential oversights is entire­
ly ours. Wo distributed the work involved, thus Sections
III and IV were preparod by S. M. Kozcl, Sections II, VI,
and VII by E. I. Rnshba, while Sections I, V, and VIII
wero prepared by S. A. Slavalinskii. Many of tho problems
in the collection were suggested by Professors M. D. Galanin, A. D. Gladun, L. L. Gol’
din, B. G. Erozolimskii,
S. P. Kapitza, L. A. Mikaclyan, and D. V. Sivtikhin, ami
by Assistant Professors D. B. Dinlroplov, A. P. Kiryanov,
L. B. Luganskii, A. B. Franzcsson, and 1. F. Shchyogolov.
We are especially grateful to Assistant Professor D. A. Zaikin
for his help.
We have freely used several problems books which spe­
cialised in tho different branches of physics and which were
published in small quantities at the MPTI between 1976
and 1977. A large amount of the work of .preparing these
problems books was performed by L. P. Bakanina, N. S. Be0
ryuleva, A. V. Stepanov, M. A. Tulaikova, and I. A. Fomin.
The preparation of the manuscript for publication was main­
ly performed by T. A. Seliverstovn and L. N. Shustova.
We express our gratitude to all those mentioned. We are
also grateful to our editor B. S. Belikov, whose remarks
greatly improved the manuscript.
When compiling the collection weproforred the moro prac­
tical problems originating from the outcomo of experiments.
Such problems involve real physical objects and not ideal­
ized schemes. Many problems require approximations and
should assist students in developing a clarity of physical
thought and a sense-of-scale of physical quantities and phe­
nomena. In order to solve the majority of the problems both
a formal knowledge of the laws and a broad physical out­
look are required. Some problems make no pretence at being
original, they are rather of a general physical interest.
We hope that this collection will be useful for physics stu
dents, postgraduates, and physics instructors at higher edu­
cational institutions.
Since the publication of the Russian edition of this col­
lection we, our colleagues, and our students have dotected
a number of errors. All the errors so far identifiod have been
corrected and we should like to express our warm gratitude
to all those who drew them to our attention.
PROBLEMS
a
I. Mechanics
j
1.1. Is it possiblo to measure the gravitational constant y
with a relative error of 6 = 10% by rolling a lead ball to
a test weight suspended from the pan of ah analytical spring
balance whose maximum rolaliVo sensitivity is T| = 10"7
(0.1 mg/kg)?
1.2. A binary star has a period T — 3 years, while the
distance L betwoon its components is two astronomical units.
Express the mass of the slur in terms of the mass of the
Sun.
;
1.3. Find the tension in a 63 m cord used to connect a 70 kg
astronaut to a satellite whose mass M is tnuch larger than
that of the astronaut. The satellite closely orbits the Earth.
Which arrangement of the astronaut, satellite and the Earth
causes the tension to be at a maximurp?
1.4. Determine the mass of the planet Mars in terms of the
parameters of the elliptic orbit of the Soviet space probe
"Mars-2" around the planet. Tho maximum! distance between
the probe and the planot surface at apogee is a — 25 000 km,
the minimum distance at porigeo is b — jl380 km, tho re­
volution period I’
m being 18 h 00 min. >Mars's diameter
Du is 6800 km, the necessary parameters of the Earth are
taken to be known. In your calculations, take the Earth's
mass to be unit mass.
1.5. By considering the Earth to be a uniform ball of radius
R e with density p, find the gravitational pressure p as a
function of the distance from the Earth's centre. Calculate
the pressure at the centre assuming that, R e = 6370 km,
p = 5.5 g/cms.
1.6. Estimate the difference due to tho rotation of the Earth
between the distance from the contrc of the Earth to the
sea level /?p at the pole and that ot the equator 7?c.
1.7. Determine tho direction and magnitude of tho escape
velocity v at which a spacecraft will leave the solar system
if launched from the Earth with a velocity v0 whose di­
ll
lection is perpendicular to the lino connecting the Earth and
the Sun, and in the direction of the Earth’
s rotation around
the Sun.
1.8. Find the minimum velocity with which an uncontrolled
spacecraft launched from the Earth along an Earth-Moon
trajectory strikes the surface of the Moon. The motion of
Ihe Earth ond the Moon should be neglected.
1.9. Evaluate the mean time t belwoon the collisions of two
Suu-like stars in a cluster of stars if their average relative
velocity u is 60 km/s and the number of stars per cubic
light year is N « 10. Define more accurately your concept
of collision. The mass of the Sun and its radius are M q —
2 X 10s0 kg and
= 7 x 108 m, respectively.
1.10. Two satellites A and D are 45 km apart and move in
a common circular orbit of the Earth. In order to dock, the
satellites must approach each other and yot continue to move
along their common orbit. Which simple sequence of short
bursts of the lagging satellite’
s engine can accomplish
this manoeuvre, if the engine is orientod at a tangent
to the orbit and each burst can change tho satellite’
s velocity
by a Au not exceeding 8 km/h?
1.11. At a certain altitude the density of the atmosphere
is p ft « 1.6 x 10"11 kg/m3. Evaluate the aerodynamic drag
experienced by a satellite with a cross-sectional area S —
0.5 m* and mnss m = 10 kg orbiting at this height. How
will the velocity and height of the .satellite change per re­
volution?
1.12. In a science fiction story the Earth rotational volocily
was changed by launching a projectile along u tangent to
the equator. What should the difference (c — v) be between
the velocities of light and tho projectile in order to slop the
Earth rotating around its axis? The Earth radius is B e =
6370 km, and its mnss is /WE = 6 X 1024 kg. The Earth’
s
moment of inertia relative to its rotational axis, with its
density nonuniformity properly incorporated, is quite ac­
curately given by the formula / - Me
Gompnro
kinetic energies of the projectile and tho Earth’
s rotation.
The projectile’
s rest mass can bo taken to be m = 108 kg.
1.13. Determine the power needed by a photon rocket to
move outside the solar system with a nonrelativistic speed
and a constant acceleration of a = 10 m/s2. Tho rocket's
mass is M = 103 kg. Compare the power of such a rockot
with that of the Bratsk hydroelectric power station (4.5 X
108 kW).
12
1.14. A toy rocket is filled with water,; which occupies a
small part of its internal cavity. The remainder of the cavity
is pumped full of air until tho prossuro is p. Estimato tho
rocket's final altitude h%assuming the mass of water m is
much smaller than that of the rocket A/, that tho time taken
by the water to discharge is much smaller than the duration
of the Oight, and that tho cross section of the rocket's
nozzle is much smaller Ilian that of cavity.
1 ^15. A satellite revolves around a planet (or other body with
a spherically symmetric mass distribution) in the immediate
vicinity of its surface. Show that the period of revolution T
depends on the planet’
s average density p. Calculate the
period of revolution of a satellite around a neutron star as­
suming that latter’
s density is the same as the density of
matter inside atomic nuclei (p « 10u g/cms).
1.16. A satellite of mass m moves along a circular orbit
around a star of mass Af, m being much less than M. At
a certain instant the star explodes like a supernova and ejects
a mass qM. Assuming that this mass goes into orbit instan­
taneously, describe the possible kinds of motion the satel­
lite may have subsequently.
*1.17. A spaceship nears the Moon along a parabolic tra­
jectory that almost touches the Moon's surface. In order to
transfer into a circular orbit a retro engine fires at the
instant of the closest approach. Tho engine ejects gas at
a spoed of 4 km/s relative to the spaceship in its
direction of motion. What fraction of total mass should
the fuel burnt mako up? Estimato the temperature of combus­
tion if the specific heqt capacity of the ejected gas is cp =
2.2 X 10s J/K-kg.
1.18. A rocket is launched from tho top of the highest moun­
tain on the Moon. Tho angle between the direction of gas
jet and tho horizon is maintained at <p = 0.1 rad. The speed
of the jet relative to the rocket is u = 4 km/s. How must
the rocket's mass M be changed in time for the rockot to
roovo horizontally? llow long will it take to reach orbital
velocity? By how many times will the rocket’
s mass be re­
duced during this interval? What will the astronaut’
s over­
load be? The Moon’
s radius is R m = 1700 km, and the freefall acceleration near the Moon’
s surface gM = 1.7 m/s*.
1.19. A satellite flies along a circular orbit close to the Earth’
s
surface. Tho satellite’
s mass is 50 kg. A small meteorite
which has a mass of 0.1 g and is moving towards tho centre
of the Earth at 80 km/s hits the satellite and adheres to it.1
3
13
Assuming that Ihe impact is central, ostimnlc the iliflcronco
between Ihc upogoo a and the perigoo h or the new orbit.
1.20. A" spaceship lakes ofT from the North Pule at a tangent
to the Earth’
s surface. Kiml the rotation angle of tlio Kur til's
axis due to the m issile’
s launch. Tlio Earth’
s muss is 6 X
lO14 kg and its radius Jt^ 6400 km, the m issile’
s mass is
1000 t and its launch speed 15i km/s. Tlio engine can be as­
sumed only to operate at launrli.
1 •A “
launching-pad”with a spaceship is moving along a
circular orbit of the Moon whoso radius It is Iriplo that of
the Moon /?M. The ship leaves the launching-pad with a rel­
ative velocity equol to Ihc launching-pad’
s initial orbital
velocity i;0, and the launching-pad then foils to the Moon.
Determine the angle tp at which the launching-pad crashes
onto the surface if its mass is twice that of the spaceship m.
1.22. Determine the minimum fuel reserve necessary for a
soft lunar landing of a rocket. A retro engino is fired for a
time t at a height h above the Moon's surfaco. Assume that
Ihc rocket’
s velocity at infinity is much less than the velocity
at height h and that
v0 is gained by tlio rocket duo
to the gravity of the Moon.
The jet velocity relative to
the rocket is u and the mass
I
of the fuelled rocket before
the engine is started is m0.
Fig. 1
Assume that
where
??m is the Moon'sa radius.
ramus.
✓I.2J. A pendulum clock is located on tho Earth’
s surface
over a subway tunnel. The tunnel’
s diameter is D = 2R =
10 m and its axis is at a depth h = 15 m. Taking the average
soil density to be p = 2 g/cms, estimate the relative chango
in rlock s period duo to the presence of the tunnel.
1.24. By what angle cp will a car (Fig. 1) be tilted as it
brakes? The centre of mass is equidistant from tho front
.ind rear wheels and is 0.4 m abovo the ground. Tho sliding
friction coefficient / is 0.8 and the distance between the
whecd axles / is 5 b. The elasticity of all the suspension
springs is the same and such that if the car was resting
on a horizontal platform the deflection would bo A = 10 cm.
1-2;>. A gymnast falls 12 m into an clastic net. By how many
limes will the maximum force, which is exerted upon the
gymnast by the net, exceed his initial weight if tho net
s.ig> A0 — 1 rn under tho initial weight of tlio gymnast?
14
1.26. An a-partlclo flying at a volocity u0 is olastically scallered from a resting nuclous through 90°relative to its ini­
tial directum of motion. For which ratio bolwoon tho mass
of tlio a-purlielo m und that of the nucleus M is this possible?
Dclormine tho velocities of the a-purlielo v uud nucleus u
aflor the collision. Dolormino also tho anglo cp between tho
direction of the escaped nucleus and the initial direction
of motion of tho a-pnrliclo.
1.27. By how muny limos is the nuclous recoil enorgy in the
a-decay of M#Ra larger than in its ydecay if tho total energy
roleased in a-decay is E x — 4.9. MoV and that released in
tho y-dccay is Et = 0.2 MoV?
1.28. Is it possiblo for u cesium-133 atom to ionize by col­
liding with an oxygon-16 atom having a kinetic energy T%
of 4 oV? Cesium-133's ionization energy is 3.9 eV.
1.29. At which kinetic energies T of a-particles can they
be inelastically scattered from UN nuclei if the onergy of tho
first excited stato of this nuclous is U — 2.31 MeV?
1.30. In what mannor will the speed of a pocket watch bo
changed if laid on a horizontal, absolutely smooth tablo?
Assume that the axis of the torsional pendulum passes
through the watch’
s contre while the watch's moment of
inertia 70 is 500 times larger than that of the pendulum /.
1.31. In an earlier classical model; an electron was consider­
ed to be a rotating ball of mass m uniformly^ filling a sphere
of radius r0 = eVmc*, where e is the electron charge. The
electron's intrinsic angular momentum (spin) was hJ2,
Show that this model contradicts the special theory of rela­
tivity, since the linear velocity of the electron's rotation v
exceeds the velocity of, light c.
1.32. Estimate the value of the angulart momentum L
of a bicycle wheel (adult size) moving at tho 30 km/h speed.
1.33. A man sits on a swing moving with a small nngulnr
amplitude-q)0. When the swing passes through tho equilib­
rium position, the man stands quickly nmj at the instant
of the swing's maximum deviation ho sits again. How will
the angular amplitude of the swing bo changed during a pe­
riod? The man's centre of mass moves up and down through
a height h. The swing's ropes are I m long. Assume in tJho
calculations that
and that mass of thej swing is neg­
ligibly small compared to the mass of the ujian.
1.34. A tennis-ball hits a hoavy rackot at 6Q°to the normal
ond rebounds olastically. Tho ball's mass is liegligibly small
compared to that of tho rackot. How must tho rackot movo1
5
15
to make the ball rebound at right angles to its initial tra­
jectory?
1.35. A bar is firmly fixed to a vertical axle at an angle cp
to the latter. Two identical masses m aro located at distances
I from the fixing point. The axle is rotated at an angular
velocity to. The distances from the fixing point to axle’
s ball­
bearings are a and b. Find the reaction forces on the ball­
bearings.
1.36. A horizontal disc of radius R and mass m is suspended
in a gas by an olastic thread with torsional modulus / at a
distance h (h<ti R) from a horizontal fixed surface. Find tho
internal friction coefficient r\ in the gas as a result of meas­
uring the logarithmic damping factor d of torsional oscilla­
tions. Assume that the gas’
s motion between the disc and
fixed surface is laminar and neglect edge effects.
1.37. A hoop of radius R is thrown with a linear velocity
v0 and simultaneously is given an angular velocity o)0. De­
termine the minimum angular
velocity at which the hoop
will roll backwards after mov­
ing forward and slipping. Find
the final velocity v if co0 >•
(i)0min* Rolling friction can
be neglected.
1.38. A ball of mass m and ra­
dius R (Fig. 2) lying on a
Fig. 2
horizontal table experiences a
short horizontal impulse which
imports n momentum p. The height of the impact point above
the centre is kR {k^ 1). Find the energy of translational
and rotational motions of the ball. At what k will the'ball
roll without sliding?
1.39. A stool is fitted so that it rests with two legs on the
floor, then it is freed and falls on all four legs. Estimate how
far the stool will move across the floor.
1.40. A wheel has fallen off a car and rolled along the ground.
The wheel was observed to have travelled in a circle
of radius R. Determine the angle of inclination between the
wheel axle and the horizon. The car moved at a velocity v.
Assume all the mass of the wheel is concentrated at its pe­
riphery. It is known that R is much larger than the radius
of the wheel.
1.41. Determine the maximum gyroscopic pressure of a rap­
id turbine that can be installed in a ship. The ship pitches
16
frith an angular amplitude of cp = 9°and a period T = 15 s
around the axis perpendicular to tho rotor axis. The tur­
bine rotor, whoso momont of inertia is / = 1260 kg*ma,
rotalos at v = 3000 rpm. Tho distance holwcon tho rotor
bearings is I = 2 m.
1.42. A motor-bike moves along the inner surface of a cy­
linder. To be able to move in a horizontal plane, the bike
has a flywheel. This has a moment of inertia I ami spins
around an axis that becomes to be vortical wlion tho bike is
standing on tho ground. The mass of the bike and a driver
is m, while the centre of mass is locatod a distance I from
•Vi <
the wall (the radius of the cylindrical wall is much
larger than the dimensions of the bike). What must the
coefficient of rotation transmission from tho wheels to the
flywheel k bo for the biko to move without tho whcols slip­
ping? The coefficient of friction betweon the wheels and the
wall is / and the wheel radius is r.
1.43. A rod of length 2 I is inclined at an angle <p to the ho­
rizontal. It falls without rotating from a certain height (Fig.3)
to a horizontal table and hits its surface initially on the
left-hand end and then on the right-hand end. The impact
on the rod is elastic and friction can be neglected. When
struck on the right-hand end the rod will again be at an an­
gle q> to the horizontal. Find the height h. For what angles
can this situation occur?
.44. A method of determining absolutely the energy radi­
ated by quantum pulse generators is to measure the mechani­
cal momentum due to the reflection or absorption of the ra­
diation light by a solid surface. An appropriate instrument
is a torsion ballistic pendulum. Calculate the oscillation
period of a pendulum that has a sensitivity of t| =
10mm/(J *m). The pondulum’
s momont of inertia can be taken
as / = 2 X 10"6 kg*ml , and the distance between the rota-
T
2-0408
17
tional axis and reflecting mirror is r = 50 mm. Calculate
the diameter D of Lho quartz suspension thread of length
I = O.t m that will ensure this sensitivity. The shear modu­
lus of quartz is G = 1011 N/in*.
1.45. Two identical steel bars of length I — U.l in and cross
section 10 x 10 mm2 (density p = 7800 kg/in3, Young's
modulus E = 2 x 10u N/m2) collide end to end. Estimate
the duration of the collision by considering elastic waves. At
what velocities will inelastic phenomena arise if the steel's
limit of elasticity is p = 2 X 10* N/m2?
1.46. Estimate the duration of a collision of a football
kicked gently against a wall.
1.47. A rapidly rotating uniform ball (angular velocity
(D0) is placed onto a horizontal board so that its rotational
axis is at an angle cp with the vertical. Determine the ball
velocity v and angular velocity to of iLs rotation established
as soon as the ball stops slipping on the board. Rolling
friction is to be neglected. The ball's radius is A.
1.48. A billiard ball rolls without slipping across a horizon­
tal table and hits a vertical wall. During the impact both
the ball and the wall deform elastically. The coefficient of
friction of the ball against the wall is /. Neglecting both grav­
ity during the impact and the rolling friction, determine at
which angle to the horizontal the ball will rebound from the
wall. Investigate the dependence of the result on the magni­
tude of the fiction coefficient /.
1.49. A cylindrical tank with a base area S is fillod with wa­
ter to a height h. A small hole of area o appears in the bottom
of the tank. Determine how long it takes the water to run
out from the tank.
1.50. A cylindrical can of height h is immersed in water
to a depth h0 (Fig. 4). A small hole of area o appears in the
bottom of the can. Determine how quickly the can sinks.
1-51. Determine the maximum pressure that can be pro­
duced by water during freezing. The density of ice is p =
0.92 g/cms, Young's modulus is E — 2.8 X 1010 N/ma,
and Poisson's coefficient is p = 0.3.
1.52. A 1 : 100 scale model of a ship is being tested in a
basin. The design speed of tho ship is 36 kra/h. At what
speed must the model be lowod to havo gravitational
waves like those of tho full-scale ship?
1.53. Estimate by tho dimensional method the phase ve­
locity of waves over the surface of a liquid by neglecting Uio
effect of surface tension and finite depth.
18
Ij; Thermodynamics and Molecular Physics
2.1. Calculate the boat capacity C (V) of one mole of an ideal
gas undergoing the process showh in Figi 5. Assmno
value of the parameter y = CJCv is known. Whnt is the
maximum temperature that will bo reached by ljjc Kas dur­
ing this procoss? Draw tlic heal capacity plot C [V). indi­
cate the polytropic processes
whoso plots in the p-F-planc
art tangential to the straight
line from Fig. 5 at the points
corresponding to C (P) = 0
and C (V) = oo.
2.2. An ideal gas is compressed
by a piston in a cylinder so
that the heat dissipated into
the environment is equal to a
change in the internal energy
of the gas. Determine the work
,
expended by compressing one mole of the gas to half its vol­
ume. What is the heat capacity in this process? The ini­
tial temperature of the gas is 7*0.
2.3. Prove that for a substance having an arbitrary equa­
tion of state the temperature coefficient of volume expansion
p, the temperature coefficient of pressure y, and the isothormnl coefficient of omnidirectional compression x are
related thus p = yx.
2.4. Find the velocity of the adiabatic exhaust of an ideal
gas jet from a vessel through a small hole into a vacuum,
given the velocity of sound in the gas is s.
2.5. Calculate how many times greater than the final mass
M should the launching mass of a single-stage rocket M0
be in order for the rocket to reach orbital velocity v4. As­
sume the combustion products of rockol engino exhaust into
a vacuum via an ideal adiabatic process with adiabatic in­
dex v = 1.2. The average molar mass of the combustion
products is ll = 30 g/molo, and the tomperature of combus­
tion is 3000 K. Noglect tho effects of gravity and air friction
during tho acceleration of tho rocket.
2.6. Calculate how much greator the efficiency of an inter­
nal-combustion engine will bo if its compression ratio is incroasod from 5 to 10. Roplnco tho roal cyclo of the ongino by
an idonl closed cyclo compnsod of 2 isochoros And 2 ndinbnts,
and assume tho working substnneo is n polyatomic idoal gas.1
0
10
2.7. The atmosphere of the Earth may bo considorod lo be
a giant heat engino whoso lioat sourco and boat sinks are
the equator and poles respectively, while solar radiation
serves as the enorgy source. Assuming that the total Jinx
of solar energy arriving at the Earth is Q = 1.7 x 1017 W
and the efficiency of the “
engine”is an ordor of magnitude
smaller than the maximum possible, cstimato the avorage
power available for forming winds por squnro kilomotre
of the Earth's surfocc. Discuss tho physical reasons that
cause q < q inax.
2.8. Prove that for a substance with an arbitrary equation
of state given by a single-valued function T = T (p, V),
two polytropes cannot be encountered in more than one
state.
2.9. During the adiabatic compression of a liquid tho rela­
tive volume change is LVIV = 0.1% and tho temperature
is raised by LT = 1 K. Find CP!CV from these data, if the
temperature coefficient of the liquid’
s volume expansion is
p = 10”
4 K "1. By how much does the pressure in a liquid
whose temperature coefficient of pressure is y = 106N/(ma X
K) change?
2.10. A silver wire 1 mm in diameter is adiabatically loaded
at room temperature by a force F = 10 N. Assuming that
the specific beat capacity is c = 234 J/(K*kg), the density
p = 10 g/cm3, and the temperature coefficient of linear ex­
pansion a = 1.9 X 10"5 K “
l, determino the chango in the
temperature of the wire.
2.11. A heat-conducting piston of mass m is hold inside o
vertical thermally insulated cylinder of radius r and subdi­
vides the cylinder into two equal parts. Each part contains
v moles of the same ideal gas at a pressure p and a tempera­
ture T. When the piston is released it falls due to gravity.
Determine tho change in the system's entropy AS at tho
instant of establishing equilibrium. Neglect the heat ca­
pacity of the cylinder and piston and take n r^p^m g.
2.12. A volume F, = 3 1 contains v, = 0.5 mol of oxygen
(Oj) and a volume V% = 2 1 contains v9 = 0.5 mol of nitro­
gen (N,) at the temperature T = 300 K. Find the maximum
work that could bo performed by mixing these gases in tho
total volume Vx -f V% both isolhermolly and adiabatically.
Assume both gases are ideal.
2.13. A closed tube of volume V0 contains a mixture of two
ideal gases, v mol each. The initial prossuro is p. Two pistons
arc located at the extremities of tho tube (Fig. 0) and each2
0
20
is transparent to ono or llio gases. When the pistons are
moved to the middlo the gases are completely separated. Cal­
culate directly the work A performed by the external forces
{luring an isothermal quasi-static displacement of the pis­
tons and compare the ratio AIT (where T is the absolute
temperature) with the change
in the total entropy of the
gases.
2.14. An absent-ininded physicist - l i r
- q leaves in his lab a thin-walled
rubber bag Ailed with helium
201 in volume. The physicist
returns to find that all the helium has diffusod outside. Find
ithe;change in the helium’
s entropy. Air contains 107
molecules of gas per helium atom. What minimum work must
be expended to retrieve from the atmosphere and into the
bag the same amount of helium using an isothermal process?
2.15. In a shock wave a gas is heated so rapidly that as the
temperature rises from 7’
0 to a high temperature T, only the
external translational and rotational degreos of freedom are
Involved. Gradually the internal, viz. vibrational, degrees
of freedom are thon excited by a process called “
vibrational
relaxation”
. Find Lho temperature T2 of a diatomic gas af­
ter vibrational relaxation and the change in gas’
s molar en­
tropy during this relaxation given that AT, » Aw and kT0 ^
hia, where <o is lho frequency of intramolecular vibrations.
Assume that the relaxation occurs at a constant pressuro and
under the conditions of thermal insulation.
2.16. The saturated vapour pressuro of water at 2i)0 K
is 0.02 atm. Some vapour lakes up a volume of 10 1. Find
the chango in the free energy AF and entropy AS after isothormally compressing it to 5 I. Assume the vapour is an
idoal gas. The specific heat of vapourization at this lompernture is X = 2460 J/g.
2.17. Depict on the p-F-plnne and using thermodynamic
identities analyze lho Carnot cyrlc for a boat ongine working
with a substance for which (dplOT)v < 0 throughout its
working range. Indicate the “
hot" and “
cold”isotherms, mark
in tho signs of the heat flows on both isolhorms, etc.
2.18. A molo of a Van dor Wnals gas, whoso constant b
is known, is contained in a vertical cylinder under a mas­
sive piston of area S. Find lho frequency of the smnll-nmpliludo oscillations of the piston around its equilibrium posi­
tion ussuming that lho compression-rarefaction process is2
1
21
isothermal with T — 27’
0t. Sol tho equilibrium gas volumo
under the experimental conditions to bo oqual to tho criti­
cal volume. Noglccl external prossuro.
2.19. Find the adiabatic velocity of sound at the critical
point of a Van dcr Wauls gas. Assumo the constants a and
b of tho gas and its molar mass p are known. Tho heat ca­
pacity C v is givon and indopondonl of toinperuture.
2.20. A fissure 6 = 10~4cm wido and / = 5 cm long has boon
round ul the bottom of a container of liquid holium. Tho
thickness of the bottom is d = 0.5 mm. Find tho maximum
velocity of the helium
in the fissure and tho total con­
sumption ofliquid per unit time m if tho holium column
height is h = 20 cm. The helium's density and coefficient of
dynamic viscosity aro p — 0.15 g/cin* and q - 3.2 X 10“
6
g/(cra*s), respectively. Assumo the flow is laminar.
2.21. A wire of radius
= 1 mm is pulled at a constant
velocity i>0 = 10 cm/s along tho axis of a tube of radius 7?3 =
1 cm that is filled with a liquid having a viscosity q =
0.01 P. Determine the friction forco F per unit longth
of the wire anil find tho velocity distribution of tho liquid
across lube's radius, assuming the flow is laminar.
2.22. A long vertical lube of length I and radius H is tilled
with a liquid whose coefficient of kinematic viscosity is v.
In what time t will the wholo liquid flow out of the tube
due to gravity? Surface tension is to bo uogloclod and assuino
that the establishment of the liquid's velocity is instan­
taneous and the flow is laminar.
2.23. Water flows out of a wide, open vessol through a hori­
zontal cylindrical capillary. The capillary has a radius R —
1 mm and a length /
10 cm and is located near tho bottom
of the vessel, llow much power N is lost in boat release if
the water’
s height in the vessel is h = 5 cm? The flow is
laminar and the coefficient of dynamfc viscosity is q «
10"* P.
2.24. A rocket with a cylindrical thermally insulated cabin
that is h metres long is accelerating along tho cylinder's
axis. The air inass inside the cubin is m. IIow will the air
temperature and entropy he changed inside the cubin if tho
engine is switched off? Air is to be considered an ideal gas
with molar mass p. Assume that pahlRT0<^ 1, where T0
is the air temperature in the rahin. Consider two cases:
(1) the engine is switched off slowly so that the processes
occur qu.isistatically, and (2) llm engine is swilchod off
instantaneously.
22
2.25. One molo of nn ideal gas is contained in a thermally
insulated cylindrical vessel of radius r. llow will the entropy
of the gas bo chnngod if the vessel is rotated until it has on
angular velocity of
o i < y ]/" — - i where p is the gas's
molar mass?
2.26. Estimate tho radius R of nn aluminium particle sus­
pended in a liquid of density 1 g/cma and viscosity 1 P if
tho viscous fall velocity of the particle is equal to the veloc­
ity of its thermal motion at room temperature. State wheth­
er such particlos would be precipitnted from an aluminium
paint.
2.27. A motionless thin-walled vessel of volume V is filled
with a monatomic ideal gas and maintained at a constant
temperature T. The vessel’
s wall has a hole of area S through
which gas molecules escape into a vacuum. The hole's di­
mensions are small compared with those of the vessel and
with the gas’
s mean freo path length. Derive the relation­
ship botween time and the heat flow Q that must be sup­
plied to the vossel to maintain it at a constant tem­
perature.
3.28. A monatomic ideal gas fills & motionless thin-walled
vessel of volume V provided with thermally insulated walls.
Tho vessel’
s wall has a hole of area S through which moleculos flow out into a vacuum. Tho hole’
s dimensions aro
small compared with those of the vessel and tho gas’
s mean
free path length. Derive the relationship between time
and the gas’
s temperature in the vessel. Assume the initial
temperature is Tn and all the gas’
s parameters are known.
Neglect the heat capacity of the vessel's walls.
2.29. Estimate the time needed for water to evaporate from
a tube I = 10 cm long that is soldered shut at one end. The
temperature is T = 300 K and initially the tube was half­
full of water. The relative air humidity is 50% and the sa­
turated vapour pressure psat « 27 mm Hg. The mean free
path A in the air-vapour system is about 10~*cm. Assume
the vapour near the water's surface is saturated and neg­
lect capillary phenomena.
2.30. Knowing that the moan free path length of a singly
chargod 40Ar ion in a gas is A = 10"5cm, find tho averngo ion
drift velocity Udr in tho gas duo to a uniform clcclric fiold
of 300 V/cm. The gas is nt room tomperaturo.
2.31. A vessel is ovneuntod through n long tubo by an ideal
vacuum pump (viz. it traps ovory molecule entering it).2
3
23
Because of leaks in Iho vessel’
s walls the internal prossuro
does not fall to zero. Aftor prolonged pumping the pressure
stabilizes such that the mean free path length A w h e r e
d is the tube's diameter. In what way would the final pres­
sure be changed if the tube’
s diameter is halvod?
2.32. Air is pumped from a largo vessel at a pressuro of 10~4
mm Hg and a teroperaturo of 300 K and pumped through a
pipeline of length I — 2m and radius R = 10 cm. The pump
has a capacity of V, = 1000 1/s. What would the pump capa­
city be if the vessel is evacuated at the same speed and the
pump is attached directly to the vessel?
2.33. From the spread of radioactive gases after nucloar oxplosions it is known that turbulonco causes tho Earth's at­
mosphere to intormix enliroly within about a year. Mow
many limes faster does atmospheric turbulcnco mix the at­
mosphere than molecular diffusion?
2.34. Gaseous helium leaks into o vacuum through a holo of
area 5 = 0.1 mm3 from a thin-wallod vessel in which the
pressure and temperature are maintained at p — 10"° atm
and T = 273 K, respectively. The total numhor of atoms N
that leak through the holo is measured for a time interval
of < = 10"3 s. Find the relative root-mean-square fluctua­
tion of this number. Also, find the probability that no atoms
leak out during an experimental run.
2.35. Calculate the relative root-mean-square fluctuation of
the potential energy of intramolecular vibrations for: (a)
a diatomic molecule, (b) an TV-atomic molecule, (c) a molo
of an ideal gas composed of diatomic molecules. The vibrutioDS are assumed to be harmonic.
2.36. Two identical communicating vessels are filled with a
gas under normal conditions. What should the volume V
of each vessel be to ensure that the probability of a stato
in which the pressure in the vessels is isothermally changed
by 10"7% is e100 times smaller than the probability of the
initial state?
2.37. To measure temperature fluctuations in a helium-fillod
vessel of volume V, a small control volume
V is iso­
lated. At what value of ywill the root-mean-squaro tempera­
ture fluctuation amount to 10“
4%? The helium is under nor­
mal conditions.
2.38. A vessel of voliimo 1 1 at room Icmporaluro contains
the same number N of the atoms of two different gases.
Estimate the value*of N for which the probability that the
gases will separate at least once during the lifetimo2
4
24
of the observablo Universe ( « 10'° years) is comparable
2°.39™A vessel contains a Van tier Wanls gas such lluil tbo
average molar volume is equal to the critical volumo, and
the temperature T exceeds the critical temperature Tcr.
Find the rool-mean-square isothermal volumo fluctuation
(AF)* of a small olomont of tho gas having tho equilibrium
volume V.
.
2.40. The thcrmorcgulnlor of a car ongino is a cylindrical
vessel with corrugated wulls (bellows) and filled with ethyl
ulcohol and its vapour (Fig. 7). At a low 1cmporn tore the
pressure in tho bellows is low so it compresses and shuts a
valve, thus attenuating tho circulation rato of tho water in
tho engine's cooling systom. As the wator temperaturo risos,
tho bellows expand and open the valve so that the water
starts circulating more rapidly thus cooling the engine,
etc. What must the pressing force F of a valve spring bo
to open the valve at T = 363 K? Tho bellows' diameter is
D = 2 cm, the boiling point of alcohol at 1 atm is T0 =
351 K, alcohol's specific boat of vaporization is X =
850 J/g, and its chemical formula C2H5OH.
2.41. Geysers can be considered to originate from largo un­
derground reservoirs filled with ground water and boiled
by subterranean hoot (Fig. 8). Tho roservoirs aro connected
to tho Earth's surface via a narrow chnnnol lhat is filled
with water during a “
quiet”period. Assuming that an “
active”
period starts when tho water in the underground reservoir
boils and that during tho oruption tho chnnnol is filled
25
sololy with vapour which does roach tho outsido, estimate
what fraction of water is lost from a geyser reservoir during
one discharge. Tho channel’
s depth is h = 90 m,
and the latent heat of vaporization of water is
A = 41 kJ/mol.
2.42. At normal pressure holium-3 (3lIo) remains a liquid
down to absolute zero. The minimum pressure undor which
it is solidified is p min = 28.9 atm. The entropy of liquid
holium-3 per molo over tho tomperalure range of interest is
described thus S\ = R (770), whore R is the gas constant
and 0 = 0.22 K (cf. Problom 7.30). Tho onlropy of solid
holium-3 over the temperature rango in question is independ­
ent of temperature and is S a = R In 2 per molo (cf. Prob­
lem 7.9). The difference between tho volumes of liquid and
solid helium is AF = Vl — F a « 1.25 cm3/mol. Find the
temperature on the melting curve corresponding to p raim
the pressure at which helium-3 solidifies at T = 0 and tho
temperature dependence of heat of melting.
2.43. According to one model, the Earth’
s centre (called
•
the core) consists of iron. The outer layers of the coro are
molten while the centre itself, R « 1200 km in radius, is
solid. The core cools at a rate of u = 10"7 K/year, tempera­
ture differences within the coro being neglected. How far
will the radius of the solid part expand in 109 years? Assume
that tho specific heat of iron molting undor tho conditions
s surface is q « 125 J/g, tho temperature at this
at tho core’
surface being T « 3700 K, and the chango in iron's density
during solidification being Ap = 0.3 g/cm3. Because Ap <£ p
it is possible to disregard the change in the pressure dis­
tribution p (r) as the core solidifies.
2.44. A closed vessel contains water in an equilibrium with
its vapour at 100 °C. The vapour/water mass ratio p is
0.1, the specific heat capacity of water c0 is 4.2 J/(K-g).
Find the specific heat capacity c of the system, assuming that
the vapour is an ideal gas. The specific heat of vaporiza­
tion is X = 2200 J/g.
2-45. Find the specific heat capacity at constant volumo
c v of a rarefied fog (viz. a saturated vapour with water
droplets whose total mass is much less than that of tho va­
pour) at 100 °C. The molar heat of vaporization of water
at 100 °C and constant pressure is A = 41 kJ/mol.
2Ait. Due to a fall in external pressure a soap-bubble dou­
bles its radius. Assuming the process is isothermal, find the
change in soap-bubble’
s entropy. The initial pressure is p 0,
26
tho surface tension coefficient of the soap film is a, ami the
specific heat of forming a soap li Iin surface is q. Assumo that
2c / r < p 0.
.i
|
III. Electricity and Magnetism
3*1* Electrons are omitted us a rosult of thermal omission
from a negatively charged plate into n pornllel-plate capac­
itor whoso field strength E = 103 V/m. By taking into ac­
count the action of tho oloctric image field, estimate how
far from tho emitter plalo tho electron volocity will bo tho
slowest.
.
3.2. An ideal gas whoso inolcculos have a polarizability
of a = 4 X 10'*° m3 is contained in a largo vessel at 300 K.
E______^
C
D
Fig. 9
Fig. 10
A charged parallel-plate capacitor with a field strength
E =»3 X 10* V/m is pluced inside the vessel. Find the rolative difference between tho concentrations of the inoleculos
Inside and outside the capacitor.
3.3. What is the electric fiold in a cavity formod by the in­
tersection of two balls (Fig. 9). Tho balls carry charges with
densities p and — p uniformly distributed throughout
their volumes. The distance between the centres of the
balls is a.
3.4. How should charges a (0) bo distributed on tho surface
of a sphoro of radius R for tho field inside tho sphere to bo
uniform and equal to E'f Wluil will bo the field outside tho
sphere?
3.5. How much does the pormittivity e of an uideal gas"
composed of a large number of conducting balls of radius
R differ from unity? The concentration n of tho bolls is
small enough for nR*<g. 1.
3.G. An uncharged conducting spherical spock of dust has
entered an external uniform electric field.
How
27
will Iho field strength at points ,1, /J, C, and /) (Fig. 10)
be changed?
3.7. Under normal conditions, gaseous argon lias a permitti­
vity e — 1 ^ 6 x 10"4. Making use of this fact, calculate
the displacement of the ucentro of mass" of the olectron shell
of an argon atom from the nucleus in a sialic electric field
with strength E — 300 V/cm. Argon’
s atomic number
Z = 18. (In the absence of an external field the electron
distribution around the nucleus is considered spherically
symmetric).
3.8. In the Thomson model of tho atom tho positivo charge
e is assumed to be distributed insido a sphero of radius
10~8 cm. How should the positivo chargo density dopond on
the distance from the sphere’
s contro if an olectron (a nega­
tively charged point particle carrying a charge —e) insido
the sphere oscillates harmonically? Find tho frequency of
the electron oscillations assuming that the charges do not
interact.
3.9. Is there any possibility in principle for determining a
rocket’
s acceleration by purely electrical methods (ways)?
Do numerical estimates for an acceleration of 10 g and a con­
ductor of length Z = 10 ra. Neglect external electric and
magnetic fields.
3.10. A point charge q is placed inside an uncharged spheri­
cal conducting shell, a distance a from its centre. The radii
of the internal and external surfaces are r and /?, respective­
ly. Find the surface density a of the induced electric charge
on the shell’
s external surface and tho shell potential tp.
Determine the surface densities o L and a, of the charges in­
duced on the shell’
s internal surface at points lying on the
same diameter as the charge q.
3.11. A dielectric plate of thickness h having a “
frozen”
polarization P = const is placed inside a parallel-plate
capacitor, the plates of the capacitor being connected to­
gether. The polarization vector is perpendicular both to the
side faces of the dielectric plate and the capacitor plates.
Determine the electric field strength and induction
inside the plate. The distance between the capacitor
plates is d.
3.12. Given a thin dielectric cylinder of length 21and radiusr
with a “
frozen”polarization P = const (Fig. 11), find the
field at the point A. By how many times is this field strong­
er than that at tho point B?
3.13. Due to inhomogeneous deformation, a piezodielectric
28
plate 2d thick is polarized so that the polarization in the
middle of tho plate is /'« and olscwhero is P ~
p (i^-xV d2), whero x is tho distance from Hie plate s
midplanc. The polarization vector points along tho j-axis
(Fig. 12). Detormino tho oloctric field slrongth both insido
and outside the plato as well as tho potential difference be­
tween its lateral surfaces. Edge effects should be neglected.
I
*
'/>
<i<
'//
*
h
---- P
t
P
i
Fig.
Fig. 12
3*14. A pulsed discharge from a capacitor through a rare­
fied gas (hydrogon) causes it to heat to a temperature T.
Estimate the value of T assuming that the entire discharge
onergy was spent on healing the gas. Indicate the possible
causes why the temperature may be lower. Make numerical
estimates for U = 3 X 104 V and C = 1.8 X 10"6 F assum­
ing that the hydrogen is at room temperature T0 = 293 K
and occupies a volumo V = 10"2 ms at the pressure p =
1.29 N/m2before the discharge. Estimate what the temperature
of the capacitor plates is if the entire discharge energy was
dissipated. Tho specific heal capacity of copper is c = 4.2 X
102 J/(K-kg), and the mass of the plates m is 0.1 kg.
3.15. Find the attractive force between a point electric di­
pole, which has a dipole moment of p = 4 X 10"10 C*m,
and an infinite metallic plato L — 1 cm from the dipole.
The dipole axis is perpondicular to the plate. DoLormino the
work needed to movo the dipole 2 cm from the plate
surface.
3.16. Derive an expression for the energy of a small dipole
in an external electric field E. Consider tho case of a rigid di­
pole with a dipolo moment p and that of an elastic dipole with
a polarizability a (p = ae0E).
3.17. Estimato tho potential difference between the head
and tail of tho steel core of an armour-piercing projectile
29
when Ihc projectile ponclralos an obstacle. Assumo that the
core is 2b cm long and that it lost speed on piercing armour
plate (/ — 5 cm thick. The projectile’
s speed at the instant
it hit the armour plate was 1000 m/s.
3.18. A dielectric with a permittivity e fills a half-space.
A point charge q in a vacuum is a distanco L from the plane
boundary of the dielectric. Find the charge distribution
o over the dielectric’
s surfaco
and the force F acting on the
charge.
v/3.19. A parallel-plate capaci­
tor C is connected to the battery
of o.m.f. % in parallel with a
resistor. The capacitor plates
are rapidly brought together
so that the separation distanco
is halved. Assuming that dur­
ing tho movement of the plates
the charge on the capacitor remained practically unchanged,
find the Joule heat dissipated by the rosistor at tho in­
stant the recharge ceases. Estimate the order of magnitude
of the resistance for which the hypothesis could practically
be fulfilled if the time needed to bring the plates together is
At -1 10-* s and C ^ lO ’
10 F.
3.20. Estimate tho forco acting on an atom 200 A. from tho
lip surface of a metal needle having an edgo radius U - 100 A.
Tho potential at the needle is cp - 10 kV. Tho polariza­
bility of tho atom a is of tho same ordor as its volumo.
3.21. Two cylindrical conductors of radius r0 are soldered
to a large metallic sheet of thickness a, a distance b from
each other (see Fig. 13). Find the resistance between tho con­
ductors, if a<£ r0<g.b. The conductivity of tho conductors
is much larger than that of the sheet X.
3.22. When the voltage across a kenotron reaches 500 V, the
anode temperature is 800 °C. What will the anode tempera­
ture be when the anode voltage is 1000 V when: (1) the
anode current is saturated at 500 V and (2) there is no satura­
tion even at 1000 V?
A'ole. When there is no saturation the anode current J
is proportional to l P a.
3.23. A device consists of two concentric conducting spheres,
the internal being made of a radioactive material that
emits fast electrons. The velocity of the electrons in tho gap
between the spheres and hence their ionizing action are con30
stant. After traversing the air gap, tho electrons are absorbed
by the external sphere. Whon the battery is turned off an
equilibrium sets in between the charge flux carriod by the
fast electrons and tho conduction current in the ionizod
air. Find the electric field in the gap between the spheres
if the element e.m.f. is £ and the radii of the spheres are
/?! and /?t, respectively.
3.24. Determine the conductivity of tho isolation in a spher­
ical capacitor filled with a lossy dielectric.
3.25. The space between two concentric spheres is filled
with a dielectric whose conductivity depends only on the
distance from the point to the spheres. What must the con­
ductivity relation be for the
volume density of the Joule
losses when a current is flow­
ing to be the same at every
point in the dielectric?
3.26. In a Van de Graaf gen­
erator, shown in Fig. 14, the
charge is transported by a di­
electric belt to charge a highvolta^o spherical electrode.
The surface-bound charges are
transferred to the belt from tho
source locatod near tho lowor
pulley. Estimate the maxim­
um potential and maximum
KiK. !'■
current that can bo producod by
such generator if the radius of
the high-voltage electrode, is R = 1.5 m, tho belt moves nt
v — 20 m/s, and the belt’
s width is I — 100 cm. The bolt
and high-voltage electrode are placed in a gas in which
the break-down occurs at an electric field strength of £|, =
30 kV/cm.
«>3'.27. A spherical capacitor with sphere radii Rx and 7?a
is filled with a weakly conducting medium. Its capacitanco
is C, while the potential difference across the capacitor af­
ter it was disconnected from a battery halves after a time
t. Determine the permittivity e of the medium and its re­
sistivity p.
3w28. Modern data indicate that the experimentally permis­
sible allowable difference between the absoluto values of
the charges on an electron qe and a proton qp is such that
I (?p — ?.V?p I <
31
Could this difference account for llio existence of the
Earth's obsorved geomagnetic moment? Tho gooinagnotic
hold is B b « 3 \ It)-'1Tnnd its density p « 5 X 105kg/m3.
For the atoms constituting the Earth, the ratio of relutivo
atomic mass A to atomic itumbor Z is about 2.
3.29. A long thin-wnllcd duralumin tubo is charged and
rapidly rotated about a longitudinal axis. What form will
the resultant magnetic held liavo? The uppor limit of tho
speed at which tho tube can rolato is governed by duralumin's
mechanical strength (/,nax = 5.9 X 10s N/m3). What is
maximum possible ratio botween the volume densities of
the energies of tho magnetic
held inside the tube and the
electric field on the tube's
external surface? Duralumin's
density p is 2.7 X 103kg/m3.
Pig. 15
3.30.
ethylene him, a wide band is
pulled over rollers at v = 15 m/s (Fig. 15). During the proc­
ess the film's surface acquires a uniformly distributed
charge o mainly because of friction. Estimate the maximum
values of a and the magnetic hold's flux density B near the
film’
s surface taking into account that an electric held
strength of E = 20 kV/cm can cause a discharge in air.
3.31. How many fixed singly chargod positive ions must oxist
within a parallel homogeneous electron boam of a circular
cross section so that the radius of the beam remains constant
if the beam moves at u velocity v? The concentration of the
electrons in the beam is n0 and collisions betweon electrons
and ions should be neglected.
3.32. A current J = 10s A flows in the surface layer of a
long plasma cylinder of radius R = 0.05 m. The pressure
inside the plasma is p = 10* N/ma. Determine tho force
acting per unit area of the lateral surface of the plasma cy­
linder. Is the plasma compressing or expanding here? Find
the current required to balance the radial forces.
3.33. A plasma hlament is conhned by a magnetic hold
parallel to the hlament’
s axis because the held does not
penetrate the plasma. Estimate the magnetic field’
s flux
density needed to confine a plasma with a particle concentra­
tion of n = 10,# cm"3 and temperature T = 10® K.
3.34. A steady axially symmetric beam of electrons is accel­
erated by a potential difference U and formed along a long
32
To ma
ovacuated cylindrical lubo of radius R. Find tho oloctron
concentration distribution in torms of the radius r in a cross
section of tho boom if the magnolic hold R measured as a
function of r in tho cross section can bo doscribod by tho ex­
pression B = B0(rlR)* for r < R (B0 and q > 0 being
constants). Determine the electric field E (r) assuming that
the beam parameters do not change along the beam's axis.
3.35. A square frame with current .% = 1 A is near a long
straight conductor carrying a curront Ch =»10 A (Fig. 16).
The frame and tho conductor lie in the same plane, a frame
Fig. 16
side is a = 6.8 cm, and the distanco between them is b =
4 cm. What work should be done to move the straight
conductor into the position indicated by tho dashed
line?
3.36. N turns of a wire are wound onto an iron toroidal core
with a magnetic permeability \i. Tho toroid's radius is /?,
the radius of the core cross sec­
tion being such t h a t r < R. The
toroid is cut into halves sepa­
rated by an air gap x (Fig. 17).
Determine the force of attrac­
tion between the two halves
when a current Cl is flowing
through the coil. Consider the
case of x = 0.
3.37. To determine the mag­
netic susceptibility x of a dia­
Fig. 18
magnetic material, a pair of
scales is used to measure the
force pushing a small sample out of the gap between the
poles of an electromagnet (Fig. 18). If the magnetic held
in the gap is described by the relation B — J?0e"ar*, where
y
8—0408
33
r is the distance in Iho radial direction from the axis of sym­
metry (in cm), U0 is the held strength along the axis of 103Cl,
and a is a constant equal to 10-2 ciu'2. How far from the axis
must the diauingnolic sample he for the expulsive force to
bo at a maximum and what is Iho maximum forco tor a sam­
ple iu the form of a small thin disc of volume V — 0.1 cm3?
The magnetic susceptibility of the material is x = —1.4 x
10s (bismuth). The disc is placed perpendicular to the
magnetic held.
3.38. Superstrong magnetic holds can be produced by the
explosive compression of a conductive lubo in which there
is an initial magnetic held B0. Determine the hnal magnetic
held B and the tube’
s radius R , if during the compression
the pressure of the magnetic held is balanced by a pressuro
of 10* atm caused by the ex­
plosion. Consider a case when
B0 = 5 T and initially the
tube's radius R0 = 0.05 m.
Mechanical
and electrical
resistances should be neg­
lected.
Fig. 19
3.39.
drodynamic generator consists
of a parallel-plate capacitor with a plate area S and a sepa­
ration d which is put into a flowing conducting liquid with
a specific conductivity of X, the liquid moving parallel to
the plates with constant velocity v. The capacitor is placed
in a magnetic held with a flux density D and directed per­
pendicular to the velocity of the liquid (Fig. 19). How much
power will be dissipated iu an external circuit containing a
resistance /??
3.40. A compass is placed above a wire carrying a direct
current at a distance R = 0.1 m from the wire’
s axis. Fiud
the current needed to lift the compass needle over its peg.
The residual induction in the needle steel is equal to the
saturation induction B0 — 2 T. The steel’
s density is p =
7.8 x 10s kg/m3.
3.41. Determine the period of the small oscillations in a
thin magnetic bar of length / = 0.1 m freely suspended from
the middle in a geomagnetic held of B e = 2 X 10'6 T.
The steel density is p = 7.8 x 103 kg/ra3, and the residual
induction B0 = 1 T.
3.42. After a small steel ball is magnetized to saturation in
an external held it is switched off. Estimate tho residual
34
A simp
magnotization of Iho ball if B and // aro rclalod by tlio equa­
tion B = B0 (1 + II/nc) and for the steel IIc = 4 X 10s A/m
and B0 = 1 T. The demagnetization factor of Iho ball is
P = 1/3.
3.43. The magnetization I as a function of the held strength
H for a magnetic material is shown in Fig. 20. This mate­
rial is usod to make tlio core of a thin toroidal coil with N
turns. The coil length (perimeter) is L. There is a narrow
transverse gap I in tlio core. Determine Iho current .70 at
which the core will be saturated. lIo>v will the magnetic flux
density B be changed in the core’
s gap for J > J 0? /0 and
Hi are assumed to be given.
3.44. In order to impart an angular velocity to an Earth salollito the geomagnetic held con be used. Find Iho angular
velocity gained by the satellite if a storage battery with a
capacity of Q = 5 A*h is discharged suddenly through a
coil of N = 20 turns wound around tlio satellite's surfoco
along the circumference of a large circle. The satellite has
a mass of m = 10s kg and is a thin-walled sphere. The geo­
magnetic field is parallel to the winding plane and its flux
density is B — 0.5 G.
3.45. What current should bo passed through a long thin
single-layer solenoid with winding density of n turns per unit
length to have a flux density B throughout the solenoid so
that it is equal to that of a permanent magnet of the same
dimensions? The magnotization I is constant and directed
along the axis.
3.46. A thin long permanent magnet of length 21 and
radius r has a constant magnetization I (Fig. 21). Find the
flux density B at point A. How much is it moro than the flux
density at i4'?
3*
35
3.47. Determine the frequency of the trnnsvorso oscillations
of the protons trapped in a relativistic olcclron heum with
cross-sectional area n/i* = 3.14 X 10"* in- and current
d = 103 A.
3.48. Two slits 6’
, and *S’
a, each d — 0.1 cm wide, have been
placed in an ovacuntcd vessel and sopnrntn out a “
shoet”
Fig. 22
Fig. 23
electron beam with onergy W = 400 oV (Fig. 22). How far
x from slit S %will the electron beam's width have doubled
due to the Coulomb ropulsion between the electrons if tho
olcclron current per unit length of tho slit (downstream of
the slit S |) is C! = 10"4A/cm?
When calculating, assume the
slits are infinitely long.
3.49.
of the residual gas untwists
along a spiral in n crossod
a.c. electric (amplitude E =
1 V/cm) and d.c. magnetic
{13 = 3 X 10a G) fields (Fig.
23). Find the frequency at
which the NJ ions will reach
the collector A. At this frequen­
cy, the radius of the spiral
increases until the ion reaches the collector with a radius
R — 1 cm. If we slightly chango the frequency, the ion will
untwist for some time and then begin to twist toward the
source. Estimate how the current frequency must be changed
to cut the current off the collector.
3.50. An electric dipole moves in a uniform magnetic field
B at a velocity v perpendicular to B. The dipole moment p
forms a small /.cpwith the direction [v*BJ (Fig. 24). Find the
angular frequency o)0 of tho small amplitude oscillations of
the dipolo assuming its moment of inertia 70, velocity n, and
dipole moment p, and the flux density 13 are known.
36
In an o
3.51. In an electron accelerator, a botalron, tho accelerating
voltage is the o.m.f. inducod by changes in tho niagnotic
flux that pierces the electrons’orbit. The electrons move
along an orbit of approximately constant radius. Assuming
the radius of tho orbit is constant, determine the relation
between tho average magnetic field B ny piercing the orbit
and the magnetic held B in the orbit (at a given instant of
time). The magnetic held runs parallel to the axis of sym­
metry of the betatron's vacuum chamber.
3.52. In a cylindrical proportional counter a beam of parti­
cles produces volumetric ionization. Estimate tho lime du­
ring which the ions collect in a counter filled with argon at
normal pressure. The cathode
radius is R = 1 cm, the anode
radius r = 2 x 10”
* cm, the
potential difference between
the anode and the cathode
U = 2500 V, and tho mobili­
ty of positive argon ions is
p+ = 1.4 cmV(V-s).
3.53. A proton beam with an
energy W = 4 MeV is extrac­
ted from an accelerator and travels 4 ra through a vacuum
before hitting a target. The beam expands due to Coulomb
interaction botwoon tho protons. Estimato the maximum cur­
rent density / of the beam if the bourn radius is ullowed to
expand by 10% of its initial value. The proton distribution
in the beam is axially symmetric and their initial transvorso
velocities may be neglected.
3.54. In a device for separating the isotopes 235U and 23iU
(Fig. 25) a beam of singly charged accelerated uranium ions
with an energy of W = 5 keV moves from a source through
a slit S into a homogeneous magnetic held perpendicular to
the plane of the figure. Ions with different masses move along
different circumferences and after transversing a semictrclo
thoy hit collectors. Tho design of the collectors should be
such that the 235U and 238U beams at tho exit are further
than A = 5 mm apart. What magnetic held B will satisfy
this condition? Also, find the time t needed to completely
separate M = 1 kg of natural uranium if the ion current
produced by the source is .7 = 5 mA.
3.55. In a direct actiou (electrostatic) accelerator a proton
moves inside a vacuum lube in a practically homogeneous
oloctric hold. Extranoous magnetic fields bend the trajectory
37
or tlio prolon so llial it may liil tlio wall boforo arriving at
tho end. Estimate the permissible lovol of an oxtornal homo­
geneous magnetic field for such an accelerator if tho tube
length is I — 2 m and the protons ore accelerated to an ener­
gy IK = 4 MeV. The permissible deviation of the protons
from the axis at tho end of the tube is b = 1 cm. The initial
proton velocity should bo neglected.
3.56. A proposed technique to produce the high temperatures
needed to realize thermonuclear reactions involves what
is called “
magnetic thermal insulation*', the fast particles
being prevented from escap­
/
ing from the high temperature
zone by a magnetic held. Es­
timate the current in a gas dis­
charge column of radius R =
3 cm such that electrons with
an average speed of random
motion corresponding to T =
10° K cannot move further
than 3 X 10‘
3 cm from tho
column's surface.
3.57.
charge e moves along an equi­
librium circular orbit of radius r0 in tho horizontal plane of
a magnetic gap (Fig. 20) in which tho magnolic Hold falls
ofT along the radius such that R z (r) = A/rn (0 < « < 1).
The orbital cenlro coincides with the zz symmotry axis. De­
termine tho frequency of tho vortical oscillations of tho purliclc for small deviations from the horizontal piano.
3.58. Determine the frequency of radial oscillations of tho
particle in Problem 3.57 for small deviations from tho equi­
librium orbit.
3.59. A superconducting ball of radius R is placod in a ho­
mogeneous magnetic field with a flux density /?„ . Find both
the magnetic field outsido tho ball and tho current surface
density i.
Hint. The magnetic field strength and flux density vanish
inside a superconductor.
3.00. Ily how much dues the magnetic permoabilily |i
of an “
ideal gas”composed of a large number of superconduct­
ing halls of radius It differ from unity. The ball concentra­
tion n is small enough for /f3/<<^ 1.
3.61. The currents flowing over the surface of a supercon­
ducting body always result in a zero magnetic field inside
38
A pa
tho superconductor. However, if the magnetic field around
the body is too high the superconductivity breaks down and
the metal returns to the normal stale. The critical fiold B ct
for lead at 2 K is 750 G. Estimate how large a ball can be
suspended on a “
magnetic cushion”at 2 K. The density of
lead is p = 11.3 g/cma.
3.62. A superconducting ball flies towards a solenoid along
its axis. The flux density of tho solenoid’
s magnetic field
is B0 = 0.1 T. What must be the initial velocity v0 of the
ball for it to penetrato the solenoid? The ball’
s radius is
R = 0.02 m and its mass is m = 10-3 kg.
3.63. A superconducting solenoid is deformed so that its
magnetic field is adiabatically compressed. What equation
(analogous to the adiabatic
equation in the kinetic theory
-r—
of gases) can describe the behav­
a
iour of the magnetic pressure
in terms of the change in the
solenoid’
s cross-sectional area?
a
M
3.64. A current J 0 is induced
Fig. 27
in, a closed superconducting
electromagnet's winding. The
steel magnetic circuit (length
L) has a magnetic permeability p and a small gap of width
I such that tho magnetic fiold lonkngo in tho gap can be neglocted. How will tho current through the winding bo
changed if the gap is halvod by deforming the core?
3.65. A thin straight conductor with a diroct curront flowing
through it is positioned above a superconducting plane. As­
suming that the linear density of the conductor p is 2 X
10"* kg/m, find how high above the plane the conductor
carrying a current of ,7 = 20 A will hover?
3.66. Estimate the resonance frequency of a toroidal kly­
stron cavity whose cross section is illustrated in Fig. 27.
The centre of the cavity is a capacitor through which a cur­
rent is flowing, while the toroidal cavity is filled with the
magnetic field produced by the current. The cavity’
s di­
mensions are a = 10 cm and d = 1 cm.
3.67. An inductive sensor is a radio-frequency device to
delect small variations in inductance. A commonly used
sensor consists of an olectrical oscillatory circuit with a vary­
ing inductance (Fig. 28). Estimate the minimum measura­
ble relative variation in induclauco AMI* if tho circuit is
tuned to resonance. The power supply voltage U = 100 V,
30
the minimum monsurablo variation in voltago across the re­
sistor is AF = 10 \iV, nntl tho circuit quality factor (Qfactor) is Q = 100.
3.68. A capacitive sensor is a very sensitive device used to
detect small mechanical displacoments. Tho common capac­
itive sensor consists of an electrical oscillating circuit with
Fig. 28
Fig. 29
an air capacitor (Fig. 20) one of whose plates is at rest.
Estimate the minimum measurable displacement of tho ca­
pacitor plate Ad if the circuit is tuned to resonanco. The pow­
er supply voltage is U = 100 V, the minimum measurable
Fig. 30
Fig. 31
variation in voltage across the resistor is AF = 10 p.V, the
circuit's ^-factor is Q = 103, and the intcrplalo width is
d = 1 mm.
3.69. There is an a.c. e.m.f. described by the relationship
t (0 = £o cos* Q* in the circuit shown in Fig. 30. Deter­
mine the currents 3 and 3X if the circuit's parameters aro
known to satisfy the relation £2* = 1/4 LC.
3.70. The a.c. e.m.f. in the circuit shown in Fig. 31 is de­
scribed by relationship £ (t) = £0cos2 Qt. Determine the
currents 3 and J, if the circuit’
s parameters aro known to
satisfy the relation Q* = 1/4 LC.
40
3 71 Given the RC parameters of the circuit shown In Fig.
32, what will be the frequency Q of tho output voltago
V00t when it is in phase with the input voltage V,.? What
will be the ratio between the amplitudes of Jout *nd V\xJ
3.72. At a certain instant of time switch K in the circuit
in Fig. 33 is on and a capacitor C with an initial charge q0
starts to discharge through inductance L. When the discharge
current reaches its maximum valuo,/£ is again swilchod off.
Fig. 32
Fig. 33
How much charge will run through the resistor /?? Tho re­
sistance of a diode D in tho forward direction is much lowor
than that of R whiloin tho backward direction it is infinite­
ly high.
3.73. Near the coil of an oscillatory circuit with L fiR par­
ameters there is located a second coil with inductance La.
The mutual inductanco betweon the coils is M. What will
be the resonance frequency of tho circuit, if the terminals of
tho second coil aro shorted? Tho inductive resistance of the
second coil at the circuit’
s oscillation frequency is assumed
to bo significantly larger than its active resistance.
IV. Optics
4.1. A spectrograph has a collimator objective of diameter
D with a focal length Fx and a camera objective of the same
diamoter with a focal length Ft. A light source B is projected
sharply onto the exit slit of the spectrograph using a conden­
ser in the first experiment magnifying the source (the distance
betweon tho condonsor and slit is bx) and in tho second ex­
periment reducing tho source (the distance between the condensor and slit is b2). What must the condenser’
s diameter
be for the illumination intensity on tho phototube to be tho
same in both cases? What is the illumination intensity if the
reflection and absorption losses are neglected?
4.2. A radio-frequency emission from a cosmic sourco with
on angular size
is being received by a centre-fed dipole
aerial. The dipolo is siluatod on a smooth seashore at a height4
1
41
// i'iliovi* soa-lovul. Treating water surface as if it wore a plane
mirror, determine how Ihc signal’
s intensity will ho changod
in terms of the elevation a of Iho source abovo tho ho­
rizontal. For which values of tho source’
s angular size will
the signal intensity be inde­
pendent of a? To simplify your
working, carry out the calcu­
lations for small values of a
and t|>. Tho wavelength X is
given.
4.3.
placed a distance / = 1 m from
a thin mica plate of thickness
h = 0.1 mm and with refraclivo index n = 1.4 (Fig. 34). At the same distance from
the plate, a small screen S is placed at right angles to tho
reflected rays to observe the interference fringes. Tho angle
of incidence is i = 60°. (!) Determine the order of the fringe
Fig. 35
Fig. 3G
located at the screen's centre. (2) Find tho width of the inter­
ference fringes. (3) Estimate the admissible size of tho source.
(4) Estimate the admissible nonmonochromaticity AX.
The experiment involves light with a wavelength X =
5C00 A.
Using a telescope set to infinity, intorfcronco fringes
arc observed in a thin glass sheet of thickness h — 0.2 mm
and with refractive index n = 1.41, the angle of observa­
tion i being from 0 to 90° (Fig. 35). (1) Find the maximum
and minimum orders of the fringes observed. (2) Eslimato
the admissible soorce nonmonochromaticity for which all
the interference fringes will be fairly distinct. (3) What is
the admissiblesi/e of the light source in this interference ex­
periment? Green light (X — 5000 A) is used.
4.5. In a Rayleigh interferometer (Fig. 30) a plane wavo is
A sour
«liffraclo«l by two slits. A diffraction pallorri is observed in
tho focal piano of a Ions whoso focal length F is 100 cm.
One of tho slits is closed by a plato made from a dispersive
substance of thickness d ~ 0.01 mm and for which n (X) =
A — Z?X, where A and B are
Constants. Here the white (ach­
romatic) fringe is shifted 4 mm.
Determine the constant A , if
the slits are 1 cm apart.
4.6. How far are the interfer­
§ h " ,= '
ence fringes produced by a
Lummer-Gehrcke plate shifttod when the temperature is
Fig. 37
changed by 1 °C? The answer
must be given in relative units
(with the distance botwoen adjacent fringes taken to be uni­
ty). The plato thickness is d = 2 cm, the rcfractivo index
n = 1.5, and tho tomporaturo coefficient of linear expansion
of glass a = 8.5 X 10~6 K "1,
tho wavelength X = 5000 A.
Tho temperature dependence
of n must be neglected.
4.7. A Fabry-Perot interfero­
meter (Fig. 37) consists of two
plane mirrors with reflection
coefficient (in intensity) p =
09% separated by 10 cm. A
plane monochromatic wave is
incident on this interferome­
ter used as an optical cavity.
Estimate the width of resonance
IW
curve (in MHz) and deter­
Fig. 38
mine the frequency interval be­
tween two adjacent resonances.
4.8. Two boams of while light produced by a single point
source ore brought together onto the entrance slit of an opti­
cal spectral dovico. Tho path-length difference is A = 300 m.
Estimate the resolving power of n spectral dovice capablo of
delecting the interference of llicso beams.
4.9. Two beams of whito light from one source arrive at a
point of observation P (Fig. 38o) with n path-length difference
A. Using a high resolution spectroscope tbo energy distri­
bution in tho oscillation spectrum at P is investigated by
superimposing tho boams. It has boon provod that there are
43
alternating maxima ami minima of intensity I (v), whorons
the distance in froquoncy bolwoon adjacent maxima
Av is 10 MHz (Fig. 386). Determine tlio path-length dif­
ference A.
4.10. An electron moves in a vacuum with a volocity v
near the surface of a diffraction grating with period d.
The electron’
s velocity is in parallel to the grating's sur­
face and perpendicular to tho grating’
s rulings. What
Fig. 39
Fig. 40
wavelength may be emitted at 0 to the normal to the grating
due to an interaction between tho electron and the grating
(the Smith-Purcell effect)?
4.11. A parabolic mirror 1 m in diameter is used as an
aerial for 3-cm waves. Find the minimum distance at
which a receiver must bo placed to obtain radiation
pattern.
4.12. A plane wavc passes through a glass plato with
refractive index n = 3/2 by falling normally onto its surface.
There is a step change in the plate’
s thickness, the chango
being by 6 = 2/3 X, along the straight lino passing through
the point C and perpendicular to plane in Fig. 39. Find tho
intensity of light at the point 0 lying in the plane that passes
through C if the case of a plane-parallel plate (viz. at 6 =
0) it is /0.
4.13. In an optical cavity composed of 4 piano mirrors
(Fig. 40), light can propagate in oithcr direction along the
perimeter of a square of length /. If the cavity is rotated
with an angular volocity il around on axis perpendicular to
the plane of the figure, then the resonance frequencies for
the waves arc different in each direction. Explain the pheno­
menon and determine Iho difference between tho frequencies.
4.14. Estimate the duration of a light pulso from a faco of an
octahedral rotating mirror 200 in away from a point sourco
of light. The light pulse is delected by a photomultiplier
44
placed near tho source and behind a narrow slit (Fig. 41).
Tho mirror’
s face is 1 cm wido. Let the wavelength of the
light be X = 5000 A. Tho mirror rotatos with a frequency
v = 16 Hz.
4.15. A lens with a focal length F = 50 cm and diameter
D = 5 cm is illuminated by a parallel monochromatic beam
of light with wavolength X = 6300 A. IIow many times great­
er is the light intensity at
[the lens’
s focus Ilian the in­
tensity of tho wnvo incidont
on tho lens? Estimate tho
size of the spot in tho focal
plane.
4.16. A rocket moves away
from the Earth and ceases to be
discernible against the back­
ground of the sky using a telescopo with a D x = 80 mm ob­
Fig. 41
jective when it is 2 X 104 km
away from tho Earth. IIow far
from the Earth could tho rockot be detected using a tele­
scope with a D %= 200 mm objective and given the eye’
s
contrast sensitivity is the same?
4.17. Radiation from a continuous laser at a wavelength
X = 0.63 pm and power P = 10 mV is directed onto a sat­
ellite using a telescope having an objective of diameter D =
30 cm. The light reflected by tho satellite is collected by
another telescope and focussed onto a photodetector whoso
threshold sensitivity is f\hr = 10"14W. Estimate the max­
imum distanco at wliicty tho light reflected by the satellite
will still be detectable. Tho satellite's surface uniformly
scatters the incident light with a reflecting coefficient
p = 90%. The satellite is d = 20 cm in diameter.
4.18. How far away can the naked eye see the light of a laser
which generates a continuous power of 10 W at a frequency
v = 4 x 10l* Hz if the beam is formed using a parabolic
mirror of diameter D = 50 cm? An eye can see a source if
the pupil (diameter d = 5 mm) receives 60 quanta of light
per second with a frequency within the green part of a
spectrum.
4.19. When an aerial photograph is made of the ground an
objective with focal longlli of 10 cm and diameter 5 cm is
used. Tho photograph is takon using a film with a resolution
of 100 linos per mm. Dotormino how much detail on tho
45
ground ran )>e resolved on n photograph iT they were taken
from a height of 10 km.
4.20. A telescope objective has a Tucal length /'*, - d ill
and diameter IJ - 15 cm. Determine what tin1 focal length
l'\ of the eyc-pieco must ho in order to use completely tho
objective’
s resolving power; the eye pupil’
s diameter is it mm.
Assuming that thoro is no aberration in the Lelesoope-eyo
system, estimate how far away a hook with letters I v 2 mm
in size can he read using the telescope.
4.21. Some astronauts have landed on the Moon. In order
to communicate the fact to tho Earth, they lay a black circu­
lar canvas out over the Moon’
s surface. What radius must tho
circle have to be visible from the Earth through a telescope
with a 5-m objective? The receiver’
s contrast sensitivity is
0 .01 .
4.22. A glittering metallic interplanetary spaceship 10 m
in diameter has landed on tho Moon during a full Moon.
Estimate the diameter of a telescope’
s mirror needed to ob­
serve the spaceship from the Earth if the contrast for relia­
ble eye observations is taken to be k — 0.15. The reflectivity
of the m oon’
s surface is
- 0.1 ami that of the niclal sur­
face is p 2 — 1. The distance between the Earth and the Moon
is 4 x 10* km, and the observation is made in light of
wavelength X = 0.6 pm.
4.23. The angular aperture of an electron microscope is 10~4
and that of an optical one is about 1. What voltage is needed
to accelerate the electrons for the resolving powers of the de­
vices to bo identical?
4.24. The spectral lines emitted by a healed gas appear broad­
ened because atoms of the gas move with different veloc­
ities relative to an observer (Doppler effect). Assuming that
the velocity distribution of the atoms is Maxwellian, esti­
mate the dimensions of a diffraction grating to be used for
studying the form of the spectral lines emitted by 20Ne at
1000 K. The diffraction constant of the grating d is 1 pm.
4.25. Estimate the order of the velocity with which a space­
ship must move away from the Sun in order for an astro­
naut using a spectrometer with a diffraction grating to be
able to detect the spaceship’
s motion relative to tho Sun by
observing the visible part of the Sun’
s hydrogen spectrum
in the second diffraction order. How many rullings must
the diffraction grating have? The Sun’
s surface tempera­
ture is G000 K.
4.26. A binary star consists of two stars similar in mass ro41i
tating with a period of 10 days and 2 X 107 km apart. De­
termine how many rulings a diffraction grating must have
ill order that the relative rotation of the stars can ho detect­
ed iu tho second diffraction order hy observing the visible
part of tho hydrogen spectrum iu their rudiulion. Is it pos­
sible in priuciplo to doled the relutivo rotution of the stars
using this techniquo, if tho poriod of rotation is 10 years?
The temperature of slurs’surfaces is 0000 K.
4.27. Ruby lasor radiation isscaltorod by sound oscillations
in water. Whon tho light is scnllorod n Dopplor frequen­
cy shift occurs. Estimate how many rulings a diffraction
grating must have for the shift in tho light scattered ut right
angles to be detected in the first diffraction order. The veloc­
ity of sound in water is 1400 m/s, and the refractive index n
is 1.3. Assume that sound waves in wator have evory possiblo
orientation.
4.28. How many rulings roust a diffraction grating have in
order to detect (during a total solar eclipse) a gravitational
shift in the spectral linos of tho spectrum of solar radiation?
Tho observation is made in tho second diffraction ordor. The
data needed for a numerical calculation can be dorived from
the angular diameter of the Sun (obsorved from tho Earth),
viz. a « 0.01 rad and the orbital velocity of tho Earth,
v = 30 km/s.
4.29. Determine the resolving power of tho following typo
of infrared spectrometer. Tho radiation of an infrared source
in the range am « 3 pm is mixed in a nonlinear crystal
with the radiation of a stabilized argon laser. A sum radia­
tion with u froquoucy lying within optical rango is produced.
This radiation is analyzed using-a Fabry-Perot interferome­
ter whose mirrors are 1 cm apart and have a reflection co­
efficient of 0.9.
4.30. A beam of plane-polarized light of wavelength G280 A
and power 3 W is incident on a quartz four-wavo plate. Un­
der what conditions will the plate experience a torque and
what will be its magnitude and direction?
4.31. A plane monochromatic wave is incident on a period­
ic structure composed of thin parallel dielectric plates
(Fig. 42). The plate thickness is d0, the distance between the
plates is d, while the permittivity of the plates is
and
that of the surrounding medium e. The wavelength is much
longer than d0 and d. Demonstrate that this structure is
analogous to uniaxial crystal and determine the refractive
indices of tho ordinary and extraordinary rays.
47
4.32. A plane monochromatic wave of frequency o) is inci­
dent on a system of nll-cmaling polnroids and quart/ plates
cut parallel to their oplical axis (Fig. 43). The principal
directions of all the polaroids are parallel and at 45° to
the optical axis of the plates. Tho wave is polarized along
the polnroids’principal direction. Determine tho wave’
s am­
plitude at tho outlet of the system if at the inlet it is AQ.
The system consists of N plates
and N + 1 polnroids. The
^1 ^1
thickness of tho plalos is d,
2d, . . ., 2N~1d. The refractive
indices of quartz aro n0 and
n9. The light reflected at the
interface between the plate and
the polaroid must be neg­
lected. Is such a system a
spectral device or not?
Fig. 42
4.33.
of gas near the surface of Venus
and tho derivative*of the refractive index with respect to
height at the surface. Venus’
s atmosphere consists of C02mo­
lecules whose polarizability is 2.7 X 10'Mcm3and its surface
pressure p0 is 100 atm, and its temporaturo is 500 °C. Find
p,
?N.
1
^
i ^
1
§
! 1 11 1
1
nZ
$
P
AA
1
Fig. 43
the curvature of a light beam emitted horizontally. What can
we say about the planet’
s atmospheric optics from this result?
Note. The radius of curvature R of a horizontal beam is
governed by the relationship
4.34. How much more dense must the Earth’
s atmosphere bo
for the circular refraction of light around the EarLh to exist,
as it does on Venus? The refractive index of air at atmospher­
ic pressure is n0 = 1.0003 (cf. Problem 4.33).
48
Find the
4.35. A lens with a relative aperture of 1 : 3.5 focuses solar
light onto tho surfaco of a black boll placed in n vacuum. To
what temperature could the ball be hoatod if its diameter
equals that of the solar image, if the Sun subtends from the
Earth an angle a = 0.01 rad, and if the solar constant p
is 0.14 W/cma?
,
t
.
4.36. How much power P must a laser beam of diameter
£) — 1 mm have in order to initiate an electric breakdown in
Fig. 44
Fig. 45
a gas? Under oxporimcnlal conditions tho mean freo path
of electrons in the gas is 10~* cm and its ionization
potential is 10 V.
4.37. A mirror in tho form of an extremely elongated para­
boloid of revolution focuses soft X-rays duo to total internal
refraction at sliding angles of incidouco a on tho paraboloid
far from the vortox (Fig. 44). Estimate the angle of conver­
gence q> at the paraboloid's focus for 2 keV X-rays if tho
mirror is made from berillium (berillium's density is
1.82 g/cm3).
4.38. Determine the number of free electrons per atom of
silver if a film of the metal is transparent for ultraviolet
light starting from 5 eV. Silver’
s mass number is 108 and
its density 10.5 g/cm3.
4.39. The refractive index of the ionosphere for radiowaves
of frequency 10 MHz is 0.90. Find the electron concentration
in ionosphere, and the phase and group velocities for the
radiowaves.
4.40. A laser beam is focused by an ideal optical system with
P/D = 1. Estimate the laser power P needed to donate an
energy of about me2 to electrons in the electric field at the
system’
s focus.
4.41. A laser beam at point A (Fig. 45) with frequency o)0 is
incident on a satellite flying with velocity v. The reflected
beam is detected at point B. What will the froquency of the
4—0408
49
detected light bo? Estimate the rosolving powor a detecting
spectral device would need in order for the relativistic
correction to the frequency shift to be dotccled.
4.42. In order to lest tho tlioory of relativity an exact meas­
urement of the parameters of tlio orbit of an Earth salollile
using radiowaves is planned. However, mousuroment orrors
emerge because the radiowaves are refracted in the ionosphere,
the average electron concentration there being 105 cm"3.
Estimate the minimum frequency at which the measure­
ments should be performed.
4.43. The acceleration of a space rocket is the greater the
higher the velocity of the gas ejected from its nozzle. From
this point of view the best type of engine is a photon rocket
which emits a photon flux from its nozzle. Calculate what
velocity will be acquired by such a rocket if starting from
zero velocity half its mass is converted to photons. The
engine efficiency can be taken as unity.
4.44. After 16 orbits around the Earth a satellite is brought
back down to the space-launch complex. What will the
difference between the satellite and space-launch complex
clocks be and how accurately can it be detected if both the
stability and reproducibility of the clock readings amount to
10"13 (hydrogen maser)? The effects of trajectory curvature,
gravity, and the acceleration of tako-off and landing on the
satellite's clock must be neglected.
V. Atomic Physics
5.1. Determine the energy of the y-quanla which experienced
a Compton backscattering (<p = 180°) if the ejected electron
is ultrarelativistic (E
mrs).
5.2. What X-ray wavelength is doubled after a Compton
scattering through 90°?
5.3. A crystalline plate is used in an X-ray spectrograph
that uses Bragg interference reflection. What is the
minimum thickness of the plate D at which it will be pos­
sible to detect the Compton shift when a photon is scattered
at an angle <p = 90* to the initial direction? The X-ray wave­
length is /. = 0.7 A and the scattered X-rays are incident on
the crystal at the slip angle 0 = 30°.
5.4. The Compton scattering of a photon from an electron
at rest results in the electron being given a recoil momen­
tum p. Determine at which angles <p to the incident photon
direction the electron can be ejected.
50
5.5. Eslimato Iho anglo of doflcction duo to gravity of
a photon moving near the Sun’
s surfnco (A/s = 2 X 1033 g,
7?s = 7 X 10'° cm).
5.0. A pnrallol beam of monooiiorgulic noulrons moving with
a volocity v is incidont on a plane crystal surface at a slip
angle (p0. It then undergoes m-lh order Uragg reflection
(Fig. 46). Tho neutron source
is set in motion with a con­
stant velocity u along tho nor­
mal to tho roflocting piano. At
what anglo q> to the plane must
the neutron beam be directed
to observe the initial m-th
order Bragg reflection?
Flg. 46
5.7. Ono technique for roonochromoLisinff slow neutrons involves a solid cylinder of radius R — 10 cm and length L =
1.0 m with a spiral groove of width b =* 1 cm and pitch
anglo q> = 30°(Fig. 47). Tho cylinder rotates at v = 3000 rpm.
Determine the wavelength of tho neutrons transmitted by
tho monochromator and estimate thoir monochromaticity.
The neutron beam is directed along the cylinder axis.
5.8. Estimate the minimum diameter D of the spot produced
on a detector by a beam of silver atoms emitted from an oven
with a temperature T — 1470 K. The distance between the
oven’
s exit slit and the detector is L = 1m.
5.9. A one-dimensional well of width b and infinitely deep
contains N electrons. Determine the minimum value of the
total energy and pressure of the electrons upon the w ell’
s
walls. Electron interaction should bo neglected.
5.10. Estimate the kinetic energy of a nucleon in a nucleus
assuming that the nuclear radius is r = 10"11 cm.
5.11. Assuming nuclear forces are duo to on exchange of nuc­
lear field quanta, viz. mesons, betwoen nucleons estimate
4*
51
the range of the nuclear forces if the rest mass of the mesons
is known to be me1 « 100 MeV and thoir velocity is close
to that of light.
5.12. An olectron moves with a velocity v perpendicularly to
the bounding planes of a plane-parallel layer of a substance
of thickness d and with refractive index n. The electron
velocity v is greater than c/n so that Vavilov-Chcronkov
radiation is observed. Determine Ihc additional angular
opening Aq> of the radiation, which results from the finite
thickness of the layer (lfig.
48).
5.13. Find the minimum en­
ergy of on olectron at which it
will easily pass ovor a rectan­
gular woll of depth U = —5 eV
and width b = 10"8 cm.
5.14. The wavelength of the
Ha-line in a Dalmer scries is
X = 6503 A. Determine the
ionization potentials of positronium and muonium in the
ground state. The muon moss
is mn = 214 mt (in9 is tho
oloctron mass).
Kig. AH
5.15. Calculate tho onorgy of
tho radiation emittod during
the transition of a negalivo muon in a hydrogon atom from
= 106 MeV. In
the N to M shell. The muon mass is
the calculation lake into account that such a mass is about
a tenth the mass of a proton. How large is the Bohr orbit
radius in this case?
5.16. A positively charged muon that forms a hydrogen-like
atom, viz. muonium, together with an electron decays into
a positron and a neutrino which rapidly scatter in different
directions. What is the average value of the kinetic energy
of the remaining electron if the muon decay occurred when
tho muonium was in the 25-state? The oloctron wavo func­
tion is
\y = ----!---- /1 ----- — ) e-r/2r,
V1
2r , r
where r, is the radius of the first Bohr orbit.
5.17. Atomic energy levels are usually found by assuming
a point nuclear charge. In fact nuclei have finite dimen­
sions and nuclear radii are described thus
= 1.3 X
52
10Ll33/ X c m , where A is the relativo atomic mass. Estimate
the sign and’
order of magnitude of the relativo correction
AE/E to the energy of a muon in the /(-shell of a neon raesoatom (Z = 10, A = 20), which correction is due to the
muon being afraction of time inside the nucleus, viz. in
a field whose potential differs from Za*/r. The normalized
expression for the wave function of an electron’
s ground
state in a hydrogen atom is
ur---- !— e-r,ri,
1^5
whoro rx is tho radius of the first Bohr orbit.
5.18. In Compton scattering of quanta from electrons, the
phenomenon is complicated by the^olectrons not being at
rest in the atoms. Estimate
tho relevant spread in ejection
2 3 jr.cm
nnglo of the recoil electrons
knocked out of tho hydrogen
atoms if the X-ray ^quanta or
a wavelength X = 1 A are elas­
tically scattered strictly back­
wards.
5.19. A parallol beam of at­
oms which are all in tho same
excited stale movos along a va­
cuum tube axis at v =10® cm Is.
Tho tube walls havo tho win­
dows for detecting the radiation
from the atoms as a function
of the distance covered by the
beam in the tube. The results of llieso measurements are
shown in Fig. 49. The abscissa is the distanco covered by
the beam along the tube and measured from the first window,
while the ordinate is the natural logarithm of the ratio be­
(x) and the intensity measured
tween the intensity of light
by tho detector in the first window. Detcrmino the natural
width of the spectral line emitted by tho beam atoms.
5.20. Estimate the minimum size of a speck of iron needed
for the Mossbauer effect to be observed with a transition ener­
gy E = 14 keV and a lifetime x = 10-s s if the speck recoil
results in a Doppler shift equal to the intrinsic line width.
Note. The Mossbauer effect occurs when at a rather iow
temperature the entire crystal (in this case the speck of dust)
recoils rather than just tho individual irradiated nucleus.
53
5.21. At what height abovo a source // must an absorber bo
placed to delect the red shift predicted by tho general the­
ory of relativity? Consider the Mossbauer effect on the
•
7Zn isotope. The lifetime of the excited level with an
energy E = 93 keV is t = 10-6 s. Assume that in order to
get the needed accuracy the red shift must bo tenfold more
than the width T of the resonance absorption line.
5.22. The energies of the transitions between three successive
rotational energy levels in a diatomic molecule are measured
experimentally (Fig. 50). Find the orbital quantum numbers I
of the levels and tho moment of inertia / of the molecule.
5.23. In Shull's experiments (1968), a neutron beam was ob­
served to split into two beams when refracted at the bounda­
ry of a homogeneous magnolic held. Find tho small anglo 0
I Z-xr^eV
i
W/S/A.
L,
1* KT^eV
Fig. 50
Fig. 51
between tin* directions of IIn* refracted bourns. The m agnetic
field had an induction //
2.5 T. The neutrons of wavelength
X - :> A were incident at .'10° on the sharp boundary of the
m agn etic hold.
5.24. A beam of sodium atoms is ejected from an oven whose
temperature is T — 700 K. The beam is split in a transvorse
inhomogeneous magnetic field with a gradient dB/dx =
5 T/cm along the path of length I = 10 cm. A detector is
placed 65 cm apart from the magnet. Find the separation
distance between tho spots on tho detector screen.
5.25. A beam of lithium atoms in tho ground stale with
a maximum energy Ek = 0.1 cV passes through a Stcrn-Gerlacli magnet L, = G cm long and having a field gradient
dBldx = 5 T/cm. Two identical diaphragms D, L = 1 m
apart, stand in front of the magnet (Fig. 51). How largo can
the diaphragms be for the components of the split beam to he
completely separated?
5.26. Kstimate tho distance L between mirrors and a FabryPerol interferometer needed to observe Zeeman splitting in
S'.
a magnetic field B — 1 T. The interferometer mirrors are
silvorod in such a way that N « 20 reflections are observed
between them.
5.27. Estimate how large a magnetic field can be detected
for a Sun-like star (period of revolution t = 10* s, radius
R =s 1010 cm, surface temperature T = 6 X 10s K) on the
basis of measuring the Zeeman effect in the optical part of
spectrum (to = 10l# s-1).
5.28. Caesium is an alkali metal. During a P-S transition,
caesium emits a broad doublet composed of two lines: Xx =
4555 A and
= 4593 A. Draw a diagram showing how the
terms of the doublet’
s will be split in a magnetic field.
Is the line splitting in a magnetic field B — 3 T described
by the formulae for the normal or the anomalous Zeeman
effect?
5.29. In lithium’
s spectrum the first two lines of the prin­
cipal series pertain to the transitions 2 tPl/t - * 2 tSi/i and
2 2Ps/a -* 2 aSi/a. The wavelengths of these lines are 6707.80
and 6707.95 A, respectively. Estimate the magnetic field
that induces electron orbital motion in a lithium atom in
the 2P-state.
5.30. Estimate (in electron-volts) the splitting of the positronium 2P-stato induced by the interaction of the spin
magnetic moments of the positron and electron.
5.31. Estimate the order of magnitude of the splitting of
the loading line of tho 13aIraor sorios in hydrogen's spoclrum
due to the interaction botweon the magnetic moments of the
olectron and tho nuclous (hyporfino splitting of spectral
lines). Could this splitting bo detected by a spectral device
if the average lifetime At of excited hydrogen atoms is of
the order of 10"* s? The magnetic moment of a proton is
2.8 nuclear magnetons.
5.32. Estimate the order of magnitude of the wavelength
of the radiation of interstellar atomic hydrogen in radiofrequency range. Interstellar hydrogen is in the ground
stato and its radiation is due to electron spin reori­
entation.
5.33. Free atoms can possess a magnetic moment but have no
dipole electric moment. Under certain conditions, atoms
forming a crystal lino Inltico can also havo such a moment.
In this case paraolcclric resonance is possiblo, which is
analogous to the paramagnotic ono. Find tho atomic dipole
momont if it is known that tho resonance absorption of elec­
tromagnetic waves of wavelength X = 5mm is observed in
55
n stationary electric Hold strength 1C — 2.5 v 1U3 kV/m.
Estimate the size of tho atomic dipole.
5.34. Liquid helium possesses n negativo work function for
excess electrons. Therefore an electron introduced into the
helium repels tho atoms of the liquid and forms a sphoricnl
cavity, which i9, for tho electron, a potential woll with
practically an infinitoly high wall. Calculate tho cavity
radius It when the electron occupios the lowest quantum lovcl
in the cavity. The surface tension of liquid helium is o =
0.35 dyn/cm. Assume the external pressure to he zero.
Noglect tho polarizability of helium (due to tho smallnoss
of its dielectric susceptibility).
5.35. A region of high pressure forms around an eleclricnlly
charged particle, o.g. an Ho+ ion, in liquid helium, and in
the immediate vicinity of the chargo tho helium solidifies.
The incrcaso in pressure is caused by tho attraction of tho
helium atoms to tho eloclric charge because of tho polariza­
bility of their electron shell. Find the radius It of the ball
formed by solid helium if the helium density p = 0.145 g/cm3,
tho molar polarizability a N A = 0.125 cmVmol, and the
solidification pressure
~ 25 atm. Neglect helium’
s com­
pressibility; oxlcrnal pressure is absont and assumo that
the hall size significantly oxccods an atomic radius.
VI. Radiation
6.1. A circular disc at a temperature
is the .source of
thermal radiation. The disc is placed at a height II above
u plane blackened on both sides. Tho disc radius 7?<C // and
the disc is oriented parallol to the plate. Find tho stationary
temperature distribution across tho piano. Assumo that
the system is in vacuum, thermal radiation background is
absent, and thermal conductivity along tho piano can bo
neglected.
6.2. Find the ratio of tho energy flux donsitios W botwoou
corpuscular and wave radiations of tho Sun at tho Earth.
Assume that the corpuscular radiation is a neutral plasma
composed of electrons and protons with a concentration of
particles of oach sort n = 5 cm"3 and a flux velocity v —
300 km/s, and that the Sun radiates as an absolutely black
body with a temperature T — 0000 K. The angular dimen­
sion of the Sun is to bo taken as a — 10"3 rad.
6.3. At what concentration n of gas molecules is the gasokinetic pressure at T — 100 K equal to the pressure exerted
56
by equilibrium thermal radiation at the same temperaturo?
6.4. Gnscous neon in a closed vessel of conslnnl volume is
in equilibrium with the thermal radiation at T = 500 K. At
what pressuro will the neon's heat capacity becoinparablo
with the heat capacity of thermal radiation?
6.5. Starting from tho photon picturo derive the equation of
stato for electromagnetic radiation, p = 1/3 u. Using this
equation and applying thermodynamic laws to electro­
magnetic radiation, find the dependence of onorgy and
entropy per unit volumo (u and s) on loinporaluro T. ltolalo
the integration conslnnl that omorges in this calculation to
tho Stofan-Boltzmonn constant o.
6.6. Find the heat capacity Cv and tho adiabatic oqualion
for the equilibrium radiation enclosed in a vcssol of varinblo
volume.
6.7. An excited atom with excitation onorgy E = 1 cV is
in a fiold of equilibrium radiation with tomporaturo T —
300 K. Find tho probability ratio botwoou iuducod ami
spontaneous atomic radiations. Find an analogous ratio for
the olectron spin in a magnetic fiold D — 103 (I.
6.8. Estimate tho probability wK1> for a nioloculo to undergo
n spontaneous transition from an oxcilod level Em into
a lovel En when the molecule is insido n envity tuned to
a frequency u> = (Em — E n)!h. Tho corresponding proba­
bility of spontaneous rodinlion in free spneo is u£p. Tho vol­
umo of tho cavity is V and its quality factor is Q. Assume that
tho width of tho molecular lovcls T is always smntlor than
that of tho cavity line, viz. r < u>/(7.
6.9. A laser cavity with a ruby crystal has ono mirror with
100% reflection and another with a transmission t = 0.1 at
the wavolongth of lasor generation modo. Tho crystal length
is./y == 12 cm. It is known that the absorption indox of
jight in noncxcited ruby crystal at the maximum of the work­
ing lino is a = 0.4 cm*1. Find whnl fraction of tho chromi­
um atoms must bo excited to start laser action. Tho scatter­
ing of light in the crystal should ho neglected.
VII. Solid State
7.1. An X-ray with frequency 1.1 x 10,s Mz is incident on
a cubic crystal in the [1001 direction and experiences strong
Bragg scattering in the 11221 direction. Assuming that tho
57
crystal is a single face-conlcrcd Bravais lattice and the
observed scattering is due to first order interference, find the
minimum interatomic distance in the crystal.
7.2. Find the dispersion la w fo r longitudinal phonons in an
endless chain whose elementary coll contains two atoms with
masses M x and il/a. The distance between two adjacent
atoms is a, and the rigidity of the interatomic links is y. Con­
struct the plot of tho dopendenco obtained. Indicate the
limiting conversion to a monatomic chain at
i.
7.3. Find the polarizability (per cell) for a chain whose
elementary cell contains two single-charged ions with charges
of opposite sign. The remaining conditions are the same
as those in 7.2. The electric vector of an exciting electro­
magnetic wave of frequency a) is oriented along the
chain.
7.4. Using the conservation laws of energy and momentum,
consider the inelastic single-phonon scattering of neutrons
in an ideal crystal. Discuss the possibility of reconstructing
the phonon dispersion law from the neutron scattering.
7.5. Derive the valuable formula fra ~ (m/M)1f2 E Bt for
the oscillation frequencies of molecules and crystals. Here
m and M are the masses of tho electron and tho nuclei, re­
spectively, and Z?al has tho atomic order of magnitude. Ob­
tain the estimate x j a ~ (mlM)x/A for the ratio of the zero
oscillation amplitude x0 to tho interatomic distnneo a.
7.li. Using an analogy between photons and long-wavo pho­
nons, express the low temperaluro lallico boat capacity of
crystals in terms of the velocities of transverse and longi­
tudinal sound.
7.7. Find the temperature dependence of the lattice heat ca­
pacity of one and two-dimensional crystals in tho low tompcraluro range.
7.8. Starting from a model of a diatomic molecule with
a weak cubic anharmonicity, express the temporaturo coeffi­
cient of expansion a in the direction of the molecular axis in
terms or the parameters of tho model's potential andcstimalo
this coefficient. Assume tho motion of nuclei to be classical.
7.9. Find the low temperature entropy of crystalline holium-3. The helium-3 nuclcushas a spin / = 1/2. Assume that
the temperature
0 (0 is the Debye temperature) is such
that practically all the oscillatory degrees of freedom are
“frozen-out”, hut the nuclear spins remain totally disor­
dered. Under the same assumptions find the entropy of an
argon-37 crystal (nuclear spin / = 3/2 h).
58
7.10. Electron gas pressure is ono of tho principal factors
controlling tho compressibility of metals. Find the coef­
ficient of the uniform compression of electron gas for copper
at T — 0 K, if the electron concentration is n = 8.5 X
10aa cm"3. Assume tho effective mass is equal to tho free
electron mass.
7.11. Find the Fermi volocity vF of the electrons in a metal
with ono electron per elementary cell and with a “
one-dimen­
sional" dispersion law E — E0 cos {kza) for E 0 = 0.5 eV and
9 = 3 A (& = pIh is the wave vector).
7.12. In a metal whoso crystalline lattice has z-aais sym­
metry of order not lower than the third, the simplest
dispersion law for electrons can be given as E (p) = 1/2 m
(pi -f pj). Assume that each elementary cell contains one
conduction electron, the lattice period along the z-axis is
a = 3 A, the elementary cell volume is V = 0.85 a3, and m is
equal to the froe oloctron mass. Calculate tho Fermi velocity
of the electrons and determine llioir frequency of rotation
in a magnetic field H in terms of the angle 0 between the H
Vector and the z-axis.
7-13. With no electric current an external static electric
field only penetrates a thin surfaco layer of a metal.
Derive the equation for tho decay of tho fiold into the
metal assuming that tho total potonlinl drop is V< Eyle
(Ey is tho oloctron Fermi energy). Eslimalo the doplh of
field penetration (tho Thomas-Fermi screening length l Tv)
for a normal inotal (n0 « 1033 cm"3, Ey « 5 oV, pormiltivity e « 1) and for a Bi-typo somiraotnl (n0 « 3 x 1017cm "3,
Ey « 2 X 10~a eV, e « 100) at a f = OK. The permitti­
vity e is conlrollod by tho polarizability of tho intornal
electrons which do not participate in tho electrical conduction.
7.14. For typical values of tho parameters of metals, cstimato the temperature at which electron and lattice heal ca­
pacities become equal by expressing this temperature in terms
of tho masses of tho electrons and tho nucloi, and in terms
of tho energy E at having an atomic order of magnitude.
Make a numerical eslimalo of this temperature.
7.15. Since the moan free path in a thin wire is limited
by the wire’
s diameter, the mean free paths of electrons
and phonons are practically the same. Under these condi­
tions, estimate the temporature at which the electron and
lattico thermal conductivities of metals become oqual.
59
7.16. Find the frequency of the nnlurn) long-wavo oscilla­
tions of electron gas in conductors which induce no ningnolic
friction"
field (the plasma frequency u>p). Neglect any “
encountered by the motion of the electrons in the crystal­
line lattice and assume that the electrons' dynamics is
described by an effective mass m*. Estimate the frequency
top and quantum /i<op for metals.
7.17. Find the frequency dependence of the complox permit­
tivity i (o) for a conductor. Explain tho transparency
of metals observed in the ultraviolet region of spectrum.
Assumo that the conduction electrons are described by on
effective mass m* and o mean
free lime x.
7.18. Determine the depth of
penetration of a high-frequen­
cy electromagnetic field with
n frequency a> into a metal
whose conductivity is o. As­
sume that <OT<fC 1.
7.19. Calculate the current
produced by electrons moving
in crossed electric and magne­
tic fields (E _L H). The effective
electron mass is m*. and the mean free lime x. Apply tho
results to the calculation of resistance in terms of H for
the following cases: (a) a current runs through an infinite
plate with E and n lying in plane of tho plate; (b) a current
runs through a Corbino disc (in this disc electrodes aro
concentric circles, while a magnetic field is applied perpen­
dicular to the plane of the disc). Comment on tho results
obtained.
7.20. Thermoelectrons emitted from a metallic cathode move
in a vacuum due to an external field E and mirror imago
forces. The combined action of these forces results in tho
work function at the metal-vacuum interface being lowered.
Find the dependence of the work function (pE (Fig. .r>2)
and the thermoemission current .7 on the field E.
7.21. Derive the equation for the dissipation of space charge
in conductors and the characteristic lime of this process, viz.
the Maxwellian time of relaxation xm- Determine xm for a cry­
stal of pure germanium at room temperature (a = 0.014 S/cm,
r = 10).
7.22. In both semiconductors and metals (cf. Problem 7.13),
an external field is screened by conduction electrons. Tho
GO
difference between the two is that in semiconductors the
electron gns is generally nondogcncrale, viz. this gas fol­
lows a Boltzmann distribution. Under Iheso conditions, as­
suming the external held is weak, derive the law of the field
falling off deep into a nondcgcncrale semiconductor. Esti­
mate the field penetration depth /D (Debye screening
length) for a semiconductor with permittivity e « 15 and
electron concentration n « 10u cm “
s.
7.23. Calculate tho electron rotation frequency e>c (cyclo­
tron frequency) in a stationary, uniform, random-wionted
field H whose dispersion law is
F ,nv
b
Pi
, K
| ^
“ 2mT + 2my + 2m, *
7.24. Calculate tho shape of the cyclotron resonance curve
i(viz. the. frequency dependence of the conductivity) for
Electrons with an isotropic offoclivo mass m* and a mean
•free time x. The electromagnetic wave is circularly polemized
in a plane perpendicular to a stationary magnetic field.
7.25. Anelectronwho.se dispersion law is# = ? 0 cos (kKa)
moves in a stationary uniform electric field E directed
along the rr-axis. Solve the equations of motion and give
h physical interpretation of the result.
7.26. Solve the provious problem, given a frictional force
proportional to the velocity (F = —yv). Investigate tho possi­
ble modos of the electron's motion and calculate the con­
ductivity of such a semiconductor.
7.27. Consider the scattering of a slow electron in a semi­
conductor with absorption and emission of a long-wave
acoustic phonon by using the laws of conservation of energy
and momentum. Find the angle between the phonon wave vec­
tor q and initial electron momentum p in terms of p and q.
Show that for v < s (v is the electron velocity, s is the veloc­
ity of sound) tho electron cannot radiate phonons, while
foru m s the electrons are scatlorcd almost elastically, viz.
little energy is changed during the scattering. Assume
that the electron energy is E = p*/2m, and the phonon
energy is Aw (q) = hsq.
7.28. In some semiconductors the electron mean free path
appears to be of the order of the interatomic distances. In
such a situation the electrons can be considered to move
in random “
jumps". Estimate the conductivity of such
semiconductor for T « 300 K, if the olectron concentra­
tion is nt « 1018cin":l, the average rato of the jumps is
61
v « 1013 s-*, and the interatomic distance is d « 3 A.
7.29. A medium whose conductivity depends on an electric
field E , such that o (E) = o0/| i -|- (&7£0)sl, is placod
in an external homogeneous field 15 || x. Tho space charge
density (»(j) then begins lo fliicluale in lliis medium. Find
how the charge distribution along the i-axis will be changed
in time. Assume the fluctuation lo he small.
7.30. Derive a formula for the concentration of hand elec­
trons in a uondegcncralo semiconductor, given the chemical
potential £ (Fermi level).
7.31. Find the relation between the concentrations of tho
electrons n and the holes p in a nondcgcncrato semiconductor
having an arbitrary concentration of impurities, and the
concentration of carriers nt in the same semiconductor but
without impurities (viz, in an intrinsic semiconductor).
7.32. Investigate and show schematically the temperature
dependence of the concentrations of the electrons n and
holes p in a semiconductor with shallow donor levels. The
electron binding energy on donors is 2Alf and the forbidden
gap width 2A»2Aj.
7.33. Find the Hydbcrg constant, llohr radius, and effective
mass M for an cxcilon, viz. the hydrogen-like formation
constructed from an electron and a hole. Estimate tho values
of these quantities assuming that the effective masses of the
electron and hole are
0.1 m0 (m0 is the electron
mass), and the permittivity is e « 10.
7.34. The holes produced by illuminating an electron semi­
conductor surface diffuse inside, where they recombine with
conduction electrons. Determine the effective penetration
depth of the holes if their lifetime is x = 10° s and mobil­
ity is (i = 2000 cm2/(V-s) while the sample temperature
is r = 300 K.
7.35. Assuming that the electrons in a superconductor movo
without resistance, calculate the depth X to which a low
frequency electromagnetic field penetrates a superconductor
(London length). Estimate its value for typical parameters
of metals.
7.36. At moderate pressures helium remains liquid down to
the temperature T = OK. Determine the temperature depend­
ence of liquid helium's heat capacity (isotope 3He) at low
temperatures and estimate the numerical value of the coef­
ficient in this dependence by neglecting interatomic inter­
action. The molar volume of liquid helium at normal pres­
sure is V = 37 cm3/inoJ.
62
VIII. Nuclear Physics
8.1. In 197C the Nobel prize in physics was awarded for the
discovery of a new elementary particle, viz. the /-meson.
Tho discovery was made practically simultaneously and in­
dependently in two different experiments. One used colliding
beams of electrons and positrons accelerated up to the same
energy ^cma* Determine the mass and estimate the lower
bound of the J-particlc lifetime.
In another experiment tho /-meson decoy products wore
detected. Find the mass of the /-meson decaying into an
electron and a positron, if it is known that their energies
are identical (Ex = Et = 3.1 GeV) and the divergence angle
between them is q> = 60°.
*
8.2. The quark model of hadrons assumes that mesons are
composed of a quark and an antiquark while baryons aro
composed of three quarks. In a simple version of the model
there are four quarks and the same number of antiquarks.
The quantum numbers of the quarks arc given in the fol­
lowing table.
Spin,
h
Electric
charge*. Q
Daryon
number, U
Strangeness,
S
Clinrm,
1/2
+2/3
1/3
0
0
u
1/2
— 1/3
1/3
0
0
U
1/2
-1/3
1/3
-1
0
9e
1/2
+2/3
1/3
0
+1
Quark
type
y---------
C
From these quarks construct a proton, Q-hyperon, posi­
tively charged pion, kaon, and a /-particle. Also, find the
orientation of the spins of the quarks constituting these
particles.
8.3. In 1976 a new baryon was discovered which was inter­
preted as tho first member of the group of charmed anti­
lambda-particles. The baryon decayed according to the
scheme Ac-*-A0 + ji+ + Ji" + ji~. The baryon mass was
63
Me- — 2.25 fii»V. Find what fraction of the baryon rosl
mnss is transformed into Ilie kinetic energy '/' of llio decay
products.
K/i. Tho annihilation of an antiprolon sloppod in liquid
hydrogen produces three pious:
n+
+ ji°
.
Determine the onorgy of each if one of the pions has tho
maxi mum possible cnorgy.
S
8.5.
liding beams of electrons two
identical storage rings aro
used in which beams of ultrarelativistic particles move in
opposite directions and col­
lide with each other along an
interaction length I = 0.5 m
(Fig. 53). A system of coun­
ters surrounding the interac­
tion region is positioned
so that ono in ton particle
S
interactions (a detection ef­
ficiency e — 0 .1) is delcclod.
Kig. O.’
i
Find the circulating currcul ,)
that must bo accumulated
in each ring in order for the counter system to observe not
less than K — 10 counts per second. The cross-sectional area
of the circulating beams is S — 5 mm2 and the effective
cross section of the interaction of two colliding particles is
o - 1(1 5 barn. Assume that the number density of tho
particles along the orbit is constant.
8 .G. Determine the kinetic energy of a proton beam which
starts to emit Cherenkov radiation in nitrogen at 50 atm.
At normal pressure the refractive index is n0 = 1.0003.
8.7. A Cherenkov counter is installed in a beam of kaons
and pions whose momentum is pc = 500 MoV. Wlial must tho
refractive index of the counter substance n be for the coun­
ter to detect only kaons?
8 .8 . By virtue of the decay
+ v„ a beam of pions with
energy En = 10 (JcV gradually transforms into a flux of
muons and neutrinos. Estimate at what distance from the
production region the number of muons in the beam will be
double the number of pions. Neglect the escape of muons
from the beam due to deviations from the pion trajectory
and the decay of muons.
8.9. At what energy of a proton incident on a resting proton
O'.
In experim
in the reaction p -f p -*•d + ji+ may the kinetic energy of
the pion vanish in the laboratory frame? The deuleron
mass mdc2 = 2m»c2 = 2 X 0.94 GeV, and that of pion
mnc2 = 0.14 GeV.
8.10. To produce meson beams in accelerators, a thin target
is installed in a beam of accelerated particles. The mesons
are generated in the target. Determine how many mesons aro
produced per second if at the distance L = 5 m away from
the target a detector records n = 5 x 10* mcson/s, and the
detector area is S = 100 cm2. Perform the calculation for
pions with an enorgy T — 500 MoV, assuming that the pions
are omitted from the target isotropically (viz. with the
same probability at auy angle).
8.11. Estimate the fraction of cosmic protons reaching the
Earth without undergoing nucloar interactions in flight. As­
sume that tho cross section of the proton-nucleus interac­
tion is equal to tho goomclric cross section of nitrogen
nuclei.
8.12. Nuclear reactions on the Sun can be studied by mea­
suring the solar neutrino flux via the reaclion
v + pCl -► *“+ JJAr
Tho average (over solar neutrino spectrum) reaclion cross
sectioniso = 1.4 X 10~42cm2. Assuming that tho Sun emits
N *s 3 X 1033 neutrinos per second dotermine how much CCI4
(natural mixture of iaotopos) is uoudod to produco 100 faAr
atoms per year. The naturul mixture of chlorine isotopes
contains 25% (by mass) of *JC1 nuclei. Tho radius of the
Earth's orbit is R = 1.5 x 10* km.
8.13. Determine the maximum number of pions produced in
the p + p reaction if the incident proton momentum
p a 5 GeV. Assume that prior to the reaction the target
protons are at rest.
8.14. Find the lifetime of a muon in the laboratory framo if
the muon was produced from the decay of a stopped kaon.
8.15. In the radioactive decay of the °°Co nucleus an elec­
tron is emitted whose spin is parallel to its momentum.
Estimate the angle <p through which a disc suspended by
a filament will rotate if the *°Co preparation is applied over
one of the disc faces. The disc is thick enough to
absorb all the electrons emitted towards the disc. The
preparation’
s activity N is 10 Ci and tho filament torsion
modulus is / « 10“
* dyn>cm.
5-0408
65
ft.16. A specimen of toflon (a polymor whose choinicnl formu­
la is (CKa),.. whore n is an integer) of mass M = 50 g is mag­
netized in n magnetic field H — 20 kOe at T = 0.05 K. The
magnetization is due to the splitting, under the action of
the magnetic field, of the ground state of the l5F nucleus
(nuclear spin s =*1/2) into two sublevels. Upon switching
on the field the specimen gains an angular momentum L —
12.1 x 10"# erg*s (the analogue of Einstein-De Haas
effect in a ferromagnetic). Determine the value of the mag­
netic moment p, of the fluorine nucleus.
8.17. Estimate the half-life TUt of an even-even radio­
active nucleus emitting 1 MeV a-particles if tho ”$Th
nucleus has Tx/i = 1.4 x 1010 years and omits 4 MeV a-par­
ticles, while a *J{Po preparation emits 8.8 MeV a-particles
and has a Tlt% = 3 X 10" 7 s.
8.18. Determine the effective cross section of the reaction
* + E A l - H S i + P.
if it is known that np = 8 protons per 10* a-particles are
produced when a thick aluminium target is irradiated by
8 MeV a-particles. The a-parlicle rango in air under normal
conditions is R tlr = 7.0 cm.
8.19. I n a l MW uranium reactor (/u,) « 6 antinoulrinos
are produced, on average, per fission event of a uranium
nucleus (mainly from the decay of radioactive fission frag­
ments). The mean antineutrino energy is E~ « 1.5 MeV.
The reactor is surrounded by a thick biological (concrete)
shielding. Estimate both the antincutrino flux <b~ L = 5 m
beyond the biological shielding of the reactor and the
energy fraction taken from the reactor by the antineutri­
nos.
8.20. What is the kinetic energy Tn of the neutrons if, as a
result of the neutron irradiation of a liquid hydrogen target
and the reaction n + p-+ d 4 - n° the v~<IuaDt& ^rom the
decay ji0-+-y -f y are known to he emitted in opposite direc­
tions in the laboratory frame?
8.21. In an evacuated vessel of volume V = 1 1 ultracold
neutrons are reflected from the walls with a reflection coeffi­
cient practically equal to unity. The vessel is provided with
a window of area S covered with a foil that is completely
transparent to the ultracold neutrons. What must the
window area be if the neutrons are observed to bo stored in
the trap for holf as long as the mean lifetimo of free neutrons,
66
t « 10* 3? Assumo tho velocity of all ullracold noutrons is
the same and equal to v = 5 m/s.
8.22. Proceeding from Weizsackor’
sformulu, ostimate the sur­
face tension of nuclear matter.
fe.23. For neutrons to enter most substancos from a vacuum
they must overcome an energy barrier. Therefore, suffi­
ciently slow neutrons can be trapped in a closed cavity and
may be accumulated. Determine what fraction of a flux of
thermal neutrons with a Maxwellian velocity distribution
will be found trapped in a cop­
per chamber. For total in­
ternal reflection, the limiting
slipping angle for neutrons
moving with a mean thermal
velocity will amount toi = 10'.
The collisions betweon the neu­
trons and the chamber walls
may be
considered
as
elastic.
8.24. After entering water,
Pig 54
last neutrons rapidly slow to
thermal velocities and diffuse through the water unless they
are caught by hydrogen nuclei (capture by oxygen atoms can
be neglected). If the capture cross section is a = 0.3 barn
(1 barn = 10~24 cm1) estimate the neutron lifetime in
water t.
8.25. A slow neutron is scattered from a free nucleus with
spin / = 3/2 in the L = 0 state (orbital moment is zero).
Determine the scattering probability in states with parallel
and antiparallel orientations of the nuclear and neutron
spins.
8.26. Estimate the kinetic energy of a neutron Tnt and a aLi
nucleus TN produced in the photodisintegration of a 7Li
nucleus induced by 15 McV y-quanta. The neutron is emitted
forwards, viz. in the direction followed by the y-quantum.
8.27. A well-collimated beam of y-quanta with energy
500 MeV is incident upon a deuterium target. The secondary
beam contains nucleons duo to the photodisintegration of deu­
terium. Estimate the angle of ejection q> of the nucleons.
The deuterium radius is ; d = 2 x 10‘13 cm.
Hint. Take into account that the binding enorgy of the
nucleons in deuterium is small compared with y-quanta
energy.
8.28. In reactors operating with thermal noutrons very slow
s»
67
ultracold neutrons are produced. The feature of ultracold
neutrons is that for velocities v < vb (the usual boundary
velocity is vb « 10 m/s) the neutrons will he elastically
reflected from the walls for any angle of incidence. The ultracold
neutrons are extracted from the reactor using hollow tubes,
viz. neutron guides. Figure 54 shows a reactor R and a
shaped neutron guide with a neutron detector Z> at its end. As­
suming that the neutron velocity spectrum in the reactor is
Maxwellian (T « 400 K) find how the neutron flux reaching
the detector depends on its elevation K? Estimate the height
at which the flux vanishes. The effect of the detector on the
neutron distribution should be neglected.
ANSWERS AND SOLUTIONS
I. Mechanics
1.1.
T h e E a rth
a cts u p o n a s a m p le w e ig h t o f m a ss
F = [ym — - peRe ) / R e =
m w it h
Y P e ^ e " 1-
T h e c h a n g e i n th e f o r c e d u e t o a l e a d b a l l o f r a d iu s
R\ i s
bF = - ^ y p ]R\m,
w h e r e p j = 11 000 k g / m s i s t h e l e a d d e n s it y . T o m e a s u r e th e
c h a n g e w it h th e g i v e n a c c u r a c y r e q u i r e s
4 £ - > 4 - = 10-',
o
r
or
m,
Pi
v iz . t h e m a s s o f t h e le a d b a ll w o u l d e x c e e d 10s to n .
1.2.
L e t mx a n d m2 b e th e m a s s e s o f th e c o m p o n e n t s o f
a d o u b l e s ta r , lx a n d l2 th e d i s t a n c e s f r o m t h e ir c e n t r e s t o
t h e c e n t r e o f m a s s , a n d vx a n d v2 t h e ir v e l o c i t i e s , r e s p e c t i v e ­
ly . T h e n , a s s u m i n g f o r s i m p l i c i t y t h a t t h e y m o v e i n c i r c u l a r
o r b i t s , w e o b t a i n f o r th o f r e q u e n c y o f r e v o l u t i o n
II
lx
v2 __ Vl+Vt
l2 ~
L
L = lx+ Izi
T h e c o n s id e r a t io n o f th e fo r c e s a c t in g u p o n
s u l t s in
mxv\ _
h ~
m2vl
h
mxm2
w =
u* , i;«___ m2
L* ’ h~*~ 11
-y-
mx a n d m2 re ­
L>
*
i t c a n b e e a s i l y o b t a i n e d t h a t y (mx + m2) =
^n2L*T-2. S i m i l a r l y , f o r th o S u n - E a r t h s y s t e m (th e m a s s o f
t h e E a r t h MR i s n e g l i g i b l e c o m p a r e d w it h t h e s o l a r m a s s Ms),
w o o b t a i n yMs = ^ R e Te , w h o r e 7’
E = 1 year, w h ile
R e i s b y d e f i n it i o n o q u a l t o a n a s t r o n o m i c a l u n it. T h u s ,
F u r th e r ,
G‘
J
1.3. If on astronaut and a satellite inovo in the same orbit
then the cord tension force is F = 0. The force is greatest
when the astronaut, satellite and the Earth’
s centre lie on the
same straight line (Fig. 55). The period of revolution T of
the satellite-astronaut system near tho Earth is related to
the radius R of tho orbit of the system’
s contro of mass by
4n> „
- fT
vM e
r T~
whore Me is the mass of the Earth. Accordingly, to secure
tho astronaut on un orbit of radius R -f A/? neods an ad­
ditional force F conditioned by the relation
/d
i a
1
yMn
|
F
7*a ( f l+ A / f ) — (/?4-A/?)- *" m ‘
Whence, with an accuracy up to the terms of the order of
(AR/R)z wo have
6 yMKm A R __ 3//i/:AH
’
W ~~ R ~
R
n
« 2 x 10-* N.
Kig. 55
(here we took into account that
the satellite and astronaut
masses, .'1/ and m, arc much less than M u, and since M
m,
AR ss L). The same force will be needed to secure an astro­
naut between tho satellite and the Earth. To summarize, we
may say that tho gravitational force attracting the astro­
naut to the satellite is six orders of magnitude less than F.
1.4. Tho third Kepler’
s law suggests that the ratio between
the Mars and Earth masses is
where Tu is the period of revolution of the Earth satellite
found from tho relationship mco*/?E « mg (m is the satel­
lite’
s mass), L h = 1/2 (Z?M -f a + b).
1.5. p (r) = 4^- p*v(flfc—/■
*), p(0) » 2-10" Pa.
l.G.
— /?p « 10 km.
1 .7 . v — {vu -h V u\ — 2gRz )- —
E.
where uE and
/?e are the orbital velocity of the Earth and its radius,
respectively.
70
1.8. The potential energy of the apparatus U can be written
in terms of its distance x from the Moon’
s centre
U
_
ymM}L_y Myim ^
where AfE = 6 X 10” kg, Mw = 7.3 X 10” kg and m are
the masses of the Earth, Moon, and apparatus, respectively,
while L — 3.8 X 10* m is the distance between the centres
of the Earth and the Moon. The function U (i) has a maxi­
mum (dUfdx = 0) at x = 1
u)iW hein£ e(Iua^ t0
^m.x = — ym L^Myz +
-If the kinetic energy T
of the apparatus at this point vanishes then it is T =
Umax — U (/?M) near the lunar surfnco, wlioro /?M =
1.7 X 10* m is the lunar radius. The velocity of tho appa­
ratus near the lunar surface is
_ fo..r Mm . «E
” 1 M. RM + l - rm
(*1 T + *IP
l
1/2
■]>
« 7.5 kin/s.
1.9. A collision is a process during which the centres of
two stars pass each other closer than the s u m of their radii
(in our case, closer than 2RS).
To determine tho collision cross section, let us turn to
the centre of mass system of the colliding stars. Let b be
the maximum impact parameter (relative to tho centre of
mass) of this collision. Then, as it is seon from Fig. 56, the
collision cross section is a = n (26)*. It follows from the
71
laws of conservation of energy and momentum Unit
o Ms ; u \ * _ 0 M
$v%
2T
VM\
, u
\T ) ”
”
2 2---- 2/?r ’
,, „
“M s R *u '
where y is the velocity of each star relative to a fixed centre 0
whon these stars are at their closest, eliminating y, we
obtain
L. = 1+ ZrMs
S
J?s«*
2yMs
10*.
/fsu*
Hence, tho collision cross section is
o = 4nb'=Hnl±2!!i.
U*
Since the mean free path length L — 1/(no), where n =
N!{cTf (c is the velocity of light, T = 1 your = 3.2 x
107 s), thon the time interval between collisions for ono
star is
’- T 3 T "T S F ffPvVs = 2 2 y 10“" = 7 * 10”
vix. one collision por cubic light year occurs once every
t=
=■1.4 x 1013 years.
1.10. For motion in a gruvitnlionnl fiold, tho following
relations are valid:
P b ”3 = const, bE = const,
where T is the period of revolution, b the raojor soraioxis
of tho orbit, and E tho total onorgy. Honco wo have T*E* =
const, or for small variations of the period and velocity y
we obtain
A7*
T~
3
2
bE
F.
3 m
uAu
2
E
*
where m is the satellite mass. Therefore, because E is nogotivo, to decrease the period of revolution wo must also
decrease the satellite velocity (Ay < 0). As a rosult, tho
satellite will pass into an olliptical orbit with a smaller
value of b and in n turns it will travel from its initial
position on the circular orbit by a distance
L= —w A r= -|
72
n - ~ r
Ai> =
— 3n T bv
(Here wo have taken into account that for circular orbits
£ = —nw*l2.) I3y increasing the satellite velocity by
_ A e it will again be possible to transfer the satellite back
to its initial circular orbit so that the relative velocity
of the satellites will vanish. Determine n niul Sv at L ~
45 km and T = 1.5 h:
At/ = ---~
= ——
3nT
n ' km/h.
At n = 1, |Ao| > 8 krn/h, and at n = 2. |Ai»| = 5 km/h.
Thus, nmiu = 2. To dock the satellites, the velocity of tho
satellite that lags behind H must be decreased by 5 km/h
and two turns later restored to tho initial value.
1.11. The thermal velocity of atmosphoric molecules is much
smaller than tho satcllito velocity, r = 8 km/s. Under such
conditions, the collisions between molecules and tho satel­
lite ore mainly inelastic. It might bo, thereforo, thought
that after such a collision a moleculo gains n velocity ik
Accordingly, tho atmospheric molcculos that collide with
tho satellite acquiro a momentum «Sp*e* por second, vi/.
the frictional force acting on tho satollito is
P=
= 5 X 10'4 N.
Tho total onergy of the satullito on n circular orbit of
radius It is
P
mi'*
Y"»^fK
--- 2~ = ----W ~ •
whore AfK and m aro tho masses of tho liarlh and satollito,
respectively. Whcnco, upon changing tho onorgy by A£, tho
orbit radius changes by AIt - 2litj\E/{yinJ\f K)% whereas
the velocity is changed by Au = — A/iVniu. Tho satullito
onergy will chango due to tho frictional forcos by AE =
— 2nRF per revolution. Thcrohy tho orbit radius will ho
decreased by
4nfl»
3P
Afl
ymME
ymp 0.4 km.
Horo p = 5500 kg/m* is tho moon donsily of tho ICorlh. The
satollito volocity is incromontod hy
Av as OIL p ~ 0.5 m/s.
73
1.12. The projectile velocity u can he obtained from the
law of conservation of angular momontum:
invRe _
V 1—v*/c* ’
where u = 7.3 X 10’
s s " 1 is the angular velocity of the
Earth’
s rotation. Whence
rn«v«cl J?|.
C V
(c+ e) /5(0* '
Since v is only slightly different from e, a good approxima­
tion is
c —v
m3c3/ijj
2/2(o2
_ 0 /
me
y c = 1.56x10-” m/s.
~~2 \ MkRE(»
Since the projectile moves with an ullrarelalivislic veloc­
ity, its kinetic energy T (which coincides with the total
energy E) can he calculated from the formula
T~
~
mc%
Y 1— »l/c2 ~
The ratio between this quantity and the kinetic energy of
the Earth’
s rotation, Te = /co*/2 is
T
2c
= 1.3 in8.
Tv. mil
1.13. The change in the electromagnetic field momentum per
second is equal to tho‘thrust F = nM. Accordingly, Iho
power consumed,
N = Fc = aMc = 3 X 10l»W
(c is the velocity of light) is 670 times greater than that of the
Bratsk hydroelectric power station.
1.14. h = -^57;, where p is tho density of water.
pgM1
1.15. Sec the solution of Problem 1.2. For a neutron star,
T=
(3JI/YP)1/* = 1.2 X 10-3 s.
1.16. On a circular orbit of radius /?, the total (E), poten­
tial (£/), and kinetic (T) cnergios aro related through the
relation
r. V _
rp
ymM
E~ 2 ” T~
V< *
74
Aftor the explosion, the total energy E' will be
£ ' = r + £ / ( i - 9) = - ^ (
4~ f) .
At q > 1/2 this energy exceeds zero and the satellite leaves
the star via a hyperbolic orbit. At 7 = 1/2, E = 0 and
the satellite’
s orbit will be a parabola. At q <Z 1/2 the satel­
lite will follow an elliptical orbit, one of whose foci will
contain the remainder of the central star, while the major
semiaxis may be found from the well-known condition:
2E'b = —ymM (1 — q).
Taking into account tho expression for E' we obtain
1.17. in the process of braking, the spaceship velocity
changes from vt = 2.4 km/s (tho oscape velocity of the
Moon) to pj = vtl Y 2 =1.7 km/s (tho orbital velocity
of tho Moon). As a result, the spaceship mass should bo
decreased from its initial value A/0 to M
M = Mt e x p ( -
as 0.84 M,
(see the solution of Problem 1.22). Here u is the gas exhaust
velocity. Thus, the burnt out fuol should amount to 16% of the
spaceship mass. The combustion lomperaturo T can he esti­
mated from the formula for the exhaust of gas velocity
into a vacuum, i.o. u = ]/ 2epT, whence
T = u*l2cp « 3600 K.
1.18. The equations for the vertical (i/r) and horizontal (ox)
components of a rocket velocity have tho following form:
Mvt = —Mu sin q>— guM, Mvx = — Mu cos 9 ,
where M is the rocket mass at the instant t. According to
the statement of the Problem vt = 0, therefore, we have
from the first equation
■V_ __
m
u sin 9 “
M
T ’
where T denotes a time-dimensional constant equal to T =
(u sin 9 )/gM« 235 s. Whence
M = M,ex p
75
Substituting M and M into liio second equation, wo havo
v , ^ £ N C o t ip «
g M/<P
or
y , = - / ? M « co l( p «
1.
The lime t needed to gain orbital velocity v, — VtinSu =
1.7 km/s is
i/i<p
T
100 s.
rucotq)
This expression for t is also valid as q> -► 0. This results
from the condition ux = 0. As (p -► 0 the mass consumption
rises infinitely and hence the time needed to boost up to
the orbital velocity tends to zero.
The relative change in the rocket mass duo to the burnt*
out fuel is
M (0)— M (t)
----= 1* - exp /
M
.
a oc
0.35.
The overload of astronauts is
g.n = / t i t +
« -f- = 17 m/s*.
1.19. a — b « fiE ( ^ ) « 3 mm, where 7?e is tho IDnrtli
radius, and E the total satellite energy.
1.20. ^ ^ 1.3 X 10-17 rad.
1.21. The conservation laws result in ton ip = vtlvr = l/2 j/ 3,
where pt and v, are the tangential and radial components of
velocity of the launching-pad.
1.22. The equation of motion of a rocket in the gravitation­
al field of the Moon has the form
( dv \
,
dm
-3r) = m** + 7 T uIts integration over time t yields(gM is the free-fall acceler­
ation on the Moon)
m W ^ m , e* p ( - * ± i3 L ) .
where m (t) is the rocket mass after tho engine has fired for
a time interval t. Thus, tho fuel supply should be
™lv'l = m 0 [i ~® *P ( -
•
f.23. Since the oscillation period of a pendulum is T
(1/1f!)' then — =
w*iero Ag is the mognitudo of tho
7G
gravitational field strength duo to the debris oxlractod from
llio lunnol. Assuming the Umnol is .sufficiently long, ono
can easily find Ag from Gauss’
s theorem. As o result,
A^ = 2nw7P
where y is the gravitational constant. A numerical eval­
uation loads to the result: ATIT ~ 10~7.
1.24. ip = 8/A/25h « 0.00 rad.
1.25. A/A0 « 5 times, where A is the net sag for a fall
from a height h.
1.26. The laws of conservation of energy and momentum
suggest that
+
+
mv = MuL,
where u± and u\ are the transverso and longitudinal ve­
locity components. Whence
., ^
M >m ,
„
Ums
mvo -I/
M
V
v = vt y
M +m '
f
M—m
.____
U nV
, /* M—rn
V
M +m 9
1.27. According to tho law of consorvation of momentum in
the case of y-decay we have
Here M = 226 X 1.66 X 10“
" kg = 3.75 X 10"*5 kg is the
mass of a **#Ra nucleus, Ey the y-quantum energy practi­
cally coincident with the total energy E t released in the
y-decay, and c the velocity of light in vacuum. Whence for
the kinetic energy Tt of the recoil nucleus we have
El
Tz = 2 M c * = 0.095 oV.
In the case of a-decay tho laws of conservation ot onergy
and momentum yield
where va and vx are the velocities of the a-particle and
recoil nucleus, respectively, and Ma tho a-pnrticle moss.
Whonco, by eliminating i/a, for tho kinetic energy Tx of the
77
recoil nucleus wo obtnin Ilie following
r, =
f , - 87 kuV.
Finally, Ilie ratio of the kinetic energies is
7•
2« ^£ l
= 9 x 10s
1.28. The maximum energy which may bo spent on the ioni­
zation corresponds to the kinetic energy lost in an absolute­
ly inelastic head-on collision. (This result can be obtained
in the centre of mass frame.) The lost energy is
A 7 » _____^ 1M
T
where jl/ 133 and Ml9 are the masses of the cesium-133 and
oxygen-16 atoms, respectively. AT1,^* = 3.5G oV < 3.9 oV,
vi/. ionization is impossible.
1.29. T =
+ M«U >2.97 MeV, where
and ,V/.
are the masses of the nitrogen atom and a-particle, respec­
tively.
1.30. ~
27" « —10"3, viz. the watch runs faster.
1.31. i> w A /. lO1* cm/s^>c (v is the volocity on the
“
surface”of the ball).
^.32. L = /•(!)« 10s g-cmVs, where I is the moment of
inertia of the wheel.
1.33. Let the mass of the man be m and the swing velocity,
as it passes the equilibrium position be v0% which is related
to the initial amplitude <p0 by the relation vl « glyl (as­
suming that the angles of deviation of the swing from the
equilibrium position are small). The total energy of the
swing at this instant is E0 = mv\l2. After the man has risen,
the total energy is
E, =
+ mgh *
+ mgh.
where <p, is the new amplitude. The change in the total
energy has occurred due to the work performed by the man
against the centrifugal and gravity forces: Ex — E0 «
mgh -}- (mvl/l) h. Comparing this expression with the differ78
once E x — E0 found earlier wo gel
i > ; ( i ( i h -
2t)
°r f i- ' i’
c* 4
,,,ot
’
which represents the increment of angular amplitude per
half period. Hence
the sought incrcmont is (per
period)
« 3<p0 y .
1.34. Let the coordinate x-axis
be directed along the normal
to the surface of the racket
while the y-axis be parallel to
the surface (in the plane of
incidence) (Fig. 57). In the
Fig. 57
coordinate system related to
the Earth let the ball velocity
be V! (vlx = vx!2, vX9
*V2) nn(l the racket velocity
be v. The racket is assumed to move translationally. Then
in the coordinate system related to the racket, the velocity v{
will have the following components
V\x = V\x — V
*.
IV= Vly —Vy
Aftor tho collision, tho hall volncity in tlio moving coordi­
nate system will bo
*>!*. V2y=v\V
(mirror reflection). Returning again to the rest system the
expressions for the components of the ball velocity va after
!the reflection from tho racket arc
Vtx = —Vix + 2ux,
v%v = vly.
Under the statement of the problem, (vLv,) = 0, viz.
vix l—vx* + 2vx) +
= 0.
The solution of this equation has the form of
vx — —ulx, where vy is arbitrary.
It should be stressed, however, that when the racket hits
the ball the condition of mirror reflection is strictly ful­
filled only in the case of a normal fall (in the coordinate
system related to the racket). This is realized at vv = vlv,
70
theroforo the I'mnl answer has to he wrillcn in iho form
v — vlt ip = 60°.
t ..to.'(..'insider the coordinate systoin nssocinled with the
rotating liar. In this system, equilibrium can bo written
down as (Fig. 58)
y
Fi - Ft - 0 ,
Fxn -f- Ftb — 2 FI c o s (p.
Hero F = ml sin oja is tho
centrifugal force of inertia,
Fxand F%are the reaction forces
of the ball bearings. From
these equations we find
o!
i
4- ,
n
r,
m/1©* sin 2©
^ =
= — T+l— •
Fig. 58
1.36.
elastic forces is M el = —/q>f
where <p is the angle of rotation of the disc. The moment of
the internal friction forces in the gas is
R
Mu — — J q ~lp~r ’
2™*dr,
o
where v (r) = cpr is the velocity of a point in the disc at
a distance r from the centre. Elementary integration yields
M tT= — 2fc<p, where A=
As h increases, the moment M\t falls off rapidly. This allows
us, in the first approximation, to neglect the friction on
the upper disc surface.
The equation of motion is
7<p -f* 2/t(p+ /<p = 0.
Here / = mR2/2 is the moment of inertia of the disc. Lot us
seek for a solution of this equation in the form
q>=
Then we obtain
«=7 - o> = - y - = / 4 - * ®
80
/ t*
The m
whore T is tho oscillation period. The logarithmic decre­
ment is
2nk
d~ 6T
V n
~
h Y W >'
hence
dh Y 2/m
«---1.37.
■>=“*”
1.38. Tltm = & ,
r rol = - ^ i ,
* = 2/5.
1 .39 . x&h/f,
where h is the height the stool centre of
gravity is lifted and / is the friction coefficient.
2i/>
1.40. 9 =sarctan-^-.
1.41. The force moment is M = FI = QLt where Q = y <p
and the angular momentum is L — 7*2jiv. Whence
F=
« 1.3xl0‘N.
mlrf
1.42. * = I '
9 I (1 + 3 cos* <p)a
1.43. *=•12 c o s 9 (1— 3 c o s 1 9 )
3 cos29 < 1,
1.44. The angle of rotation 9 of the pendulum satisfies the
equation 7*9 -f /9 = 0, where / =
15 the loraion
modulus of the thread. According to the law of conservation
of angular momentum, the initial conditions have the form
4><0 )= 0 ,
/q>(0) = 2 -fr,
where E is the energy radiated by the quantum generator.
The solution of the equation of motion may be represented
in the form
9 = 90sinG>fx
where 90= -^^- and to =
.
For the sensitivity q = 2yJE we obtain
4r
^ — (1*/ *
6-040*
81
Whence Ihc period of oscillations,
i L = «rni= 1 9 s.
2r
cj
Furtherraore, since T = 2jiV HU we have
0 = 4( ^ r ) ‘
/‘= 22»
‘
",_
1.45. The velocity of elastic waves s is VEtp = 5 km/s. For
estimation, wo assume the collision time t to be the time it
takes a wave to travel from one end of the bar to the
other and back again, viz. x « 2Us = 4 X 10”5 s. Inelastic
phenomena will emerge when at some points along the bor
the pressure reaches p, viz. the potential energy per unit
volume reaches the value pV2E. Such points aro those where
the baft come into contact. At the instant of impact all
the kinetic energy (per unit volume) at these points will
be converted into potential energy.
Thus,
and hence
v
P
V?E
_
P*
E
5 m/s.
1.46. For simplicity, let us neglect the rigidity of the
cover and the bladder and assume that the ball deformation
can be characterized by a single parameter x (Fig. 59) which
is much smaller than the ball radius R (a soft kick). Then
in a good approximation one may assume that the excess
pressure p in the football during the kick remains constant.
The force acting on the football from the wall, F = pS
(S is tho area of the contact between the football and wall),
82
can be represented as
F — jipr%« 2jipRx,
Since the force F is proportional to tho football deforma­
tion x, the equation of motion of tho football nonr tho wall
will have the form of an oscillatory equation:
•
*.
,
n
,
4nf
2npR
x-| u»* = 0 , o>J = -71- = —^ - ,
where m is the football mass. The collision time x is then
half an oscillation period, viz. t « (2nm/pR)1'2. Taking
for estimates m = 0.5 kg, p = 106 Pa, and R = 0.1 m, we
obtain i « 2 X 10"2 s.
1.47. The moments of tho oxlernal forces (thoso of gravity
and friction) relative to point A (Fig. 60) vanish. Hence,
7(d0 cos <p = const, /o>0 sin <p = /a)hor + muR
= (/ + mR2) (Dhor,
where v = (i)hor7? is the velocity of the ball under steadystate pure rolling. Whence
/ah sin q> 2
«T® rt=-® bCOS(p, uhor= — 2—^ = - _ t o 0 su i 9 ,
2
V—*y(i)0/?sin <p, 0)= (wSor+tO^en)1/2.
1.48. In the problem at hand one should distinguish botween two cases.
(1) The slipping persists during the impact time x. Then
"H W = / j
0
Where A is the reaction of the wall. On tho other hand, tho
change in the horizontal momentum component is
2mv0—j N(t)dt.
Hence,
o
y vart — 2/y0,
tan a = 2/.
It may be easily shown that the condition of slipping at tho
l io n / < hl/.7mPaCl’WZ' “ (r> > U”'i/r' lends 10
•
*
condi83
(2) Pure rolling sols in before the ond of iho imparl, viz.
al t, < t, Ibis taking place for / > 1/7. Then
(0 (Tl) =
, mv„ rl ■
= jjV (t)dt.
0
The chango in the angular momentum of the ball during
this period is
/ < o ( t, ) - / u 0= - r /
^vert__ j^o
r
^Ndt,
r/mvvert
r~
I
whence it follows that
2
o
i W “— Vo> * viz. tan a = — .
Thus
tan a = 2/ at
/ < 1/7,
tana = — at / >
1.49.
1/7 .
(^ r
a
S h—hQ
a (2**„ )»/»*
1.51. The relative change in volume is
AV 3(1—2|x)
P.
V ~
E
1.50.
where p is the pressure, AT = m
~
of water and m is the moss). Hence
AV _ Pw—P
E (Pw—P)
V
Pw ’
pw-3(1—2p)
(pw is Iho density
2 X 10° Pa.
1.52. For gravitational waves, the velocity c and tho length I
are related as c « Y gl. Therefore the wave patterns will
be similar if all the dimensions are changed in proportion to
the squared velocity of motion. Hence, tho velocity of the
model must be 3.6 km/h = 1 m/s. Notice that in the given
problem the dimensionless similarity parameters are the
84
ratios civ and l/L, where v is the velocity of the ship and L
its linear size.
1.53. The phase velocity of surface waves depends on the
liquid density p, the froe-fall acceleration g , and tho wave­
length \\ v ~ pwgmXp. Comparing the dimensions of the
left- and right-hand parts of this equation, we obtain n — 0 ,
m = 1/2, p = 1/2. Thus, v = A\f g\, where A is a nu­
merical dimensionless factor.
II. Thermodynamics and Molecular Physics
2.1. Using the first law of thermodynamics, tho MondolcovClapeyron equation, and tho equation of the process (see
Fig. 5), we get
T m -P iV JA R .
At the point C (7) = 0 tho straight lino in Fig. 5 is tangent
to an adiabatic curve, and to an isothermic curvo at C (V) = oo.
2.2. According to the problem dQ = —dU, where U is the
internal energy. Therefore the heat capacity C = dQfdT =
—Cv* Using this relationship, tho first law of thermody­
namics, and the Mendeleev-Clapcyron equation we find the
equation of the process:
7T(V' ‘)/2 «COnst,
and the expression for the work of tho oxternal forces ex­
pended in compressing the gas from a volume V0 to V is
At V = VJ2 the work is
—A = 2CVT0 (2<v-n/2 _ i).
2.3. With an arbitrary equation of stalo p = p (V, T),
d^ [ ^ - ) r dV+{%-)ydTIn an isobaric process dp = 0, tho ratio of differentials
dV/dT is equal to the partial derivative (dV/dT)p and
therefore
85
By introducing in tho usual fashion
wo obtain p = y x .
2.4. The convonlional formula for the velocity of an adia­
batic jet exhausting into a vacuum is u0 = |/*2Cp7,/nl where p,
is the molar mass, and Cp the molar heat capacity. Using the
equation of state p — pBT/\i and tho formula for tho
velocity of sound s* = yp/p, we obtain v = s\f 2!(y — 1).
2.5. According to tho Tsiolkovskii formula M0/M =
where v0 = Y 2CJtTlp is tho oxhaust velocity of the gas in
n rocket system (see the previous problom). Therefore
2.0. A direct calculation shows that the efficiency of a cycle
composed of two adiabats and two isochoros is q - 1—a i-v,
where ct is the compression ratio. As a result, givon y = 4/3
2.7. Taking a value AT » 50 K as an estimate of the
difference between the temperatures at the tropics and the
poles we obtain from Carnot’
s theorem q,nax = A7’
/7’
1»
0.16. The power N converted per second into wind energy is
where B e is Earth's radius. Causes for the difference be­
tween q and q ra, x are the absorption of solar energy at
high altitudes, excess thermal radiation (viz. exceeding
the average level) from low altitudes, thermal conductiv­
ity, and the incomplete absorption of solar radiation by tho
atmosphere.
2.8. Let us assume the opposite and arrange for two polytropes A and B to go between two adjacent states (1 and 2)
(Fig. 61a). We then form a closed cycle from A and B , and
by integrating over the cycle and taking into account that
(IQ
C (IT on Ihe poly I ropes wo ohlnin
$ <IS = (CA- CB) In
86
,
A = § dQ= (CA - Cu) (T,- Tt).
Hero CA and C n are the heat capacities on the poly tropes A
and B. The first integral vanishes since entropy is a func­
tion of state and therefore an integral of it along a closed
contour also vanishes. We can immediately see that the
second integral vanishes automatically too, irrospectivo of
what causes the first integral to vanish (CA = C D or Tx =
Tt). But since the work A is equal to the area on the p-V-plane
enclosed by the cycle, the curves A and B must coincide.
Fig. 81
If the function T (p, V) is not single-valued, this proof be­
comes invalid since the situation shown in Fig. 616, is possible.
Here at the point (p', V') the polytropes are associated with
different temperatures, while ^ p dV vanishes because tha
two hatched regions cancel each other out. As to the contour
integrals covering the hatchod rogions, both ^ dS and
dA
are nonzero since these contours are not a closed cycle.
2.9. It can easily be seen that the first law of thcrmodyna
mics can be represented as
<*? = (C„ -Cy) {■*£■)f dV + CydT.
In fact, for V = const the right-hand side cancols to Cv dT
while for p = const and when dV = (dVldT)pdT, tho righthand side becomes Cp dT. Whence it follows that for an
adiabatic process
therefore
Cp_
Cy
1 ’P AV
IV ~
87
Changing the variables in the first formula from V, T to p, T
and taking into account that
we obtain
dQ = Cpd T - ( C p- C v) ( % . ) v dP
whence the pressure increment
A71« 1.1 x 10* N/m*.
2.10. Using tho standard expression
TdS = d W - V d P = \ { ™ - ) T- v ] d p + ( ° £ ) pdT,
where IF is the enthalpy, and the fact that dS is a full differ­
ential we obtain tho Ihcrmodvnarnic identity
W r - v — rW .For an adiabatic process (dS = 0) we have,
/ dT \ ____ r i dv \
\ dp )Sep \ dT l pFor stretching the wire, in an analogous fashion, wo lind
that
/ dT \ _ _ alT
\ dF I s C »
where C = cp/a (a is the area of wire’
s cross section and I
its length). Finally
AT =
-0.03 K.
epo
2.11. The energy balance that includes the potential onergy
of the piston and the condition governing the piston’
s
mechanical equilibrium will be
0/7 IT'_AF
0 T'
hV/V _ mg
l ) - n r t AV,
j — (AF/F)* — nr*p
where T is the gas’
s final temperature, C„ the heat capacity
of v gas moles, 2F the total volume of tho cylinder, and AF
the volume increment in tho upper section during a downstroke. The equations demonstrate that, given that ////,'<C
nr3/;, wo will have AF<^ F and (T‘— T)<ti T. Assuming
88
in the second equation that f w T and omitting (Al'/V»)
in tho denominator, wc obtain
T'—T
—T
l
vf? / m \2.
Cp 1 nr*p } ’
~
V ~ 2 nr*p 9
Using the same approximation we get for the change in
entropy:
2.12. For a reversible isothermal process (T — const), the
second law of thermodynamics yields dA = dF, where
F = U — TS is tho frco energy. Therefore Amtx = —AF.
Since for an ideal gas in an isothermal process U = const, we
have Ammx = T AS. Whence it immediately follows that
4 5 S°n“= * r
(v.ln - ^ * - + v2ln
« 18 kl.
For a reversible adiabatic process (dQ = 0), the first
law of thermodynamics yields Amti% = — At/. This process
is isoentropic and condition S = const results in the fol­
lowing oquation for the final tomporaturo T*\
iT*=con«t
(vfiv, +
Vt) 1n~f~ +
‘mas
T
= 0.
Whenco T « 225 K. Tho adiabatic work is
jS—
conat
)
^max —U{T) — U ( n = (vtCVl + v2Cv#) (T — T‘
T— T‘ at—const _ T-\-T' ^T—con a t^ ^ £ i,T
~ 1,0 KJ*
= r In(r/rf A m tx --- 2~ Am*x
2.13. Each piston performs work isothermally compressing
the gas for which it is not “
transparent" from tho volumo V0
to Y = V0/2; the second gas passes freely through the piston.
The total work of the external forces in this process is
A = 2vBT In (-£?•) = 2 In 2-vRT.
A direct calculation of the change in the total entropy (as
a sum of the changes in the entropies of both gases) shows
that AS = — AIT, according to the general theory (see the
previous problem).
2.14. Let us encloso u bag of volumo f/, containing vt
helium moles inside a larger volume V2 containing va lieli89
um moles and v air moles at Iho snmo temperature T and
to la I pressure p (for simplicity we do not take into account
tlial the air itself is a gas mixture, viz. tlio rosull is inde­
pendent of this fact). Tho entropy in tho initial state, whon
helium is in the hag, will he
S, = flv, In —
'
'
+
1
flv,* I n -V^ j- + flvl»-^,
V
while in the Final state, when all tho helium has diffused
out, the entropy is
Sz =R (v, + v2) In -£±£- + flv In
The change in entropy, AS = S 2 — S x should be calculated
in the limit when v,, v2, V2 oo for which V2/vl and VJv2
retain their liuul values. A simple calculation yields
AS = flv, In
1
v i/Vi
m 110 J/K,
Tx In —
n2
where Kj and n2 are the molar concentrations of the helium
in llio hag and that in the surroundings, respectively. Tho
minimum isothermal work needed to collect the same amount
of helium logether again is (see problem 2 .12):
Mlmin —^ A5 = pK, In
32 kJ.
2.15. IT dQ — 0 and p = const, enthalpy will ho preserved.
B e f o r e relaxation Cp = 7/2R, and after it C'v — 9/2fl, there­
f o r e 7*2 = 7/1) Tx. The change in entropy AS is tho sum of
A5exl and A$int, the entropies due to the external and
internal degrees of freedom respectively. In an isobaric
process
T.
I n A S , „ , = J C0(T)
T*
where C 0 (T) is the vibrational specific heat. However, the
specific heal of a harmonic oscillator for quantum region
(viz. for IcT <§
Chm) is exponentially small and for this
reason Ilie main contribution to the integral comes from tho
integration region T ^ hwlk, where C0 « R. Hence,
AS,ot * R In (kT2/h(a)'^> A.Se*t and because of this AS «
R In {kTJhia), with the same accuracy. Physically, tho
predominance of the vibrational contribution to AS means
lliiit during the relaxation a maximum relative temperature
change occurs in the vibrational subsystem.
90
2.16. Tho derivation of the Clausius-Clapoyron oquation
shows that the heal of vaporization X is T (sv —
where s
is tho specific ontropy. Whence
Hero the Mendoleev-Clapoyron oquation was used and m is
the vapour’
s mass, and p the molar mass. Since the condi­
tion for the liquid and vapour to be in equilibrium is the
invariability of the potential cf> = £/ + />F — TS = F +
pV during a phase transition
(Ot = Oj), we have
AF = — p AF « — 10 J
2.17. First of all, it must be
demonstrated that adiabats are
always steeper than isotherms.
Along an adiabat dS = 0 and
V
Fig. 62
According to a well-known
thermodynamical idonlily, tho
bracketed expression is equal to T (dp/dT)v. Substituting
in tho full differential of the function T - T (I7, p), laking
into account the validity of the above relation along an
adiabat S = const, and using the idontity from Problem 2.3
we obtain
viz. an adiabat is steeper than an isotherm for an arbitrary
sign of (dp/dT)v . Hence, as a whole, the cycle has a conven­
tional form (Fig. 62). The (clockwise) direction in which
it is followed is dictated by the sign of the work, A > 0.
From (dp/dT)v < 0 it follows that the “
hot”isotherm 7\
lies below the “
cold” one 7*2. Along an isotherm dS =
(dpldT)v dV and this is why the clomontary hoat change
leftwards along the lower isotherm {dV < 0) is dQ — Tx d S >
0, viz. the system roreivos heat. When moving along the
upper (“
cold") isotherm, the s y s L o i n r e l e a s e s h e a l .
2.18. Using the Van der Waals gas equation we find that p
is 5/>cn and th® isothermal compressibility x is l/(6pCr). Tho
91
equation for llio oscillations of a piston of mass M is
(o* = V>S*per/MV.
Sine© V = Vcr = 36 for a Van der Waals gas and the piston
equilibrium condition is Mg = 5pcrS wo find that
toa = 2gS/56.
2.19. The adiabatic (Laplacian) velocity of sound is s* =
(dp/dp)s. Differentiating the Van der Waals oquation and
substituting the critical parameter values (Tcr = SaJ27bRt
the molar volume Vor — 36) for coefficients, we assure
ourselves that the coefficient of dV vanishes and dp =
{Hi2b) dT. That this coefficient vanishes means that
(dpldp)r = 0 , viz. at the critical point the isothermal
(Newtonian) velocity of sound is zero. Differentiating the
entropy at the critical point and then assuming dS = 0 (an
adiabatic process) we find that
CydT + ^ r d V ^ O .
Sinco the density p is \dV, whero p is tho molar mass, wo
finally obtain for the adiabatic velocity of sound
8*=V 2aR/Sb\iCv.
2 .2 0 . Let us consider a flat layer in the liquid of thickness
2x lying symmetric to the central plane. For stationary
motion the flux of the momentum through sido faces must
equal the difference in the forces acting on tho ends, i.e.
— 2dlt\ -£-=2zlpgh.
Integrating this equation with a boundary condition v (6/2)=
we obtain
0
w
= - ^ - * 2-3 cm/s-
By integrating over the cross section wo determine tho liquid
flow rate
m = |« 1.2 x 10-‘g/s.
2.21. We take a cylindrical layer of liquid with arbitrary
inner and outer radii and whoso axis is that of tho tube.
92
Since the motion of the liquid in this layer is stationary
while tho pressures oil both ends aro tho samo, tlio total
fluxes of momentum through its in nor and oulor surfaces are
nl«n the same. Therefore the flow of momentum 11(r) through
a cylindrical surface of radius r, if taken per unit length,
n - — 2nfT]
« const
Which is fndependent of r. Hore v (r) is tho velocity of the
liquid. By integrating this relationship we find
* ' + ( ' ^ r ) l n r = c°nst
land using the boundary conditions v (/?,) = i/0, v (/?,) = 0 ,
we finally obtain
‘
^ - n - T S T W “ 0'27 dyn/cm2.22. We take a cylinder of liquid of radius r whose axis
coincides with that of tho tubo. The motion of the liquid
inside the cylinder is stationary if the momentum flux through
|the cylinder’
s surface equals tho difference in tho forces
acting on its ends, i.e. if
dv
n r'p
- 2* "1 -3 7 = —
.
where p is the pressure differential between tho ends of tho
tube, r\ — vp is the dynamic viscosity, and v (r) is the
flux velocity. By integrating this equation with a boundary
condition u (R) = 0 we obtain
By integrating py (r) ovor the tube’
s cross section we find
the mass flux (Poiseuille's formula)
whero h (t) is the height of tho liquid column, and p = pgh
is the pressure differential. Integrating this equation yields
=
jg - n m ,
which, in conjunction with an initial condition ro„ = nR‘
lo.
leads to f = 8vl/gR*.
93
2.23. Suppose that n narrow ln.vor of thicknes Az between
two planes is filled with n liquid whoso dynamic, viscosity
isq, and that the pianos themselves move towards Ihn x-axis
with velocities vx and vx }- Auv (Fig. (>3). Then tho viscous
friction forces acting upon the planes will he qAut/Az ami
—qAuv/Az (per unit area), respectively. Tho total power
dissipated in tho layer is
- *1 T p
t t (v* + &vx)
Since the pressure difference between the ends of the capil­
lary is p = pg/i, tho velocity distribution in it (see the
previous problem) is
By taking a thin cylindrical layer of liquid and applying the
above formula for dissipation we obtain the following for
the total power dissipated in
the capillary
{*£■)'***
__ j i
8
(pghR'2
)* ^ j q-3 yv
T)/
2.27i. When the ongino is run­
ning, the spatial gas distribu­
tion in the cabin is heterogeneous because of tho inertia in
the coordinate system fixed with respect to tho rocket. After
the engine is switched off, the distribution bccomos homo­
geneous.
(1) During a quasistatic transition to the homogeneous
distribution the gas performs tho work against tho inertial
forces. The work due to a displacement of the gas under on
infinitely small acceleration change Aa is
Ai4 = — 2 Ft &xt = S mia Axj = ma AX = ma
Afl,
<
i
where the summation over i is over all elementary volumes
and the mk are their masses and the Axt their displacements;
X(a) is the coordinate of the centre of gravity of tho gas.
94
For \iah/RT0C 1 wo cnn easily soo that
y ~ h [i
1
A~ T L 1
Using tho same approximation, wlien calculating dXIda wo
can neglect temperature changes, which only yield very
small corrections. Therefore
dX
da “
The total work is
\ih»
i2BT0•
,
P dX j
m\ih*a*
A=m)as r da= s w r
a
which, given thormal insulation (Q = 0), govorns tho chango
in the internal onergy, \CV (7\ — Tc) + ^ = 0 , whoro v is
the number of moles. Finally,
T
'*
t
lo ~
___ 1 FT* (1^L\2
Vi Cv \ RT0 I *
Since the process is reversible, tho onlropy is unchanged:
Sx = S 0.
* (2) Given a rapid chango in acceleration, tho gas does not
move at the instant the engine is switched off and thoreforo
does not perform the work against oxtornal forces. As a re­
sult, Q = A = 0 and hence tho internal enorgy and the lomperaluro also do not chango. Tho ontropy may ho calculated
from the general formula
where n (r) is the molar volume at the point r. Yot it may be
simpler to use the fact that in both cases the initial states
of the gas are the same while the final ones differ in tomperature by T0 — Tx. Therefore tho entropy change is
52 _
So= ^
(r #- r , ) = ^
(-*£ )’.
2.25. AS = — I t ( ^2RT )2 (see
Provi°us problem).
2.26. Applying Stokes’formula for viscous drag we obtain
R1/2~ - ; 2t1— r l/ " —
tf(P—Po>
Y
P
and R ~ 10"3 cm.
95
i
The particles precipitated are those for which the average
column height evaluated using the barometric formula is
IcT/nig <: l( {in is the mass of the particles), which is
equivalent to
R ^ (w ) 1/4 ~ 0-5 x 10’4 cmHence a sediment does form in the paint.
2.27. The hole is so small that it is possible to assume its
presence does not influence the distribution of the gas atoms
(in terms of volume or velocity) and that the flow is a molec­
ular effusion. Therefore the flow of particles N is Snvl'k,
where n is the concentration and v the average thermal veloc­
ity. The mean energy carried away by each escaping atom is
$ j dvxdvy j dvt (mv2/2) vtt“mv^ 2hT
;----a
— -------------------- 2kT
$ $ dvxdvv j dvt vte-m"l*hT
- OD
o
and exceeds the mean thermal energy 5 = 3 kT/2 by kT/2.
Therefore to maintain a constant gas temperature a heat
flux Q of {kT/2)N must be supplied. The concentration
decrease over time is governed by the equation
£,.C=
± {nV)=-.-N
wluMire
n{t)
x= 4^ » Q=
Sv
0
2.28. This problem is similar to the previous one. The laws
for the conservation of the number of particles and energy are
j L (nV) = - N ,
± (- k T n V ) ---- 2MW.
We eliminate n and dn/dt and so for T (0 obtain the equation
which, with tho initial condition T (0) = T0t yields
7*__
To________
ll-MS/W)/kT9/2nmt]*
where in is the atomic mass.
06
2.29. Since the problem stales that the vapour near the surface
is saturated, the evaporation rate will be limited by the
vapour's diffusion rate through the tube. The equation con­
trolling the mass balance in evaporation and diffusion lias
the form
dh _ P«at
dt
2p
D
l—h ’
where h (t) is the height of tho liquid column in the tube,
and p the water density. Integrating this equation with the
initial condition h (0) = 111 and using D « 1/3 i;A we
find the following for tho ovapomtion limn
2
___9f?TpI— ^ 160 days,
4vAp,aiP
where p is the molar mass of water.
2.30. The drift velocity udr is p£, where p_is the mobility
related to the diffusion coefficient D « 1/3 vA by the Ein­
stein equation eD = kT\i. Therefore
Ufa «
mv?
E w 15 ra/s.
2.31. If A ^ d, then for the gas flowing in a tube, tho
role of the diameter d will be the particle range and there­
fore the effective diffusion coefGciont D eff~ vd. The flux
through the tube will be
hence, with the d value halved, the gas concentration in the
j vessel will be a factor of 23 = 8 times greater.
2.32. At a pressure of 10"4 mm Hg the free path isA ~ l m >
R , viz. theflow will be Knudsenian. Under these conditions,
the molecular flow is
where n is the concentration in the vessel, nx the concentra­
tion at the pipeline outlet, and v the mean thermal velocity.
By definition, the pumping rate N is nxVx. The rate V at
which the flow will bo tho same if pumped directly from
the vessel is N = nV. By eliminating n and nx from these
7-0408
97
Ilinn* oqnntions wo find ihiil.
v = 7 T X lZ r te3301,p2n
wt
'2.Hit. Tlio time ncodod for tlio moloculos lo diffuso n dislnnce
of the ontor of n/tn, whoro Vfo is Iho Knrtli radius, is esti­
mated to bo
~ (ji/<b)® /i>A- Givon / lj ^ 6 X 10® cm,
A ~ 10" 5 cm,
3 x 104 cin/s, tho formula yields <aifr ~
4 - 10u yours, viz. Lurr ~ 1011/turi,2.34. Sinco nl tho prossuro givon in tho problom tho free
path (viz. tho hole sizo) is A^> *S,|/a, tho flow will ho moIreulsir. Tims tho oxit of oach individual inoloculo will ho
independent and for this reason N obeys a Poisson distribu­
tion. Honco,
/ ( S a ? ____ 1
•2
_ in_a
*
~ W ~ V * 5 i~
’
w h e r e o is the moan volocity and n tho concentration of tho
helium atoms in the vessel. Given a Poisson distribution,
the probability of fixing IV particles in one run is
The probability of N = 0 is w (0) « exp (—10ia).
2.35. Using the definition of moan values and tho Boltzmann
distribution we obtain in case (a)
[+•J ( («)_&)•
0
-11/2
----------------J - - y s .
J 17 (x) ,-VM/KT
where U (j) = Kx*l2. In case (b) the fluctuation is equal to
(3 VL 5r
anc* ( 3 {N—\) ) 1/2 *or rao^ecu^es
linearly
and nonlinearly arranged atoms, respectively. In case (c)
the fluctuation is V 2/NA•
2.3G. Since the temperature remains constant, the pressure
fluctuation is duo entirely to a fluctuation in the numb or of
particles. Let us use the relation between probability and
98
ontropy. Tho ontropy of tho initial a ml li rial stales is
'Ii
5„W= 2 /fvln—
,
v »
S = R (v + Av) In v+KAv +/?(v—Av) In y_^Av- ,
whoro v is the number of moles of gas in oach vessel, and
y
A'v =
Ap is tho numbor of molos transferred from ono
vessel to the other during llio fluctuation period. For |Av|
v| we obtain
and i = o x p [ - n K ( - ^ - ) ] ^ ,
whero n is tho equilibrium cnncmitralion of tho gas mole­
cules. Finally
y = - * L / ^ _ ) 2ln
cm3.
P
V Ap /
w
2.37. When considering tho temperature fluctuations we
assume that the number of moles inside and outside the vol­
ume v (vx and v8t respectively) are fixed. Initially we assume
that the vessel is completely isolated so that Q = A = 0 and
tho invariance of the internal energy is expressed thus
AU = VjCvAr, + VjCvAfj = 0.
The change in entropy due to tho temperature fluctuation is
AS = v,Cy In T+f T'- + v,Cy In r + A7V
and for v , < v , in the region of small fluctuations this
reduces to
Thus, the probability distribution is Gaussian:
m(Ar.)
e x p [ - ± v, ^
(-^s-)’
].
But if a quantity x has a Gaussian distribution, viz.
w (x) c o
t|ien
mean-square fluctuation must be
+f x>c-*,/2X«dx
^2 = -=“
----------
+oo
Therefore, ns applied lo our problem, (AT’
,)2 = kTVx^Cy.
Since additionally pv = \\HT, wo get
u
Ji AT / T \2
Cy p \ AT )
2.5 X 10-8 cm3.
The size of the vessel drops out of the final formulas, there­
fore wo can see that it is irrelevant whether the vessel is
isolated: processes occurring far away cannot influence the
magnitude of the local fluctuations.
2.38. The ratio of the lengths of time the system stays in the
homogeneous and separated slates (t0 and f j is related to
the ratio between the corresponding probabilities, which in
turn can be expressed via the difference between the entro­
pies of the states, S 0 — 5, = 2Nk In 2, i.e.
^ oxp
= 22N.
*i
*1
*
k
The timo tl it takes for the separated state lo disintegrate
is comparable to tfie timo noedod for tho inoloculos to fly
apart, i.e.
d/v~ 10~4 s (d~ 10 cm is tho vessel sizo).
The lime t0 is the expected time it takes for a separated
state to form: t0~ 1010 years ~ 1017 s. Whence
“
log (<0/<l)
2 log 2
Oc
2.39. The energy associated with an isothermal fluctuation
volume is given by the work dono against the quasistatic
force, (dpldV)T A I'. Therefore, the energy is
£ (AK) = -j-| (-|p- )r |
and the probability of the fluctuation, given the Boltzmann
distribution, is
»(AV) O, e*p [ - - ± r |(i f )T|(AV)*] .
viz. the resulting distribution is Gaussian. Tho root-meansquare fluctuation (see Problem 2.37) is
By calculating (dpldV)T for a Van der Waals gas and as­
suming (in the derivative) that the molar volume is equal
100
to the critical one we find that
!
i
.
( MT
i<>
r
!'
~5«
3JVAV T-Tcr '
where N a is Avogadro’
s number. It can be seen that the fluc­
tuations build up rapidly near the critical point. If T «7\.r.
then N a V « 3WV, where N is tho number of gas molecules
pier unit volume. Then
1
jL
t
V% ~ 9JV T—Tcr •
$.40. The well-known formula for the temperature depen­
dence of saturated gas pressure is
p(n= P oe*p{x(T T --r)}'
where A = Xp is the molar heal of evaporation. The conI
jiD*
dition for the valve to bo opened is F —
(p — Po)*
Finally, F « 18 N.
2.41. The pressure in the subterranean reservoir p is p 0 +
pgh, where p 0 is the atmosphoric prossure. The boiling sets
in at p (T) = p, when the saturated vapour pressure is
p (T) = p„ exp {4- (-L- - -£-)}.
Tho boiling will stop when the wator tomporature drop due to
the boiling reaches T0. This will take place when the mass of
water that boils away is
whero C is the molar heat capacity of wator. By obtaining T
from the first equation we find from the second one that
AM/M « 0.14.
2.42. From the Clausius-Clapeyron equation, along the
molting curve,
dT ~ T&V ’ 7— T (S\ — Ss).
The pressure minimum dp/dT — 0 corresponds to q — 0. It
follows from the data on the entropy of both phases that the
corresponding temperature Tinln = B In 2 « 0.15 K. The
melting heat q = RT*/Q — RT In 2 is a parabola that in101
tersects zero at 7^,,,. At lower temperatures 7 < 0. This
change in the melting hoat is a unique fact called the Poraeranchuk effect. The falling section of tho p (T) curve corre­
sponds to the negative 7. We stress that AV > 0 for heli­
um-3 (as opposed to the general situation when AP < 0, as
in the case of the water-ice system). Integrating the ClausiusClapeyron equation yields
P(7’
) = Pniin + - ^ - (-§— ln2) •
Whence p{0) « 32.5 atm.
2.43. The equation of mechanical equilibrium governing
the pressure distribution inside the core is
i f = —Pff (r)= — 5-«VP*r.
wlicro y is the gravitational constant and g (r) tho accelera­
tion due to gravity. From the Clausius-Clapoyron equation
for the phase equilibrium curve,
dp
q _
dT ~ T AV
T Ap ’
where 7 is tho spocific melting heat. Dividing those equa­
tions into each other (at r = H, whore they arc both valid)
we find that
dT
4ji y Ap
dR ~ 3 q
U*
This equation relates the radius R of the solid state part to
the temperature T. Since
dR dR/dt
1 dR
dT ~ dT/dt ~
u dt '
we find that
3
qut
— 35 km.
A/? — —
YApRT
2.44. The internal energy of the systom is
U = myUy 4where uv and u, are the specific internal energies of the
vapour and liquid. Differentiating U with rospcct to T and
allowing for the conservation of the total mass (rnv + nix =
const) we get
(mv + m,)c = (uv- u 1) - ^ - + (-2-pmv + e0/n,).
102
In the first term on the right-hand side uv — uj == X —
kT/\i (X = A/p is the latent heat of vaporization, viz. the
difference between enthalpies per unit mass) and according
to the Clausius-Clapeyron equation,
Finally
dT ~ RT*
‘- T T F {«•+!>[ “ + *
\i I
v’
(l - r H ) * ,0‘
2.45. The answer follows from the solution of the previous
problem at the limit as fJ oo, viz.
J/(K,kg)2.46. The total entropy change Is the sum of the changes in
the volume and surface entropies. The volume entropy is
(AS)„ , = v T I l n =-*- - f nr*In 8 = ■4*In2*£ ,
since we can approximate the pressure inside the bubble to p 0.
The specific surface entropy s is q/T%where q is the heat
of the isothermal formation of a unit of surface, as can be
seen from the conventional definition dQ = T dS. Thereforo
(AS),urf = 2-4n I(<■')*—rl )
= 24nr*-f.
Finally,
Ac
24nr* / _ , In 2 t _ _ \
A S = - f - ( ? + - g- ! p 0r ) .
III. Electricity and Magnetism
3.1. An electron emitted from a negatively charged plate is
acted upon by the force due to the electric field in a capacitor,
Fx = eEy and the force of electric image, Ft =
—eV(4ne0‘
4x*). The total force acting upon the electron is
e*
F = e E ~ lGnegX* *
It follows from this expression that the electron slows down
unless x < V^/(16jie0£). The electron velocity is at a mini­
mum at
* = * o = x l / So «
10-
m.
103
Notice that this solution does not allow for the space charge
that may form near the negatively charged plate.
3.2. In an electric field gas molecules acquire a dipole mo­
ment p = e0a E and an energy (see Problem 3.16) W =
— pEI2 = —e 0a£ 2/2. The molecules in the force field have
the Boltzmann distribution
/
W\
k„ oE2
n — n0exp ( - — J = n0 e x p - | ^ ,
where n0 is the concentration of the molecules outside the
capacitor. Substituting in numerical values shows that the
exponent is small compared to unity. Thcroforc it is possible
to write
An
1 n0aU\ x 10-«.
2
kT
no
The concentration of the molecules (and hence tho pressure)
in the capacitor is higher than in the remainder of the vessel.
The physical reason for the higher pressure in the capacitor
is that the molecules with a dipole moment are pullod into
the region of the strong field.
3.3. It can easily be shown using Gauss’
s theorem that the
strength of the electric field of a uniformly charged ball
at r < / t is
C,
1
Q
1
e = *hi; ^ r r = ^ T P rHere r is the radius-vector from the ball’
s centre to the
point of observation, Q the total charge on the ball, R its
radius, and p the volume charge density. The hold in the
cavity may be treated as a superposition of the fields of two
uniformly charged balls:
E = E, + E, = -ji- pr, — -jJ- pr, = -gL p (r, — r,) =>
pa.
The notation is evident from Fig. 64. Thus, the electric
field in the cavity is uniform. This conclusion remains valid
irrespective of the ratio between the radii of the balls and
the distance between their centres. In particular, the field
in an uncharged spherical cavity inside a uniformly charged
ball of large radius is also uniform.
3.4. Using the solution of Problem 3.3, consider two iden­
tical halls of radii R with chargo uniformly distributed
within their volumes and charge densities p and —p. Assamo
that the balls intersect so that their centres are a distance
104
a < R apart. Within the intersection region the charges
will cancel each other out. It follows from Problem 3.3
that the field in this region will also be uniform, i.e.
Here a is the vector connecting the centres of the balls. As
shown in Fig. 65, the space charge is only nonzero in a thin
surface layer. As a —*•0 (given the condition that pa =
const) we arrive at an idoa of a surface charge on a sphere.
It can be seen that the thickness of tho charged layer at the
point defined by the angle 6 is a cos 6 and hence the chargo
per unit area is
o (6 ) = pa cos 6 = 3e0£ cos 0.
Since the field of a uniformly charged ball is the same for
r > R as that of a point chargo located at the centre of the
ball, we can conclude that in our case tho field outside the
sphere is that of a point dipole whose dipole moment is
p = — Qa= — — n/?3pa = — 4jie0/?3E.
It is possible to check that the uniform field inside the
sphere and the field of a dipole outside it both satisfy the
boundary conditions on the surface of the sphere, thus indi­
cating that this solution is correct.
3.5. It follows from the solution of Problem 3.4 that in an
external uniform field E a conducting ball of radius R ac­
quires a dipole moinonl
p = 4ne0/?5E.
105
In fact, the electric field inside the conducting ball should
vanish. The charge on the ball should be redistributed so
that the field of the charge inside the ball becomes uniform
and cancels the external field. Thus outside the ball the
field of the charge inside it will be the same as that of a
dipole.
The electric moment P per unit volume of an “
ideal gas"
composed of conducting bolls will in an external field be
p = np = 4ne0n/?3E.
Now wo can write
D = e0E 4 - P = e 0 (1 4- 4jw/?s) E = e0eE.
NVhonce it follows that
e = 1 + 4nnR*.
Because the ball concentration is small we can ignore
the interaction of the balls whon determining the dipole
momont.
3.fi. The electric field outside a sphoricnl dust speck will
be composed -of a uniform external field E and the field Ed
of a dipole:
F - 1 f (pr) r
P1
d
4ne0 L
r3J*
whoso dipole momont (see Problem 3.4) is
p = 4ne0/?3E.
At points C and D the hold E d = —E, while at points A
and B the field Ed = 2E. Whence it follows that the total
field at points C and D will vanish while at points A and B
it will be trebled.
3.7. The dipole momont of an argon atom in an external
field E can bo written as
p = Zxe = e 0a£,
where x is the displacement of the electron sholl, Z the num­
ber of electrons in the atom, and a the atomic polariza­
bility:
a = Zxe/e0E.
A dielectric permittivity e is related
larizability a thus
e = 1 4- n0a,
106
to its atomic po­
where n0 = 2.7 X 1025 m “3 is the concentration of the atoms
under normal conditions (Loschmidt number). Henco
z = (e — 1) e0E/n0Ze « 2 x 10~18 m = 2 X 10" 18 cm.
Notice that the atoms are of tho order of 10"8 cm in size,
which significantly exceeds the amount by which they are
deformed in electric fields.
3-8- P (r) - -jigs- = const,
v= ± ] /
= 2.5x10'»Hz.
This value of the frequency v is within the visible spectrum.
3.9. In the coordinate reference frame of the rocket, the
electrons in the matter are acted upon by the force of iner­
tia ma. In equilibrium, this force must be conn ter-balanced by
the force eE due to the electric field. Hence, at the ends of
a conductor of length I n potential difference U = El =
mal/e emerges during acceleration. Thus, the accele­
ration might be detected by measuring the value of U.
For a = 10 g and / = 10 m tho potential difference is U «
6 X 10" 8 V. Modern equipment can, in principle, meas­
ure such small potential differences. Note, howevor, that
in this case the standard mothod of measuring using wires
connected to the end-points of the conductor cannot bo used
because an analogous potential is set up in the connecting
wires themselves. Howover, the potential could bo deter­
mined indirectly, e.g. by measuring the amplitude of tho
current in a conductor whose orientation is periodically
changed by 180°.
3.10. The charge is uniformly distributed on an oxtornal
surface:
o = qMni?1.
The field outside the shell is the some as that of a point
chargo q at its centro:
~
_____ 1_ Q
*
7F-
In order to determine at and o 9 we need to know the elec­
tric field insido the shell. Using tho method of eloctric
images it is possible to demonstrate that the electric field
inside the shell coincides with that of two point charges q
and q (Fig. 66). The charge-image q' has the value
q'--q -r
107
and is located a distance
a
from the shell’
s centre.
The value and position of the charge-image are chosen
so that the potential of the inner surface of the shell is every­
where zero in the field of the charges q and q\ i.e.
>
9
^ — 4ne0p
i
9*
a
4ne0p' — ’
The field strength at points 1 and 2 is
E _ _ J _
9_____________ 1 _
E ^
9
1
2
4ne0(r —a)*
1
q 1 -\-a/r
q’
4*e0 (&— /■)*
, 1
q'
4ne0 (r+o)* "r 4*e0 (b + r)*
4ne0
_
9
(r—a)* ’
1 — g/r
4ne0 (r+a)« *
Now it is possible to determine the surface densities ox
and o a using the boundary condition on the conductor sur­
face, En = o/e0. The sign in this relation corresponds to
the choice of the external normal. In our case Eln = —E x
and E 2n = —
therefore
“
9
4n
l + «
/ r ____________ 9
( r - fl) * ’
*
4n
i — alr
(r+a)» *
3.11. From the condition of continuity of the normal com­
ponents of the induction vector D = e0E - fP wo have
E, = E, = E ^ E „
+ -?-.
Here £, and E , are the fields in the air gaps, and E % the
field in the plate. The potential difference bolweon the capac108
itor plates is zero, therefore
E ( d - h ) + EJi =0.
Whence
D, = t'E, + P = l> ’
L.
ZA2.Ea = ± P , % = { + ) * .
3.13. The field in the plate is
Tlio field outside the plate is
E 2 = 0.
The potential difference between the sides of the plate is
3.14. If all the energy of the capacitor is consumed to heat
the gas the atoms will all be hot enough to ionize. If N is
the number of hydrogen molecules in a given volume, the
law of conservation of energy is
££L = 4 N ± k T - fif± k T 0.
The factor 4 in the first term on the right-hand side takes
into account the processes of dissociation and ionization.
We can neglect the second term on the right-hand side (T »
T0) of this relationship. Assuming N = pV/kT0 we find
that
T=T> w
*
3 x 1 0 7 K-
A possible reason why the temperature might be lower is
that the capacitor plates got hot. If all tho discharge energy
is consumed by heating the plates, the law of conservation
of energy would be of the form
2£-=-cm6>T,
1
o
whence A7* w 200 K. Thus, only a relatively small rise
in the temperature of the capacitor plates results in a lower­
ing of the gas temperature by a few orders of magnitude.
109
3.15. Lot us npply tho motlind of oloctrical imngos. As shown
in Fig. G7, a dipole image has a dipole moment —p. The
electric hold stronglh of this dipole on its axis (tho basic di­
pole position) is
1
f
3(pr)r
p1
p _
J 2necr» *
Here r is the distance to the image and equal in our case
to 2L. The force acting upon a dipole with momont p in an
inhomogeneous olectric field E is
F = (pV) E
whence it follows that
F = p i £ = - j£ ! _ = „ 3p1.,- « 5.4 X 10"* N —0.54 dyn.
r
dr
2nt0r*
32n t0L*
*
As can be seen F > 0, which means that the dipole is re­
pelled from the conducting plane.
1
H
,
8r
»
..E
1
Pig. 67
Pig. 68
Now let us find out the work done by the oxlornal forces
to displace the dipole to a new position:
*= 1.6 x 10-*J=0.1G erg.
3.16. A rigid electric dipole can be moved from infinity to
any point in n field without any work being dono by moving
it along one of the lines of force provided that all the time
p _L E. Therefore the energy of a rigid dipole in an external
field can be defined as the work done by the external forces
to rotate the dipole (Fig. 68):
W = A = — 2q -J- sin 0*E —p E sin 0** — (pE).
Here I is the length of the dipole and q its charge.
110
In order to determino tho energy of on elastic dipole let
us consider a quasistalic change in the dipole moment when
the external held is changed. Assuming that the dipole's
length is changed by dlt electric forces will perform the work
qE dl — a e0E dE.
Expressing the energy of the dipole through the work of the
external forces we find that
3.17. U **
«14 jiV, where m and a are the mass and
charge of an electron.
3.18. The solution to this problem can be found by the
method of electric images. Tho hold of the point charge q
and that of the charge-image should satisfy the boundary
conditions at the boundary of the dielectric. As a result, we
obtain
1 E~ i
*L
°VPJ---- 2jT e+1 (p»+L*)3/*'
Here p is the distanco from the perpendicular projection
of the point where the charge q is on the dielectric surface.
The total charge Q induced on the dielectric surface is
The held of the surface charge in the half-space above the
dielectric is the same as that of a point charge Q located
inside the dielectric and a distance L from the boundary.
The force acting upon the charge q is given by the expres­
sion
p_
1
_
1 g—1 _2l
4 n e,
4L*
e+ 1
L* *
The minus sign means that tho charge q is attracted to the
dielectric.
3.19. We apply the law of conservation of energy to the
process of recharging the capacitor after bringing its plates
together:
<? = A bat- A W e.
Here Q is the Joule heat liberated in the resistor, i4bat the
work done by the extranoous forces in tho battory, and AW«
til
ihe chance in the eleclric energy of the c.npnrilor. If A7
was llio change in the cap mi lor chnrgo duo lo its having boon
recharged wo can wrilu
A7 - M 'f - (2C - C) I
67, /l„ nl = Aq'f --= Ct*.
To obtain an expression for Al-T,, wo must tako into ac­
count that whon the plates are brought together the capac­
itor capacitance becomes equal to 26’and houco the poten­
tial .difference, given the chnrgo romains the saino, is S/2 .
When the recharging process is comploto tho potontial differ­
ence will he restored to f. Therefore
AH ’
26
26 (*/2 )*
2-----X
3
Tims, <j = 4 " ClJ
In order for llio capacitor's chnrgo to romain practically
unchanged during tho timo interval At whon tho platos wore
brought together, it is necessary that
At < t = RC.
The constant RC = x determines the rale or tho process in
a circuit composed of a rosistor and a capacitor. Substitu­
ting the numerical values we obtain
/?»•—- = 10* Q.
2.2(1. T o estim ate the q u a n titie s we assume that tho Hold of
a needle lip is the same as that of a charged ball. I 11 an olectric
held of strength E an atom acquires a dipole moment p =
ae 0E. The inhomogeneous Geld acts upon tho dipole with
a force F = p
dE
. As a result we get
-?5£8- —£ - = W * 5» 10-»
»N = 10-« dyn.
r1 (4ncQ)r*
*
In this expression r = R
I = 300 A = 3 X 10"# m is
the distance between the atom and the tip centre of curva­
ture, and the polarizability a « a3, where a — 1 0"10 m is the
atom ’
s size.
3.21. Since /., ^ /., we can assume, in a good approximation,
that the cylindrical conductors have constant potentials along
their whole length and that to dclcrmino tho field wo noed
on ly consider an electrostatic prohlom. Suppose tho linear
charge density on the conductors is ±p. Applying Gauss’
s
112
theorem to ono of the rode wo find that tho hold strong Hi
of the rod a distance r from its axis is
E = dbp/(2jie0r).
i Tho potontial difference between the rods is found by
integrating over the field strength:
r»
The total current J can bo easily found if wo consider that
the field near each rod is independent of the chargo on an­
other rod (6 » r0). Assuming that the current density is con­
stant across the sheet thickness we obtain the following for
the total current flowing out of a cylindrical conductor:
J = 2nr0 aj — 2nr0akE =
Hence
d_ U_
^0
.
1 ,n b
R ~ C! ~ * U in r, *
3 .22 . In the first case,
tho current will remain constant as
the voltage increases; therefore the power liberated at the
anode will double. In a stationary state this power, when
anode temperature is high enough, is dissipated mainly as
radiation so that W oc
where T is the absolute temper­
ature (in our estimates we shall neglect the heat released
through conductors and tho othor structural olemenls of the
kenotron). Thus, the absolute temperature will be increased
21'4 « 1.2 times, viz. it will reaich about 1014 °C. In the sec­
ond case the current will bo increased 2s/3 times (threehalves power law) so that the power will be increased 2 X
23/* = 2*/* times and tho temperature will rise (25/2)1/4 ==
2*/# times, viz. it will reach some 1400 °C.
3.23. The conductivity X of an ionized gas is proportional
to the ion concentration which in turn is proportional to
the flux of the fast electrons moving between the spheres.
Thus, X oc /. It follows from Ohm’
s law (/oond = XE) that
throughout the space between the spheres the electric field
is constant. Hence,
This sort of hold can only omorge in a spherical capacitor
when a space chargo is prosont.
8-0408
113
4nX
3.24. A = TTn AnX4 ,n . where X is the specific conductivity
i//ij—i//if
of the dielectric, and Rj and R%are the radii of the spheres.
3.25. X oc 1/r4.
3.26. In a stationary mode of operation the load current is
equal to the current carried by the belt:
.7 =
olu.
The maximum current value dopends on the electric strength
of the gas:
< W = 2e0£„ ,
.7m„ = am,Jv = 2e„ E,tlv as 10'* A.
The maximum potential of the sphere is also controlled
by the electric strength of the gas. For a spherical electrode,
q> = RE,
where E is the field strength near the electrode. Therefore
q>ro. r ‘
= ^ s t = 4.5 X 10‘V.
knRlRi
3.27. e = - £ —
t
P ~ C ( R t — R1) In2 *
4ne0
3.28. The magnetic moment of a uniformly charged ball of
radius R rotating with frequency co is
Pm =
(?Jl*(O
where Q is the total charge on the ball. In fact, the magnetic
moment of a ring with current J is ;)S, where S is the
ring area. Changing to spherical coordinates r, 6 , <p we find
that the magnetic moment of the ring at r, 6 is
dpm = jipwr4 sin3 0 dr d0 ,
wherep is the space charge density. Integrating over rfrom 0
to R and over 6 from 0 to ji we arrive at the above formula
for p m.
Let us introduce the quantity y = {qp — qt)/qp and write
the charge on one atom as Z (qp — q9) = Zy qp. If the Earth
is composed of atoms with an atomic number Z and a relative
mass A, then the ratio-betwcon its charge Q and mass
where m0 is the mass of a nucleon. Using this relationship
the expression for p m can be rewritten as
Pm^
p.
where p is the Earth’
s donsity. if the Earth possessed this
magnetic moment, then the magnetic field induction at the
Pole would be given by
_ 2PflPm
*1
Whence
A
y
Z
i5w0flB
aiiopfpwflk
2.4 X
Thus, the hypothesis in the problem about the origin of
the Earth's magnetic field contradicts experimental data.
, 3.29. The tensile stress in the walls of a rotating tube can
! be determined by equating the elastic forces and the centrif­
ugal forces of inertia because they are in equilibrium. The
stress proves to be / = pn3, where p is the density of tube’
s
material, and u the spoed of rotation. Its limiting value is
uraaz = K/mtx/p = 4.7 X 10* m/s. As the tube rotates,
1 a transfer current emerges whose surface density / is given
by the formula / = on, where o is the surface charge density.
From Gauss’
s theorem, the electric field strength on the sur­
face E is o/e0 so that for the current density we have / =
e0Eu. As to the shape of the currents, the rotating tube
is like a solenoid, therefore the magnetic field outside the
tube is zero while inside it is B — p j. Substituting / = e0Eu
and e0po = c~%we
u>m _ £V2p0 _ v*
2.4 x 10-«
w9
t 0E*/2
c*
3.30. an n =•2e0£ = 3.5 x 10~® C/m3,
^max=-y-am«ri;s= eoPo^y » 3.3 X 10' 10 T « 3.3 X 10~6 G.
3.31. First we find the total force acting upon an electron
a distance r from the beam's axis. Gauss’
s theorem yields
E = —ffn0r/2e0 for the electric field, while the magnetic
field acting upon theelectron is determined from the magnetic
field circulation theorem:
it _
8*
__ en0vr
115
The totnl force F ncting upon ihe electron is llms
n0e*r
2e0
|i0n0e3t;3r
The first term represents Coulomb repulsion, and the second
mngnotic compression. This expression can be rewritten as
f - = £ r( < - - S ) If the concentration of positive ions is n, then the equi­
librium condition for the beam electrons is
( *
j/ M _
2e0 \ 1
e%)
n0e*r
whence
neV
2rc
’
3.32. Tho induction B of the magnetic Held is zero inside a
cylinder and B — p 0J/2n/? near its outer surface. The
pressure on a plasma cylinder in a magnetic field is
pm— BV2\i0—
.
Substituting numerical values we find p m « 6.4 X 104 N/ma.
In our case p m < /> and honco tlio plasma cylinder will ex­
pand. It can bo easily calculated that the condition p — p m
will bo fulfilled at the current .7 = 1.25 X 10* A.
3.33. Under equilibrium the gas-dynamic pressure in a
plasma is matched by the magnetic pressure, i.e.
whence
B = V 2p0nkT « 6 T = 6 x 10* G.
Here u = V 2eU/m is the velocity of the electrons in the
beam, and c the velocity of light.
3.35. The work performed by magnetic forces in displacing a
contour with current .7 in a magnetic field is
A = ,7Ad),
110
where AO is the change in Hie nwigtiolic flux. In our case
AO = AO|| is the change in the flux trnvorsing the d.c.
.frame. According to reciprocity theorem, AOai = AOia,
. ii J l = Cf%- Since AOia = —20ia, the problem boils down
to determining the magnetic flux in the frame in a d.c. field:
b
The minus sign depends on the choice of the positive direc­
tion of the currents in Fig. 16. Finally we got
:
A=
3-36-
ln ^
2.72 x 10-’J = 2.72 erg.
+
) ,7*=conal ~
( ^
< 7
) |(nW/tl) + x )’
>1
where Wtn is the magnolic enorgy in the coil. Tho minus sign
indicatos that both halves of the toroidal coil attract each
other. For x = 0 we got
r
( \ w N d \2
^ = - 1
AnR— ) •
,3.37. Tho force on a sample in heterogeneous magnetic field
is
where p,n is the magnetic moment of tho samplo and equal
to y.BV. This forco roachos a maximum nt rlnax = 1/2/a =
5 cm. The maximum v^lue of .the force is
=
dyn.
3.38. As the tube is compressed, the magnetic flux will be
conserved:
= nR*B,
/?=-/?„ V B jB .
The increase in the magnetic induction in tho tube is due
to the appearance of a current in its walls. Thus, the tube
behaves like a solenoid. The magnetic field pressure is
B%
Pm~ 2n„
Substituting in numerical values we get
B= V 2 i^ = 5 x lO * T
and R -
=
10‘
J m.
117
3.39. When u conducting liquid moves in a magnotic field,
an e.ro.f. of vBd is induced in the circuit. This expression
can be easily obtained by considering the action of the Lorentz force on the free charge carriers (electrons and ions).
A current of
gind
.7= -r + *
where
r--= -J- -j-
will run through tho circuit. Thus,
P = ,J2R =
( S tnd) g fl_
(r+/f)»
v'B'd'R
' f f M
'
3.40. The magnetic induction due to the current is B =
p 0 7/2xi/?. Tho force F acting on the needlo is F =
p m (dB/dB), where p m is the magnetic moment of the
needle and is oqual to B0V/\i0 (V is the needlo volume).
Thus, F = Z?0V,7/2jt/?*. In order to lift the needlo off its
peg wo must have F > pgF, where p is the steel density.
Whence
.1>*!£ **! = 2 .4 x 10* A.
3.41. The equation for small oscillations o fa bar has the
form
where I is the moment of inertia of the bar and is equal to
/*
J2" pV (V is the bar volume), and pm is the magnetic moment
equal to B0V/p0. Finally, the oscillation equation will be
12D9Bb
cp= 0.
<P +
PoPl*
The oscillation frequency is determined by the relationship
Whence
IIM
* - 12B0By
"0 ~
PoPl* *
T0~ % = 2nl
4.1 s.
3.42. We want to find the residual magnetization of the ball.
From the definition of the demagnetization factor we have
H 0 - H = p/,
118
where H 9 is the external magnetic field, H tho field inside
tfaie ball, and I the magnetization (viz. the magnetic moment
per unit volume). In our case the external held is absent,
therefore
In this relationship H is the magnetic hold strength inside
the ball due to the residual magnetization /. The magnetic
induction inside the ball is
1
*,/•
According to the problem, B = B0 (1 + /////„ ). To deter­
mine I we have the equation
-3- M =
(1
3- 777 ) •
whence
3
b9h 0
« 1.2 x 10* A/m2.
- 2poffc + *o
3,43. If we apply the magnetic held circulation theorem
we have
H (L — I) + H'l = NJ.
whore H is the held strength in the core, and H ' tho hold
strength in the gap. For a linear case (the inclined part of
the magnetization curve) the boundary condition is \\.H = H\
Thus,
H (L - /) + p/// = N.y or // \L + I (p — 1)1 = N. 1.
Whence
Allowing for (p — 1) = x = I 0IH0 we find finally that
jr. = 4 - l ^ o + i/ol.
In the case of saturation (at .J > ,70) the boundary condi­
tion should be
H' = H + /„ , viz. B' = B.
From tho magnetic field circulation theorem we got
(H' - /,) (L - 1)+ H'l — NJ,
H ' ^ ± |N.) + /„ (L— Z)J.
119
Hence
3.44. o) = - | - ^ 2 . » 0 . 1 s’
*.
3.45. .7 = I In.
3.46. IIA =
, -£ = (1 )*.
3.47. The equnlion governing n proton’
s radial motion in the
held of an electron beam enn (noglocting tho magnetic hold)
bo written
mr —eEr - 0.
Here E r is the absolute value of the radial component of
the electric held of the beam as determined from Gauss’
s
theorem: 2nrenEr = (?, whom Q = ermr1 is tho charge per
unit length of a tube of radius r (n is tho olcclron concen­
tration in tho beam). Thus, Er — nerl2e0. Since .7 =
cn vn/?*, whoro for o relativistic beam v « c, tho ex­
pression for Er can be written as
p_
r~
7
2nt0cR*
r.J
The equation of motion for the proton takes the form
mr'+-2S^S5-r = 0whence for the frequency of the natural oscillations o> we
find
*
*3
W ~ 2nticR'm *
1 ,/
0r
R V
40
ed
2ne0me *
Substituting in numerical values wo arrive at o> = 3.9 X
10* Hz.
3.48. It follows from the solution of problem 3.31 that for
i ; < c wc can neglect the magnetic component of the Lorentz force acting on the beam charge, as compared to the
electric component. The electric ftold strength at tho beam's
boundary is
E
120
ne =
2e0
3
2t0v ’
where n is the number of electrons per unit area of a sheet
flow, and v the electron velocity. The time needed for the
beam’
s width to double is
Here a is the transverse acceleration of the electrons under
tho action of the Coulomb forces. Finally wo find that
.
/ 2W \3/4 / Ae0d
*= w =(— )
W \1/2
( " 7T t )
« 17cm-
3.49. The ion will continue to spin so long as the frequency
of the a.c. electric field coincides with that of tho ions'
rotation in the magnetic field (the synchronicity condition):
eB
2nm
1
170 kHz.
In this expression m = 28 X 1.67 X 10" 37 kg is the mass of
an N\ ion.
To estimate the frequency change we must first find the
spiral pitch. Tho velocity change for adjacent turns can
be estimated from Nowton’
s second law, i.e.
c
T
mAv
y
,
where T is the period of revolution. Since R =
T*
eE
An
m
wo gel
‘
The synchronicity condition will be violated if
Av
AB
v0 > B *
Whence it follows that
Av~ w ~ 15 kH*3.50. Each charge on n dipolo is acted upon in tho direction
v
force
The equation governing tho small
oscillations of the dipole due to this force is
/„ fp+ pvBtp = 0 ,
whenco for the oscillation frequency we get
pvB
/o •
121
3.51. /?(<) =
+ const.
3.52. < = J ^
« 5 x 10-* s.
3.53. j max w * * £ i( 2 ! L ) 5/2»40 A/m2 = 4 inA/cm2.
3.54.
1.3 x 10-2 T = 130 G.
where A x and A t are the atomic numbers of the uranium iso­
topes *saU and *”U, and m and e are the proton's mass and
charge, respectively. The time required is
t=
y*,- = 8 / 107 s « 2.5 yoars.
3.55.
T = 1 0 U.
3.56. The estimate is easy for particles leaving the column
with a root mean-square velocity v = Y%kThn directod
normally to column surface. The magnetic held near the
column surface is
In this held a particle that has escaped from the column
will move along a circle whose radius is given by
mv2
evB.
As a result we get
7 = —
/ 3 kTm « 2 x 105 A.
3.57. There is a restoring force (a quasiclastic one) for
tho particles which leave the horizontal plane, the total
force depending on the magnetic held component B r. Since
B ~ H and rot H = 0 in the magnet gap, we get
dBr
BBZ
dz
whence for small z's
dB x
dr
dr
’
n D
— r-B.z.
r0
Tin* LorrnU force component is
Fz = eoBr -- —e<j)0nBzz = — wj/miz,
122
where v is the linear velocity of the particle on an orbit
and <i)0 = eBJm is the angular frequency (cyclotron fre­
quency). It follows from the equation of vertical oscilla­
tions, i.e.
m z = — no)|mz,
that
to, = (o0 Y n.
3.58. An equilibrium orbit is
stable to small radial devia­
tions of a particle if the Lorentz force changes with radius
r slower than the centrifugal force (Fig. 69). For small
deviations p = r — r0, so it s possible to write
B, (r) = Bz (r0) + ■
- i t (r - r0)« Bz (r0)- — 5, (r0) p.
The radial component of the Lorentz force is
Ft = evBt (r) = «orB, (r) •
■
=eo>,
Bt (r„ ) ( l - ^ - p )
(r0)(l— ^-) ( l - — p)
n)p.
This expression also includes tho law of the conservation of
angular momentum: o)0rj = cor2, where o)0 = eBz (r0)/m
is the angular frequency of particle rotation on an equi­
librium orbit (cyclotron frequency). We now turn to the
expression for the centrifugal force:
Fet = m tfr .-=
« miojr0 (1 - 3
.
The equation for the radial oscillations has tho form
m p = — meo* (2 — n) p.
It can be seen that an oscillating mode (an equilibrium of
a stationary orbit) is possible for n < 2 . In this case
Up = Wo V 2 — n.
3.59. It follows from symmetry that tho ring currents per­
pendicular to an external magnetic held will cross the ball's
surface. It can be easily checked that tho field of the cur­
rents outside the ball will coincide with the field of a dipole
placed in its centre. The field of a magnetic dipole with di123
pole moment Pm >s
I) _ Mo T 3 (PmO r
4n L
r»
Prn "1
r3 J *
This field along with a liomogonoousexternal magnetic Hold
B 0 satisfies the boundary condition on the surfaco of a super­
conducting ball: B 0n + Bn = 0 (the condition of conti­
nuity of normal components of induction voctor). In fact,
for any point A on the ball's surface it is possiblo to write
(Fig. 70):
Pm cos 0 1 _ MoPm cos 0
Mo f 3pm cosO
Bon = B o COS0,
R3 J“ 2nR* '
4n L
R9
At pm= — — Vf3B 0 tho boundary condition is by definition
Mo
fuliillod at any point of the
ball surfaco. Thus,
Bo + B — (1 +
-J B0
3R9(B0r) r
2r6
The current surface density
i will be found from the bound­
ary condition for tho tan­
gential components of magnet= i. Finally,
3.60. In a magnetic field B 0 each ball acquires
moment
pm = - i l / ™
a magnetic
0= - 2 n / ? H I 0
(sec Problem 3.59). Tho magnetic moment of a unit volumo
of a gas composed of tho superconducting bolls is
i = npm = —2nnIP\\0.
Whence it follows that the magnotic susceptibility x dofined in SI units by the expression I = x|i0H is
x = ——
Mo
124
nit3.
Thus,
I
M=-, 1
x - I—
Mo
nil* <
1.
An “
ideal gas" composed of superconducting balls possossos
diamagnetic properties. Tho condition nR 3
1 moansthat
the balls are spaced far apart and so their interaction may
be neglected.
3.61. In our estimates, we shall assume that tho ball is in
a non-uniform field that changes from Bor to zero at a dis­
tance of the order of a ball d inmotor. Tho magnolic mmnont
may thus bo estimated from
A,
where
^
(soo Problem 3.59). Tho forco octing upon
the ball is
Fx = p J - £ ~ - ^ R > B ' C'.
The superconducting ball is pushod out from tho rogion of
the strong magnetic field. By equating this force to the
ball's weight we find
1.7 cm.
' 8|»0pjr
3.62.
A dipolo in a magnolic field is acted upon by a forco
P* ~ Pm I T ~
(see Problem 3.59).
The work against this force.
- j F dx =
2nRR*
d lix
Mo
Ox
2nR*
nB *R *
Mo
Mo
’
is equal to the initial kinetic onorgy of tho ball:
mu* _
whence
2
“
Mo
—
=20
m/s.
Thus, at the initial velocity v„ > 20 m/s the ball will Oy
into the solenoid.
3.03. When a superconducting solenoid is deformed, tho mag­
netic flux is conserved: BSN = const, where 5 is the so enoid'scrosssection, and N the number of turns. The magnetic
pressure on tho solenoid walls is
—
p™~~
Whence,
Jil
2|i0 ‘
p mS* = const.
3.0^i. From the magnetic field circulation theorem we have
n,y
0“ H XL +
where n is the number of turns, Hi the field in the steel
and //2 the field in the gap. The boundary conditions
yield \iH, = Ht. Whence
n.'JQ = H 2 (L/p. + I).
Since the winding is superconducting, dfbldt = 0, viz.
the magnetic flux is conserved irrespective of any change
in the gap. Hence, as the gap is decreased, the field in it is
unchanged. Whence
i- f) •
where .7 is the now current in the winding. Thus,
1+ JL JL
.7= 7t
2
L
1 +P
3.65. Inside a superconductor the magnetic field vanishes.
It follows from boundary conditions that the normal com­
ponent of the magnetic field induction vanishes on its sur­
face, too. To determine the magnetic field produced by the
plane we can use the method of images, viz. wo imagine there
is a direct current at the same height over the plane, but
running in the opposite direction. The force acting upon the
current per unit length due to the image is F =
wherd B
is the magnetic induction of the field produced by the image.
This force is directed upwards. For tho conductor to hover
freely over tho plane a distance h above it will be
or £ $ r = w126
Note that the solution can easily be generalized to the case
of a flat curve of arbitrary shape, if tho radius of curvature
of the curve significantly exceeds h in all points.
3.66. Let us assume that the cavity’
s dimensions are small
compared with a resonance wavelength. Then the resonance
frequency can be determined from the formula for a qunsistationary circuit: o)0 =
LC, where C is the capacitance
of a parallel-plate capacitor in the central part of tho cavity,
while L is the inductance of the toroidal part:
When doing the calculation we use the formula (1) = L,J.
The induction B of the magnetic field in the cavity is p 0J/2ji /?,
where R is the distance from the cavity’
s centre:
p0a In 3
«/2
Hence,
£,= l!sl15L3« 2.2x 10-* H.
For the resonance frequency e>0 = \l\f LC wc find '.hat
= 2.5 X 10* s"1. The wavelength \ = 2jic/aj0 « 1 m »
a, viz. the assumption that the resonance wavelength
is much larger than the cavity dimensions was valid.
cl>0
,y ___
Go^OS
' ' ~ 2 V (/?, + /?)*+ (1/2QC,)1 '
3.71. Q flC = 1,
*
-%FL = 4-.
^ In
«
127
X73. 0 ^,=
i
Iho equivalent indue I anee of the oscillatory circuit, allowing
for the effect of the short-circuitod coil.
IV. Optics
At t|> = ■
—-m (m = 1, 2, 3, . . .) I = 2/0.
4.3. (1) The interference order m « 400. (2) The width of
interference fringes / = 2.8cm. (3) The greatest source size b «
I = 2.8 cm.;(4) The source’
s nonmonochromaticity AXs£ 14 A.
4.4. (1) mraai « 1000, /7iroln « 720; (2) AX ^ 5 A; (3) the
source may have any dimensions.
4.5. i
4.0.
4.7. A/ «
w 15 MHz,
Av = -^- = 1500 MHz.
4.8. The radiation from a white light source has a wide spec­
trum. The intensity of the interference maxima will yield
the spectral components such that A = mX, where m is an
integer.
Analogously, the spectral components for which A =
(m + 1/2) X will yield the intensity minima. Thus, as
a result of interference, the spectral instrument will re­
ceive a radiation whose spectrum contains alternating max­
ima and minima. The distance between adjacent spectral
maxima can be found from the relationship
~ 1) Xm+j = mXm, or m(Xm Xm_|) = Xm+(,
AX = Xm— Xm+1 = X/m = XVA.
The spectral instrument should possess a resolving power suf­
ficient to detect two maxima spaced AX apart:
/? > X/AX = m = A/X » 5 X 10s.
128
4.9. A = c!Av = 30 m.
4.10. An electron oxcites identical perturbations in a lattice
which follow one another every time interval t = dlv.
An interval of time between adjacent perturbations at the
point of observation, which is located at an angle 0 , is
equivalent to the path difference
A = t c — dsin 0 —
— sinOj .
In order for the perturbations to amplify one another in
the 6 direction we must have
d
— sinOj = mX,
where m is a positive integer number (m > 0 , since c/v > 1).
The possible m and X values should satisfy the relationship
|sin 0 |= | -£■—"i
4.11. It is necessary that at the point of observation the
Fraunhofer diffraction be fulfilled. The mirror size should be
much smaller than ihc dimensions of the first Fresnel zone, i.e.
d
1
nb
T < r* ~ V k 7ToAssuming that the source of the radiation is at the focus of
a parabolic mirror (viz. a = oo) we get
•y*
Xb,
or
b ^>
« 10 m.
4.12. / = I J 4.
4.13. Av = — v; the effect is due to tho differences between
the optical path lengths for the two sets of waves.
4-15- - ^ ( w ) 2*51'5 * 10’
.
cm.
.4.16* In the first case the telcscopo objective receives radia,lion whose power is
'- M W 9-0408
129
This power is distributed over an area S x of diffraction spot
produces an illumination
The background illumination in the focal plane is propor­
tional to the lens aperture:
Thus, the image contrast is
When making observations through a telescope with an objec­
tive D.y
E .+ El
.
£°
t
, P0
' a
D\
LtV *
Hence, Vi = 7 a provided that
D\
=
D\
whence
£ .= L,-%*- = 5 x 10* km.
-
1 Dx
4.17. The angular divergence of a light beam is X//), there­
fore the radiative power scattered by the satellite is
Here L is the distance to the satellite. Since the satellite
uniformly scatters the optical power P' throughout a solid
angle 2n. the received power will be
p ,nj0V4) _ P ' / D \2
~ 8 \LJ *
The condition for detecting a scattered signal is now that
« 5 x 107 km. Here h is Planck's
4.18. L
constant.
4.19. Zmlo
130
1 m.
4.20. The resolving power of the objective is fully utilized
for a normal magnification of
Whence
F. = 4 rF , = G cm.
'
D 1
The angular resolution of a telescope when being used for
visual observations is Yn times higher than that of the naked
•ye. The distance at which a text can be read using a teles­
cope is governed by the condition
Whence it follows that
L<
I « 600 m.
4.21. The limiting resolving power of a five-meter telescope
corresponds to a lunar surface area of S 0 « ji (X/?/Z))2f
where R is the distance to the Moon. This means that the
illumination of any point in the Moon's image in the focal
plane of the telescope objective is due to the radiation arising
from every element on the Moon’
s surface within an area of
S 9. If part of this area is screened by a black (nonreflecting)
canvass, the illumination of the moon’
s image at the teles­
cope’
s focus will be changed in proportion to S 0 — S, where
S is the circle area. Given the data in the problem
< 0.99.
Whence
r* > 0.01
4.22. D» r
p,
or
a
r>
4ra.
6.5 m.
°
4.23. A microscope resolves an object of size,
l _ 0.61X
n s in u *
where \ is the wavelength, n the refractive index, and u
the angular aperture. An electron wavelength is
h __ 12.24 f
N mv
YV A
#•
131
Where h is Planck's constant, nnd U Iho acccloruling voltage
expressed in volls.
The resolving powers of the oplicnl nnd oloclron micro­
scopes nre equal when
O.OIXqpi
sim iop i
_
O.OlXc _ 0 01 x 12.24
s in u fl
/ t / . s in u e
Whence
4.24. The order of magnitude of the Doppler broadening can
be found from the r.in.s. vnluc of the longitudinal velocity
of the gns molecules, i.c.
To investigate the Doppler profile n spectral instrument should
have the resolving power of
When working in first order R = N =
— , where
L
is the size of the diffraction grating. Thus
L = Rd > 40 cm.
4.25. In order for a spaceship's motion to be delected by a
Doppler shift in spectral lines, the shift must exceed the
spectral line width caused by the thermal motion of the
molecules on the Sun’
s surface (viz. tho spaceship linear
velocity should exceed the root menn-squnre velocity of the
hydrogen molecules, which is vT « 104 in/s at T = 6000 K).
Whence it follows that u > vT « 104 m/s. Tho minimum
number of diffraction grating line rulings is determined
from the relation
For m = 2 we get Ar > 1.5 x 104.
4.26. For a split line 6?. to be detected in the emission spec­
trum of a double star due to the Doppcr effect, the split
must exceed the spectral line widih caused by the thermal
132
motion of molecules. The condition is fullille.l if the ineor
velocity of the stars exceeds the root inean-sqiiaro velocity
of tho moloculcs on their snrfnco. I'or r - 10 days this condition is fulfilled if
> R = mN 5«-jgf ®
•
N > 10’
.
for t = 10 years the splitting
of the lines is much less than
their width.
4.27. An interference reflection
of light occurs only from those
sound waves for which the
Wulf relation is fulfilled (Fig.
of the scattered light is
Fig. 71
71). The Doppler shift
£
Av = 2 v —
— sin 2 *
c/n
For reliable resolution we must have
R= m * > i7.
Whence for m = 1
N > 2vn sin (0/2)
4.28. N >
1.2 X 105.
(--)* as 2.5 >: 10».
4.29. The resolving power R of a Fabry-Perot interferometer
can be estimated by considering the following. The minimum
path-length difference between rays forming an interference
pattern in the interferometer is 2L. Hence, the interference
orderm is 2£/X. The effective number of rays leaving the in­
terferometer is N « 1/(1 — p). By analogy with a diffrac­
tion grating we can write
2L___ 1_
<oL
= mN
R=
Ab)
\ 1— p
c(l-p) *
Whence
L
o
Since co = a>m + toL, where toL = const is the laser ra­
diation frequency, and tom the frequency of the investigated
source in the infrared range, we can take Ao> = Acoir.
133
Hence,
R -
1R
Ati)IR
__
Ao>
_ <»
>in n
co
. .
2nL
~ c(l--p)
Mr (1—P)
« 2 ■10*.
4.30. Linearly polarized light does not transfer angular mo­
mentum. If one of the major directions of the quarter-wave
plate coincides with that of the incident light’
s polariza­
tion, the transmitted light will also be linearly polarized
and the plate will not experience a torque. As the
major directions of the plate are rotated through ±45°
relative to the polarization plane of the incident light, the
transmitted light will be circularly polarized, either clock­
wise or anticlockwise. This light will transfer angular mo­
mentum at a rate L — ±hn per unit time, where n is the
number of photons passing through the plate per second. The
plate will experience a recoil angular momentum of
M = — L = ± hn = ± h
±
,
where P is the pow*er of the transmitted light and v its fre­
quency. Substituting the numerical P and v values into
the expression we get
M = 6.28 X 10-16 N-m.
4.31. z is the optical axis.
"'= /
n0= j / ^ ■
td ,
e£‘5 ^ 7 -
sin (2v<f/2)
wiiere <P= - x d ("* — ".)•
4.32. /4 = 4jsin (<p/2)
2-'
Such a system is a polarized interference filter acting like
a diffraction grating.
4.33. The gas’
s refractive index is
n —Y 1H 4na/V,
where a is the polarizability of the gas molecules (in the
Gaussian system) and N their concentration. Since
a '(;.) = - £ c x p [ 134
>ngyh
kT
“
1
J»
where pJkT is the molecular concentration at h = 0, we get
V ^ 1+ 4na-£r exP f — ~TT~]
« 1-f 2na TPoF (Vi
mg v ;t \
Jtr I •
The radius of curvature of n beam that starts horizontally
near planet's surface is
n
(kT)«
— 102 km.
R = dn/dh
2nap0mgy
Since R < 0, the centre of curvature is below the surface.
Thus, horizontal and near-horizontal rays cannot escape from
Venus’
s atmosphere {\R\ < R\ ). Circular refraction is
thus possible in Venus’
s atmosphere, if a light beam en­
velopes the planet at some height.
4.34. The refraction of a horizontal beam is described by
the formula
1 1
dn
~R ~ ~ ~dh '
where R is the radius of curvature of the beam and n the
gas’
s refractive index as defined by the expression
n - l = (n„
- l ) e x p [ - - ^ - ] * ( n , - 1) ( l —
The last expression is valid at small values of the height
(mghJkT < 1). Allowing for (n — 1) < 1 we get
4 -= — (n0— 1) -r=- . R = -----5 - p —
R
' ®
'kT
n0— 1
mg
« - 2 . 9 x 10‘ km.
The negative R value indicates that the beam’
s centre of
curvature is at h < 0 .
Given circular refraction the beam's radius of curvature
is equal to the Earth's radius, i.e. R& = 6.4 x 103 km. Since
;R ~
~ ~ , the atmospheric pressure (and density,
too) should be increased by | R |IRE = 4.5 times.
4.35. The lens and hence the ball receive a radiation power
of p
where D is the objective’
s diameter. The ball’
s
•
radius is equal to that of the Sun’
s image, aF/2, where F
is the lens’
s focal distance. The ball radiates a power of
oJTMn ^ ( a = 5.67 X 10'* W/(m2K‘
) is the StephenBoltzmann constant).
135
The steady-slate temperature T can be found from the
energy balance:
' ' ‘'
a*F*
p ^ = o r ‘4n 4 *
W h en ce
L - IJ L \ 2
T = \ / J4oa*
\ F )
1.7 x 103 K.
4.36. An electric breakdown occurs when an electron affected
by the electric field of a light wave gains an energy of 10 eV ==
1.6 x 10'1S J. This energy is much less than the elec­
tron’
s rest energy of m</:2 = 10~13 J, viz. the electrons’ve­
locity during the breakdown is much less than c. In one os­
cillation period the electron moves a distance of order of
nv, where v is the frequency of light. This is much less
than the light’
s wavelength, X « 5 X 10~&cm, and electron
free path A = 10"4 cm.
Thus, we can consider the action of a light wave upon a
free electron
m i = — eE0cosiot,
where £ 0 is the light wave’
s amplitude. By integrating we
get an expression for the electron velocity:
^
sin u>t.
The maximum electron kinetic energy is
Whence
mv*
m I eE0 \2
— =
=eU ■
=
2U mu2
;— •
The radiation power flux density in the beam is
-j ■
=
where S = -j-D*. E'„ = E'Jl. Finally,
p = ce<(5 - S - =
e t,S U m w '
s 2 x 101* W.
4.37. Since at sliding angles of incidence, (p « 4a, we find
from the condition of total internal reflection that
cos a = cos — = n = 1 ---p - .
t36
T h e r e fr a c tiv e
-n
in d e x
i/ V c
is
, f\
*
\
c*N
: 1 - 2e0mco* •
e'N
i 0m<a2
|For Be: Z = 4, A = 9, W = Z/VAo//l (/VA is Avogadro’
s
'number). Whence
4.38. A photon with energy 5 eV corresponds to the wave­
length X = 3 X 10'5 cm. The transparency boundary is
|given by the equation
e*N
e — 1 — t0m(ii* = 0.
Whence
e 0mo)s
N = ~ ^i
__ 4n se 0c sm
X*«*
^
|Q2fl( ,n -3
•'The concentration of atoms in silver is
/ i = - ^ e . « o x i o !!* m->.
A
' where NA is Avogadro’
s number. Thus,
N_____ t_
n ~ 6
4.39. Plasma's permittivity e
e*N
e = n2 = 1 — e0ni(oa *
where e and m are the electron's charge and mass, rcspectively, and N the electron concentration. Whence
N
(1—n*) m(i)3ea
«
»
2.4 x 10‘
‘nr».
Now we find the radiowave’
s phase velocity, i.e.
v = c/n w 3.32 x 10* m/s.
To determine the group velocity u we need to know the
function v (X). This can be easily established from the ex­
pression for na, if we take n — c/v and |<i> = 2jiv/X:
V= \ / c 1 +
Ne'X*
4n2e0»i *
137
U s in g
th e
H a y lc ig h
r e la t io n s h ip
. dv
u = i-- X -jj-,
we get
u=
= c;i « 2.7 x 108 m/s.
4.40. The focal spot’
s diameter is
± P*X .
The acceleration of the electron is due to the electric held E.
The final electron velocity is given by the relation
- ^ p . = 2m,ct. whence P2= ( - - ) 2 = j .
Thus, the average electron velocity is close to d 2. Since
the acceleration only occurs for an interval of about 774t
the distance travelled will he about
c
T
X
= -g-, viz. smaller
than the focal spot sue. Now we estimate the accelerating
field strength E :
Ac
4cs
eE
4m0ca
r
The laser's power should be
Pst (c0£2) c?.2=
w 10‘
«
W.
4.41. Using a system of coordinates in which the satellite
is at rest, we find the frequency of the signal received by
the satellite, i.e.
/ 1—v*/e*
0)| = to0
l + y cos a
The reflected signal will have the same frequency in the
satellite’
s system, but in the Earth-bound system we detect,
at point B %a signal whose frequency is
— r*/c*
(l)•
. = (0,* ■
-/-1 y
1— — c a s Q
138
1— vV e*
= (|)n
® ------------------------/
v
\I
V
\ .
y1 + y
c o s o J ( 1— y cos 0 j
The relativistic correction to tho frequency is
whence it follows that the resolving power of the spectral
jdevice must be
I
~ 10*.
I
^ I Aw /re l
4.42. The corrections due to relativity account for a {y!c)%frac­
tion of measured quantities. The plasma's refractive index
h = Y * differs from unity by
A = 1/“
A/i
K e- 4
where e and m are an electron’
s charge and mass, respectively.
;Since the ionosphere (its electron density) changes in a ran­
dom fashion, the orbital parameters must be measured with
errors less than An in order to detect reliably the effects pre­
dicted by the theory of relativity. Thus,
>:
,Whence
I V \2
e-N
( r ) > A,,S5 2^ ____
H l-
4.43. i>= | c .
4.44. At = 3 x 10-« s,
6 (Af)/At as 0.06%.
V. Atomic Physics
5.1. The laws of conservation of energy and momentum for
the Compton scattering of a y-quanlum through the angle of
180°yield
where p is the electron momentum while kta and £<i>' are
the y-quantum energies before and after the scattering, re­
spectively. Given that E » me* the momentum p » me.
It follows that Jia>' « mc2/2 = 0.25 MoV.‘
5.2. If the total energy, momentum, and angle of ejection of
a recoil electron are E, p, and 9 , respectively, the laws of
conservation for the Compton scattering, given data of the
139
problem, are
.
•»
/iw
-- l> cos cp, -^r — p si n cp,
I
//io -t- me2= - 5—+ k. o
and we also have
E- = pV* 4- wiV.
Solving this system of equations we get hm = me* =*
0.5 MeV, whence X « 2.5 x 10‘10 cm.
5.3. The rosolving power of tho spectrograph is
where N is the number of reflecting layers, m the interference
order. According to the Bragg relation:
2c/sin9 = mX,
m=
9^-n ? .
Here d is the diffraction grating constant. The change in the
wavelength is governed by the Compton formula,
6X = 2jiA (1 — cos cp).
where A = hlmc — 3.9 x 10" 11 cm is the Compton length.
Thus,
ft = N m =
a
2d sin 0
X
2nA (1—cos<p) '
X
Whence
4nA sin 0 (1— cos (p) ’
Substituting in the numerical values wc get D = 20 A.
5.4. If the initial photon energy is hu>, the final one
the photon scattering angle 6 , the electron kinetic energy £kv
its momentum p, and the electron ejection angle cp, the law
of conservation of energy and momentum is
.
»
. , . r,
W
n
— fia) -f Ek* —“ sin 0 = p sin cp,
Whence
hiI)
h<0'
a
—— c o s 0
+ p cos <p.
/ ao> \2 /hii)\2
2h(o
[ — ) = ( — ) + P l --- — peosip
and
/ico' =
140
p*cl 4- E J — 2pcEy cos cp
2{pe cos<p — Eu)
if
(pccoscp —£k)>
0.
Finally,
me*
“pc’*
5.5. Assume a photon with momontum p — h&/c and mass
m — pie flies along the z-axis (Pig. 72). The change in the
photon's transverse momentum is
cos <p >
f
Apx = ) Fx d t=
r
)
YM sm
^?s+='
_
2yMs"»
J?sc
The angle of photon's deviation can be estimated thus
sin<p« A/*= CM'f-== S x lO - ‘rad.
v
P.%
^ se*
5.6. The m-th order reflection
lated to the crystal will he
the coordinate frame re-
2d sin tp = mX = m
,
x
My *
where h is Planck's constant,
M a neutron's mass, and u its
velocity relative to the crys­
tal. In the first case the neu­
tron sourco is at rest, ami in
the second case it moves with
a velocity u. Therefore
3inq>o _ |v + u |
sin 9
|v | *
Further, we assume that u < v. In this situation we have
approximately
••
Iu + vI _ ■A . u .
|VI
and hence
~ 1
sin q) « (I +
"7"sln ^0
sin (p0J sin q>0.
5.7 . The neutron's De Broglie wavelength is
X ==JL — _*_____ht = h<p _ fop
P
mv
mL
n iL u
~
7n£v ^ 0 A anu
AX
T
=
b
= 0.2.
141
5.8. D = ( yfj* jT )' ~ « - [tin, where m is the mass of
silver atoms, k the Boltzmann constant.
5.9. The energy at the n-lh level is
„
aV,s „
Each level possesses two electrons, therefore the total energy
of all the electrons is
£=
2m b*
and the force
OE
F
SI2
2*zh- in
mfcs 2 j n
~db
5.10. Given that Ap
tion yields
n-
-
p and Ar « r, the uncertainty rela£k = pV2m«0.2 MeV.
5.11. Nucleons would be expected to emit field quanta vir­
tually within a time interval t as allowed by the uncertainty
relation, t ~ hlmc-. During this interval the quanta succeed
in travelling a distance r ~ ct « hlmc « 2 x 10 “
,s cm
from the nucleon.
5.12. The angle rp between the direction of the particle's
trajectory and the radiation's direction is given as
An electron, when placed in a layer of thickness d, has a
momentum uncertainly of Ap « hid and a velocity uncer­
tainty of Au = h,dm, where m is an electron’
s mass. The
related uncertainty in the angle <p is
A<p
eh
m d n v 8sin q> *
5.13. E n — U -f n2 *-J ^9 ; for n = 1 the energy £ j« 35 eV.
nx0
5.14. The ionization potential / i s , where m _
l + m el M
is the reduced mass. For positronium, M = m« and I =
0.8 eV, for muonium I = 13.5 eV.
142
5.15.
The position of energy level is given by the formula
(1 )
£ .= -4 ^
where the reduced mass p is defined as
(2)
and a is the fine structure constant and is equal to 1/137The energy of a transition is
'•fcH r).
(»)
where n, = 2 and n% = 3 for the M-N- transit ion. Finally,
from Eqs. (l)-(3) we get AE = 124 eV.
5.16. The mean value of the potential energy of an electron
at the instant the meson decays can be found from the equa­
tion
<£p > = w J e',/r‘(* ■~k)2( —I ) 9*™**- —
o
.
The mean kinetic energy is
5.17. The probability of finding a meson a distance r from
the centre of a nucleus is
w (r) =
The energy correction is
■
e~ 2f/ri dr.
[« - (. +
Since
2 2 )
rw = 1.3x 10-» AW « 3.5 x 10-“ cm,
t t w 2.5x10-'* cm,
r i = Z»I||*’
jUon(
we see that rx »
Since
and
, Ze* 1
/ 2 r N \2
^- '2
r\ \ r, / *
<£> = 2 <£p> = -2^z**
l,
143
where E v is the potential energy, finally we get
w 2 (^• )2«s 4 x 10"*.
5.18. To make an estimate, we shall assume that the elec­
tron's initial momentum p,t was porpcndicular to the photon's
trajectory. The value of pQ will he round from
p0r ~ h.
(1)
The electron's longitudinal momentum p 91 after colliding
with the photon will he found from the law of conservation
of momentum:
(2)
+
and Aw from the Compton formula
A). = X2 —
= 2jiA (I — cos cp),
(3)
where A = 3.9 X 10"n cm is the Compton wavelength.
In our case
A?v = 4jiA.
(4)
Using (2) and (4) we can easi­
ly see that the electron in this
lindtlein is noiindali vislic.
Fig. 73
Now we estimate the spread in
the electron's scattering angle (Fig. 73):
‘
lu)'
tan 0 = p j p 9 ~ 0.14,
5.19.
or
0 ^
10°.
The intensity variation in time t is
7 = .V - r,T.
(1)
where t is the mean lifetime of an atom and t — xlu. Tak­
ing the natural logarithms of both sides of (1) we get
From the graph in Fig. 49 we find that 1hn = 1 cm"1, whence
t =r 10' 8 s. The uncertainty in the energy is governed by the
natural linewidth, i.e.
A £ w -£- = 10-’c V.
144
5.20. By emitting n y-quanlum the dust speck acquires a
recoil momentum Mu =- E lc, where E is the Y-q»antum’
s
energy. The Doppler shift in the Y-quantum’
s frequency
due to tho motion of the cmitier (speck) is given by
We can find tho natural width of tho spectral line from (lie
uncertainty principle, viz.
AE
Aw
h
2n
T
(3)
Using (2)-(3) we determine the smallest mass tho speck can
have for the Mossbauor effect still to be observed:
E»x
2nc*h
1.6x 10“
17 kg.
Tho speck’
s radius will bo approximated by assuming the
speck is a sphorc with a density p of 8 g/cm9, i.o.
r= ( ^
■
j 1/3 « 2 X 10“6 cm.
5.21. The change in the y-quantum’
s energy as it moves a
distance H against gravity is
EE = mgH —
gH.
Using tho data in llm problem, A/s = 10 T — I()///t, where
T is the linewidth of the resonance ubsorption line. Finally,
iOhe*
H=j 400 in.
*gE
5.22. Rotational levels have energies
For each of the three levels we can write
E i = - r r H i + 1),
«
21 -(i+ jt)(i+ 2),
£a = -£(/ + 2)(/ + 3),'
whence
=
— £■
,= — (/ + !)=--10-* eV,
— Et = - j- ( H 2) = 2 y 10-* cV,
10-0408
145
or
*+ 2
o
l+ l
2'
whence it follows that Z = 0 and Z = 0, 1, and 2 correspond
to the three levels. The value of
/„ ^
t 1) = 7 x 10-“
5.23. The splitting of a neutron beam in a homogeneous mag­
netic field is due to the reorientation of the neutron spins
along and against the held. The change in a neutron's energy
when it intersects the boundary of a magnetic held A£k is
±p£, where the “+ ”sign corresponds to the case when the
neutron spin is parallel to the held, and the u—”sign when
it is antiparallel, while p = pB is the projection of the
magnetic moment onto the held direction and pB is the
Bohr magneton. The neutron's kinetic energy is related
to the De Broglie wavelength X thus E k = 2j i Wfnik2. The
refractive index is
For the angles of refraction at the boundary we have
s in 60° __
s in <p
1
n± *
For a small angle of splitting 0 = <p+ — <p_, finally wo have
e«
w (n*-',-)= - ^ Un60°
= ^ ^
« 8 x 10“
»rad
(q> is the average angle of refraction, viz. «60°).
5.24. Na atoms at the temperature given are mainly in the
*SI/, state. The projection of the magnetic moment of a
Na atom onto the held direction is
Hh =
= p Bf
since rrij = ±1/2, g = 2 (Lande’
s splitting factor); pB is
the Bohr magneton. The force which splits a beam of atoms
flying along the y-axis is
v
,
dB
which in time t = llv imparts to tho atoms a velocity
,
dB
I
Vx — ^ P B ’dx vM
and this results in the displacement of the beams across a
detector, i.e.
, ..
dB
I (£ + 1/2)
The kinetic energy of the atoms is
_
3 .T
T ““ I * 1 '
The distance between spots on the screen will thus be
A = 2,'=2,*B4
-
^
^
co,.
5.25. As in the previous problem, the force deflecting the
Li atoms is
e.
dB
f = ma = nB--JJ- ,
where a is the acceleration perpendicular to the direction
of the velocity u — Y 2 E k/m. The linear deflection of the
atoms across the screen during the time they fly through
the magnet, t2 = LJv, and through the gap between the
magnet and screen, t2 = LJv%is
The distance belwooa two booms on the screen is 2A. The
size of the image of each beam is D im = D •
--1
.
L/Z
Since D lm < A and for L%» L and Lx we finally get
D<
LL, at 0.9 cm.
5.26. The resolving power of an interferometer is
R = ^= m N ,
where m = 2L/X. Thus,
X
whence
aT —
L=
2LN
3T ’
2N AX
If we transform from tho wavelength X. lo Ilie circular fr o 2ne
, a
2nc A.
, .
Jic
quency to =
ami Aw = — j-t -AX, wo t?°l '' ~ /vaco*
The Zeeman splitting is
2mc
Finally,
,
2nmc2
“
ft
5
M e li
10'2 cm.
5.27. The Zeeman splitting should exceed the Doppler broa­
dening both because of the thermal motion of the atoms
3/2
1 /2
>/2
1/2
- 1 /2
-
-3/2
Vi
-
1/2
<■
—
- 1 /2
1/2
1 /2
F ig . 74
(ninin)y hydrogen atoms) and because of tho star's rotation,
viz.
Ao)
o)
2|t|t/?: > / Ao) \
ha i0
j
/_A® \
""V u> / Uicrin
ulherm ■^rnt
\ o> / roi
cc
Finally,
0.18 T.
5.28. The splitting in a magnetic field,
A£X»
AX. « 2ncft
>0.3 A,
where AE « \lbB « 2 X 10" 4 eV. Since AX. is much less
than X, — X2l the effect is anomalous. One line of the doublet
will split into six components, the other into four compo­
nents (Fig. 74).
5.29. The difference in energy is duo to the rotation of the
electron’
s spin in the magnetic Hold producod by tho orbital
motion of the electron. Therefore AE = 2po
whence
n_
AE _
2pn
148
ftAci) _ 2nftc AX
2X*nn
0.3 T.
5.30. The radius of Ll»o second Bohr orbit of positronium is
_
\/»
me* —8cd,
where m = mJ2 is the reduced mass for positronium, and rD
the radius of the first orbit of a hydrogon atom. The projec­
tions of the magnetic moment of electron and that of a posi­
tron are equal to the Bohr magneton pn. Tho interaction
energy is approximately
E
lli»
(8rB)3
10-fl cV.
The spins of the states are 0 and 1, respectively.
5.31. The energy of hyperfine splitting is
AE
I'pMc
10-* oV,
where jip and pe are the magnetic moments of proton and
electron, respectively, and rn = me- is the Bohr orbit
radius at n = 2. The uncertainty in the energy due to the
natural broadening of spectral line, 6E « h!A* « 10" 7 eV.
Since AE « 0.16E, the splitting is undetectable.
5.32. The onergy change during spin rooriontation is
eV,
rii
whero rB is the radius of the ground stnto of a hydrogon atom.
Finally, X = hc/AE « t m.
5.33. The dipole electric moment is p --= 2jihclXE, tho di­
pole size I = pie « 10 -s cm.
5.34. The wave function of the lowest quantum level is
spherically symmetric. Writing tho Schrodingor equation
in spherical coordinates, viz.
and making tho substitution 'K (r) = u (r)/r, wc find that
h* d*u
2m dr* — Eu.
This equation must bo solved with the boundary condition
u (R) = 0 , which corresponds to an infinitely high wall.
In addition, the wave function Hr must bo finite overywhero,
thorcforo u (0) = 0. Thoso conditions are satisfied by the
functions
Mrnf(r) co-L sin ( n j
nl
The lowost level corresponds to n — 1, i.o.
2mrt* ‘
Summing the electron enorgy and the free energy of the sur­
face, E tUrt = 4jio/?1, we get the total energy of the system.
Finding the minimum with respect to R yields the equilib­
rium radius of the cavity:
»
I nh' \ ,/4
20A.
5.35. The potential energy of an atom in a non-uniform field
U (r) is — y aEa (r) and therefore the forco acting per unit
volume is
F = -i-ArAa-£.VE*(r).
where \i is the molar mass. In our problem this force is dir.
rected radially to centre. By putting down the conditions
for the liquid equilibrium we can easily ensure that dp/dr =
F. Integrating this equation with the boundary condition
for infinity, p (oo) = 0 , we get
p(r) = 4 -jVA«-£.E*(r).
It follows from the a value given in this relationship that
the permittivity e « 1, therefore E1 = sVr4. The boundary
condition on the ball surface p (R) = p s thus yields
VI. Radiation
6 .1 .
T* = y
71;, where r is the distance from
the disc centre projection. Lambert’
s law was used to solve
this problem.
c o
150
M'corp
W'r.d
2n (me - f ntp) v*
a*or‘
~
6 X 10-® .
oscillators within a given frequency range is calculated in
the same way that the Rnylcigh-Jcans radiation formula is
derived, hence
wit x ASp, (to) = /IS
,
where A is a constant.
In the cavity, the actual frequency range contains one nor­
mal oscillation with width —
Therefore, if only the
width of molecular levels T is greater than o/Q, the density p0
is of the order of
Because of the change in the spon­
taneous lifetime, the width V in the cavity may be sig­
nificantly different from the width in a vacuum. However, it
is essential that the amplitude in resonance and the width
both vary simultaneously, as in a conventional damped os­
cillator, viz. in such a way that the area under the absorption
curve remains unchanged. Therefore
!t>sp oc /IS -5£- and
6.9.
"tot
= -L f 1- ln**~ -) 1» 0.505.
2 L
2aL
)
VII. Solid State
7.1. According to the data in tho problem, tho wave vectors
of the incident and scattered X-rays are k — /c (1, 0 , 0 )
and k' = 1/3 k( 1, 2, 2), respectively, where k — 2jiv/c.
Therefore, the wave vector change duo to the scattering is
k ' - k = 4 - * ( - i . i, i).
On the other hand, in Bragg scattering
k' — k = 2jib,
where b is the integer vector of tho reciprocal lattice. In a
cubic facc-contcrcd lattice tho basis vectors of tho diroct
lattice are usually chosen as nt = y = (1, 1 , a 0 ) and analo­
gously a 2 and n3, where a is the interatomic distance. Thus,
the basis vectors of the reciprocal lattice are constructed
according to the typo bj —
(1»
!)• Whcnco it can
be seen that in first order interference the vector b should
152
simply be one of the minimum vectors bj and lienee
Finally,
a=
2 / 2v
= 2.9 x 10-# cm.
7.2. By solving tho equations of the natural oscillations of
the chain wo get two frequency branches:
On the lower branch, for | ka\ < 1 the frequency col oc k
and hence this branch is acoustic. On the upper branch,
<oj (k = 0) = 2yl\i and the branch is called optical. A gap
exists between the branches, since <oa (ji/2 a ) > a>i (ji/2 a).
At Mx = Mt the frequencies
toJ 2 (fc) = -jf (1 ±c°s *“)
and the gap vanishes. Thus the spatial poriod of the chain
is halved giving rise to a single branch with
co2 = -jj- (1 — cos ka)
in twice the range of wave numbers, | ka |^ ji.
7.3. By solving tho equations of forced oscillations of tho
chain and calculating dipolo moment, wo can lind tho pola­
rizability
The polarizability diverges at tho frequencies of long-wave
optical oscillations, e> = (2y/|x),/2.
7.4. The laws of conservation of energy and momentum in
neutron scattering are
k = k/+ q + 2Jib,
where the upper sign corresponds to the emission and the
lower one to the absorption of a phonon. In the second ex­
pression, b is an arbitrary integer vector of the reciprocal
lattice. By measuring k and k\ it ispossibloto dotormino to (q)
directly from the first expression and q from the second one.
153
Hence the vector b can be dotorrainod if q bolongs to the
first Brillouin zone. This enablos the function e> (q) to be
reconstructed.
7.5. Using the general oscillntion oqualion
we can calculate U (x) in the harmonic approximation.
If we take into account that tho change in U during the dis­
placement of a nucleus is related to the interaction between
the electron shells, then we have to estimate U (i) oc
2s at (x/a)3. Whence <o3 oc 2?at/Ma3 and since 2?a^ oc
hVma3, we finally get
hocK{ ^ ) m-E^
The estimate for x 0 is obtained immediately if we allow
that U (x0) oc fio). Thus, the smallness of the amplitudes
(x0 <ti a), which is a necessary condition for the existence
of molecules and crystals, is the result of the smallness of
the “
adiabatic parameter”, (m/M)1'3 oc 10~3.
7.6. C = ( - ^ - 4 ”jr) ( - jrY k V ,
where
V is the crystal
volume.
7.7. At low temperatures only tho phonons with momenta
q ^ g0 — kTlhs are excited. The volume of the region in
9-space corresponding to these phonons is oc9J in the twodimensional case and ocq0 in tho ono-dimensional case.
Therefore the thermal energies are U oc T3 and T3, respec­
tively, while the heat capacities are proportional to T%
and 7\ respectively.
7.8. By putting down the oscillation equation for an anharraonic oscillator,
Mx = — yx + -i- 6x2,
it is possible, as in problem 7.5, to estimate y ~ £ at/a*
and 6 — 2sat/a3. By averaging this equation over an oscilla­
tion period (0 < t < T) we have, by definition,
<0 )1= 0 hence x = (6/2y) 1?. However, given weak anharmonicity,
we can take x* in the harmonic approximation, i.e. yx*/2 =
154
kTl2. Therefore
x=A
’
2y* * 7
and the expanBion coefficient
bk
2y*a
k
~
10"* K"!
Acat
is independent of the temperature within the whole of the
classical region, i.e. for which kT » fia.
7.9. S = R In (2/ + 1) so tliol S H«= /? In 2 , 5Ar =
2Jf In 2. As usual, the entropy of a fully ordered state
is chosen as the origin.
7 10 y «= d ^nn — 53p “
— (3n*)»
3________
« 1.5 X 10"6 atm'
7.10.
/» T O T *
7 .11. i> t = •
£*■= 2.3x10’cm/s,
u,, =
= 0.
7.12. i>r—
= 105 x ,0$ cm/s* “ = ^ cos0- To
solve the problem we use the equation of motion v =
dE (p)/dp, p = F.
7.13. The potential distribu­
tion inside a conductor is gov­
erned by Poisson’
s oquation:
A<p = 4jis (n — n0)/e,
where e is the permittivity
due to the electrons from the
filled bands (viz. all eloctrons
except the conduction-ones).'
In thermodynamic equilib­
Fig. 75
rium, the chemical potential
level £ of electrons is constant throughout the crystal.
However, because of the electric field the bottom of the
conduction band, which simply corresponds to the poten­
tial energy of an electron, changes from point to point (Fig.
75) giving Ep = E? (x). If the band bending is small com­
pared to £P, then
dn
n —n0
dEP
By oombining thoso equations we get
4ne*
dn
e<p.
The solution of the latter equation is
q>oc exp (± - ^ ) .
whore
lr r = (
)
1/2
Given the quadratic dispersion law: E P oc p \% n oz p\
(pp is the Fermi momentum of electrons), and therefore
2
3
dEy
dn
hence,
/Tf
~
Ey
n ’
I eEP \ 1/2
Un^n]
Given the indicated parameter values, ZTP « 4 X 10”
® cm
for metals and 1.5 x 10"® cm for semiconductors. In the
first case the value of ZT f demonstrates that an external
field is practically fully screened within even one lattice
constant and does not penetrate the metal.
7.14. Using the formula for an electron heat capacity ce, =
k2T
y - ^ - n and the formula fora lattice heat capacity (see
Problem 7.6) wo get
Using the estimate for s from Problem 7.5 and assuming
Ey ~ E ai wo get
whence T ~ 10 K.
7.15. T ~ B, where 0 ~
IT is 1^IC Deljye temper­
ature of the crystal. The problem has to be solved in the
same way as 7.14.
7.16. According to Maxwell’
s equations, the condition
H = B = 0 results in
cu rlE = 0,
D-j-4nj = 0.
It follows from the first relation that the electric field of
such a wave is longitudinal, viz. the amplitude E 0 in the
equation
E (r, 0 = E 0 exp [i (kr — coZ)]
is parallel to k. The vectors D and j are related to E by tho
material equations D = eE and j = oE, where e is the
156
permittivity not related to the conduction electrons, while a
is the conductivity. For hand electrons, m*v -= eE, where
v is tho average velocity of tho electron flux in an external
field. Whence
m*o) E0;
j0 = en\0
substituting this into D 4* 4jij = 0 yields the frequency
of tho natural oscillations, viz.
a)
A n e i /t
em *
For metals, (Dp ~ 2 x 10ia s"1, hfap ~ 10 eV.
7.17. In order to introduce a complex permittivity e (to)
the Maxwell equations need to be rewritten as follows
c u r lH = —
E + ^2j
c
e
e
E.
Then, in accordance with the material equations (see Prob­
lem 7.16),
e(a) = e-!-i
4no
(i)
By introducing a "friction”into the equation of electron
motion to doscribo the finilo moan free time, viz.
m*v = — ^
v-f-eE,
we get
g <*)~ »«<!-«■») •
Starting from the experimental values of the static electroconductivity of metals we can easily estimate that within
the optical frequency range (oi» 1, and therefore
e (o>) = e (1 — w*/o)a).
When a < (i)p we have e (w) < 0, viz. total (“
m etallic”
)
reflection takes place. In the ultraviolet range, for which
<d > (0P (see Problem 7.16), e (a>) > 0 and transmission
is much higher.
7.18. At moderate frequencies (cox C l ) the conductivity a
is real, and so the conduction current significantly exceeds
the displacement current (for q>£t / ci> » 1 see Problem 7.17).
157
Then Ilie complex wnve number is
k -
where 6 = ^ L _
— /
e (®) =
»
is Lho penetration depth.
7.19. The equation of an electron's motion is
m*v = e E + y V x H ~ Y v .
H||s,
E j_ n .
We need only search for the mean drift velocity v and leave
out the rotation around the H direction, which does not
contribute to the average flux. Hence we assume v = 0.
Then
Ey —^clEx
j n
7x = <T H ^W cT )*
*
7w = <7
l + (WcO*
’ 7f = U’
whero a =
is the conductivity at // = 0, o)e =
eHIm+c the cyclotron frequency.
At E g = 0 the longitudinal current is
i —
*
,x~ l+lwc*)*
F
*
and the way it falls with increasing H defines the magneto­
resistance. The transverse current,
f
__
—
<0CT
..
is the Hall current. These conditions are observed with a
Corbino disc.
The current in the plate and along the normal is absent
(Jy = 0), therefore the Hall field is E y = (i> ct E x , and j x =
aEx. It is necessary to stress the total absence of magne­
toresistance, the reason being that the current not only flows
due to the external field E x%but also due to the Hall field E~.
In strong fields ((»ct » 1) the second contribution predomi­
nates.
7.20.
=
eVlB, 0 (E, T) = 3, (T)
7.21. The spread of charge is described by the continuity ^
equation, p -I- div j = 0, and the material equation, j =
oE. In good conductors, the conductivity change asso­
ciated with excess carriers forming a space charge is negli­
gibly small compared to equilibrium conductivity, there­
fore o = const. Since both D = eE, where e is the permit158
tivity associated with tho oloclrons of filled bands, and
•div D = 4np, we finally get
The solution of this equation is
P(r. t)“Po(lr) < r",“.
where tm = e/4jio and p0 (r) is the charge distribution at
t = 0. It is clear from tbe solution that the density p, having
not been deformed, falls exponentially in time so that those
spatial regions in which the charge was absent at t = 0
remain forever neutral. Therefore the current j flows, natu­
rally, throughout the entire space. For germanium, t m «
10~I# s.
7. 2. In contrast with Problem 7.13, the equilibrium elec­
tron distribution is given by n (r) = n0e*l,<r>^r , where
U (r) = — 89 (r) and n0 is the electron concentration out­
side the space charge region. The field weakness means that
189 | < kT. Thus Poisson’
s equation reduces to
Its solutions are
where
At the given parameter values, iD ^ 5 x 10~* cm.
7.23~ By solving the equation of motion
7.24.
.
where
»e = ^ - ,
Oq= - —»
■, and n is the concentration.
[7.25. It follows from px = eE that
159
whence, by allowing for * =* vx and vx — d$ldpx,
*(*) = * . + -af co
Thus, Ihe electron oscillates around an equilibrium position
x0 with a frequency (db = eEalh and the average current
vanishes. The reason is that the kinetic energy $0 cos (k^a)
is a bounded function of /cx. As a result, the motion of an
electron with a potential energy U (x) = —eEx is limited
to a finite region (due to the law of conservation of energy).
7.2(i. In the equation of an electron's motion,
p x =- hkx — eE — yvx = e E + ya%0sin (kxa),
the variables are separated and this equation can immedi­
ately be integrated via elementary functions. The form of
the solution differs depending on the value of the parameter
T — ya %JeE.
(a) r > 1. The solution has the following form:
cos {(M—<P)/2) _
f _ i / p a _ 4 eEa (f *
sin ((*,«+ ip)/2) ® P I V 1
1
n "
*«/ ’
<p= arc sin -|r (0 <C (p < n/2).
As t -* oo the solution becomes T sin k^a = —1, which
corresponds to stationary values of kx and vx. This solution
r.sin he obtained directly by assuming that in tho initial
equation kx = 0, and so ux — eE/y and a = ePnJy. For
r » 1 we find that k^fi
1, viz. the electron remains near
the bottom of the band. It is therefore possible, as usual
(e.g. Problem 7.19), to set y — m *h and then o takes the
standard form, a = e^nxlm*.
(b) T < 1. The solution has the form:
U n -*§*- =
'-%£■(<-<,)}- r .
It follows from this expression that as t rises so does the
wave number kx\ the oscillations being superimposed there­
on. The tangent on the right-hand side goes to infinity for
the time intervals M — h/eEa Y 1 — P . For each such a
lime interval the phase of the argument in tho left-hand
side tangent is augmented by n and hence kx is increased
by &hx = 2nJa. Integrating tho initial equation of motion
100
over At we find that
HAkx = eEAt — y j vx dt = eE A t—yAx,
where Ax is the electron displacement in time At. The ratio
AxIAt is the average drift velocity while the average current
density is
/=—
In strong fields, when f < 1, we get
Y
e*nT *E
~
2y
™
E
’
viz. the final current only arises due to the M
friction”force.
This current increases with decreasing mean free time.
7.27. According to the problem, the electron is slow, viz.
ka and k'a < 1. where k and k' are the electron wave vec­
tors before and after a scattering and a is the lattice constant.
Analogously, for long-wave phonons qa <C 1 as well. There­
fore the law of conservation of momentum can be written
as k = k' ± q (see the solution of Problem 7.4, where b
must be zero due to the smallness of k, k\ and q). For the
angle 6 between p and q we get
co se = - f± £ .
where the upper sign pertains to omission, and the Jower
one to the absorption of a phonon.
7.28. a ~ - £ ^ L » iS / , r . .
7.29. By solving this problem as wo did for Problem 7.21
we get for the relaxation time
™ = - T w l E°W}'
If
> 0, the fluctuation is damped and this occurs for
E < So- It E > E 0 the fluctuation gets worse over time.
7.30. The volume V in conventional r-space and volume V*
in a wave-vector space contain VVk/(2n)s states. Since for
nondegenerate electrons tho occupation numbers n (?) =
e iP
»=2
kT* )» ^ en the electron concentration is
J
1/, 11-0408
d*k
(2nmkT)9/*
« P ^= 9 exp^.
h*
161
where £ is calculated hum llit* hoi tom of llio electron band
(£ < 0). The quantity Q is called the band’
s statistical fac­
tor.
7.31. According to Problem 7.30, the electron concentration
at an arbitrary £ is n = Qne^thT. In an analogous way,
the hole concentration is
n
/
2A + £\
p = <?„ exp(---- jjr-) ,
where Qp is the band's statistical factor for holes and 2A
the forbidden gap width. Hence
n p = Q nQ Pe ~ 2* hT
is independent of £ and hence, of the impurity concentration
too. Yet in an intrinsic semiconductor n — p = nt due to
electric neutrality. Therefore, in the general case np = n\.
7.32. At moderate tem­
peratures the electrons
are only exchanged be­
tween donors and the
electron band. For elec­
tric neutrality we must
have
Qn*/kT
= A,
[where N is the donor
concentration. To real­
ize this condition it is
sufficient to assess the number of electrons in the band
and that in the donors. Thus a quadratic equation for
et'kT is obtained, which is easily solved. At low temper­
atures, such as /i < iV, we have n = Y N Qne~*'tkT while
at higher temperatures n « N all the impurities are ionized
(impurity depletion region). The hole concentration p in
terms of n is np = n\ (see Problem 7.31). At higher tem­
peratures the excitation of the electrons from a filled band
predominates when n « p « nt « Y Q nQpe~*/hT, this taking
place at nt (T)
N. The temperature dependence of n
and p is depicted in Fig. 76.
162
7.33. Ry* =
,
a* =
, M = m* -f mj,
where m
is the reduced mass:
Ry* - 0.7 x 10-2 eV, a* ~ 10"# cm.
7.34. The diffusion hole flux is j — —
an(l tlie conti­
nuity equation for the hole Dux is
here pH is the number of recombinations per unit time
per unit volume. These relations suggest the diffusion equa­
tion
which has the solutions p oc exp (±xlL), whore L = V D t
is the diffusion displacement length. By applying the Ein­
stein relation eD = ATp, the parameter values given yield
L w 0.2 cm.
7.35. In superconductors B « H and the displacement cur­
rent can be neglected as compared with the current of the
superconducting electrons. Therefore the Maxwell equa­
tions tako tho form:
c u r l E = — j-H,
(1)
curlH = ^ j .
(2)
The equation of electron motion tnx — eE in conjunction
with j = enx yields E = ~
Substituting these into
(1) we get curl j = —
an equation for H, i.e.
El imi nating j using (2) yields
curl curl H — — 2-,
Xa
X=
V
4nean
= —.
o>p
By transforming the left-hand side and taking into account
that div H = 0 we finally get
ii*
163
For a flat metal surface this has the solution H oc exp (:fcx/X)»
viz. the magnetic, field decays exponentially with depth
into the superconductor. Since tho external field's frequency
does not enter X, the result shows that the magnetic field
is constant. The depth asked for is X — 10~M0~# cm.
7.36. The 9He nucleus contains three nucleons and its spin
is s = 1/2. Since the spin of the electron shell is zero, an
3He atom is a fermion. Using the formulas for a degenerato
Fermi-gas we get for the heat capacity per mole,
where the Fermi-energy is E? = ^
(3w ^ a j 2/3; herc yj/
is the atomic mass and N A is Avogadro’
s number. By de­
fining 0 so that C = RT/& we find that 0 ~ 1 K.
VIII. Nuclear Physics
8.1. The particle mass in this experiment is E cms which cor­
responds to the position of the maximum. The half-width
of the resonance curve at half the height is A « 2 MeV.
The lifetime can be found from the relationship z « ^ «
3 X 10"22 s. To find the mass in another experiment, lot
us use the relativistic invariance of the squared four-mo­
mentum:
mlc« = (E, + E t)J - (Pl +
= 4ET- - 3pV.
since by definition E x = E z = E and consequently px =
p 9 — p , with cos (p,, p2) = cos (p = 1/2. Since e+ and e~
are ultrarelativistic particles (v « c), we have E « pc and
me2 « 3.1 GeV.
8.2. A proton consists of three quarks, q f,qPq n > The spins
of two are parallel while the spin of the third is antiparallel
to the other two. Q-hyperon has strangeness S = —3 and
can only be composed of the three identical quarks, <7a7a(7a*
whose spins are parallel (the Q-hyperon’
s spin is 3/2 K).
The ji+-meson and /f+-meson consist of tho quarks q p q n
and qpqA> and /-meson of the quarks qcqc. The ji- and Krnesons both have a quark and an anliquark with antiparallel spins, and the /-meson has a quark and an antiquark
with parallel spins. The composition of tho Q-hyporon exem­
plifies the fact that in the simplest quark model the quarks
do not obey Pauli’
s principle.
164
8.3. T = Me2 — (m~ -|- 3mn) c2 = 714 MeV, this account­
ing for the fraction 6 = TIMe1 — 0.30. Here nu. and mn
are the A-parlicle and pion masses, respectively.
8.4. The total energy reloased in an annihilation is E =
2Mpc2. The jx°-meson energy is at maximum if the momenta
of all three pions are collinear with pn* = p*-. Thus
En+= En- and Pn* = 2pn±.
(If !/?„ ♦=/= p„ -, thon the
centre of mass moves
aloWer than in the case of equal momenta, since a fraction
of energy is expended for the pions to move relative to each
othqr.) The law of conservation of energy is
*
,
2Mpc2= En.+ 2 E k±= E„ .+ 2 V
and using (1) we get
tn •
0.92 GeV.
8.5. The number of interactions in a volume dV in time dt
is
dt]
0 U12P1P2 dV dt.
whero uia = | ut — u2 |, and the densities of the particles
are px = p8 = n/c (n is the flux, c the velocity of light).
In our case the difference in the velocities of particles,
ult = 2r, since
= —u2 and | uf ( = | u, | « c. The
number of counts (allowing for the detection efficiency,
e = 0.1) is:
K= eQ
= 2e<r —
dV.
dt
e
Taking into account that dV = SI we find that
and, finally, the current in the storage ring is
8.6. Emission starts at a particle velocity v — chi. The
appropriate kinetic energy is
f ] - ■ [ 7 3 = 1 - * ] ~ 4-8 GeV165
’
Die refractive index n will l>o found from ir — 1 H- 'iliaNt
where ;V is tlie number of atoms per unit volume, niul a the
polarizability of the nilroueu molecules. Since nt a con­
stant temperature N oc p, viz. .V is proportional to pres­
sure, and n9 — 1 ~ 2 (n — 1), wo must hnve
M— 1
,_P_
»
«
o— * ~
Po
where p 0 — 1 atm.
8.7. Itadiation from kaons in the counter only appears when
— > // > *“. where rK ami r:l arc the velocities of kaons
and pions that hnve momentum p and total energy E. Since
P
itc = pc!E, we lin«l that
where mKc9 and mncs are the kaon and pion masses, respec­
tively.
8.8. The path-length of about 70% pions bcfoie they decay
is (v « r):
L w c t 0 mnC" « 600 m.
8.9. Tp = 300 MoV.
8.10. The number of pions reaching the detector is n =
where :V is the number of all the pions generated,
ft the solid angle oT the detector from the point of pion gener­
ation, ft - S/ir, I a pious Right lime along path /,, and T
the pion lifetime in the laboratory frame,
X
T„
^-5- » T0 ( 1 \--- ~
0 m„ c*
0 \
) ,
r mne* J
where T is the kinetic energy. 1'inally, we got
.V = n -£exp (-%- -7
S
*
V ct„
/
) « 1.4x10* particle/s.
T‘
+2m,c-T I
8.11. The decrease in the number of protons due to the in­
teraction in the nitrogen layer dx (g/cma) is
—dN = N
,
where A is the mean free path prior to the interaction.
If x0is the thicknessof Earth’
s almosphcro (x0 « 1000g/cm8).
166
then the solution of the above oqualion yields
N (* = *») = /V,(x = 0)oxp ( ~
•
Here A = A/oN a , where A'a is Avogadro’
s number, A the
relative atomic mass of nitrogen, o - ji/?3, and R the radius
of a nitrogen nucleus. Whence,
N (x = x0)/N (x = 0) = 7 X 10"°.
8.12. Tho number of $Ar atoms is
n -dJo/Aci,
where t = 1 year = 3.2 X 107s, Nci the number of chlo­
rine atoms, and O =
the neutrino Dux on the Earth.
Whence
A n B 'n
* c i = Nat
and finally we find that tho experiment needs carbon tetra­
chloride in amount of
M .= J^£L J L = nNc i « 560 1,
where u is the molar mass of carbon tetrachloride.
8.13. To begin with, let us determine how much energy
could be released in tho centre of mass system (c.rn.s.). By
definition, tho system’
s volociiy is
P.= 2 P.C/S E„
wl^ere p t and E% are the momenta and energies of the protons
in the laboratory system prior to tho reaction. In our cose
p.
y
p c ______
c*
0.82.
Whence the energy of each of the colliding protons in the
c.m.s. (the energies are equal since the momenta and masses
of colliding particles are equal) is
„
m pca
E = - __ = . ==\m.,c2-- 1.8 GeV.
/ i - p*
'
167
The lotnl energy in llic c.m.s. Lltnl can he expendod to pro­
duce new particles is
2E — 2mpc* w 1.6 GeV,
since both protons (or other baryons) remain after the re­
action because of the conservation of the baryon number.
Most pious arc produced when all these particles are at rest
in the c.m.s. while nucleons remain to be baryons. There­
fore we finally have
2E— 2mpc*
11.
mne%
8.14. We find the muon's kinetic energy from the laws of
conservation of energy and momentum:
1 »”
2mKc*
*
The lifetime is
T= -
E
x0= 5.4 x 10”
* s.
muci
8.15. The angular momentum received by the disc is equal
to the total moment of tho absorbed particles, i.e.
,
1 Nh
L = ~T ~2'
The coefficient 1/4 arises because of the allowance for the
angles of incidence of tho electrons onto the disc:
—- =
n/2
n/2
0
o
j cos 0dQ = — y j cos0*dcos0.
Tho disc is rotated by an angle
q, = A = ^ L « 5 x 10”
» rad.
8.16. As the magnetization proceeds, the number of spins
of fluorine nuclei directed along the field Nx and against
the field N %is different and their difference is
A/V-JV,[l - e x p [ - * & ) ] =
= 4‘
2
#
=
^ < 1*
(where ji is the magnetic moment of fluorino nuclei, k is
Boltzmann’
s constant, and N 0 the numbor of fluorino nuclei).
When the field is switched off half the oxcoss nuclei AA72
will change orientation and the samplo will receive an ang­
ular momentum of
=
ig-N', N'-2Na,
where N a is Avogadro's number. Whenco
|i = L
-1.31 x 10-“ J/T = 2.62 ,iN.
8.17. The dependence of half-life on cnor£y (in MeV) is given
by Geiger-Nutlal rule
^i/i« C 4*
The constants C and D will be found from tho data in the
problem: D = 149 and C = —57. The auswer is 7\/a ~ 10*4
years.
8.18. To begin with, we find the ot-parliclc rouge in alu­
minium, R ai. Since the energy ionization losses are propor­
tional to N Z%where N is the number of atoms per cubic cen­
timetre and Z the nucleus charge, we have
/*a»
(A'Zlalr
tfolr "
W )A \
*
The number of nuclear reactions occurring in a thin layor
dx is d\ = naN ai o dx = np, and dx « R ai . Finally, we
get
njp_
a;^Al
= 4 X 10~2 barn.
na (A^Z)air/7nir
"a^Al^Al
8.19. The number of uranium fission ovonls is n — N/E «
3.1 X 10+1* s “
l, where E is tho avorago total onorgy re­
leased per uranium fission event. The antineutrino flux be­
yond the shielding is
<o«
^ “
4^Ii" ==6>< 1010 O™"2,8'1The fraction of ho energy carried away by tho antinoutrino
flux is .
<SvX"v>
« 0.05.
E
12—0408
169
8.20. Since Y-qunntn are scattered at 180°, the ji°-meson
is at rest in the laboratory system. Therefore
pH = Pd = P ond En + niJtcl = Ed + mncl,
or
V P2c2 + m2c4-f rripC2 = V P*c2 4- m^c4-f ra^2.
Neglecting the deutcrou’
s binding energy and the difference
between the proton's and the neutron's masses, viz. assum­
ing mn — m;i = m, md = 2m, mn — \i, wo get
-I- m2c4-}-me2= V p2c2-f-4m2c4-f pc2,
Tn = En— iic1= * " ^ - 1 0 ^ ~ mC' ” 300 MeVThe nonrelntivistic approach yields
whence Tn « 2pc2 = 270 MeV, this being less accurate!
8.21. N (f) is tho number of neutrons in the vessel at lime t.
Then the number of neutron collisions per unit area of the
vessel's internal surface is vNlAV, Tho number of neutrons
in the vessel falls for two reasons, viz. decay and leakage
through the foil. This decrease is
- d N
= N ^ - + -y- $ S d t .
The solution of this equation yields
N * N0exp [ — (1/T)+vS/kyj-i] .
Using the data in the problem,
whence
( 1 /t +
uS/4 V )- 1 =
t /2 ,
S . ^ » i0 . 8 mm*.
8.22. The binding energy of nucleus, viz. that from Weizsacker’
s formula is
U — 1\.0A— 13.0i42/*—0.584 -%rjT + smaller terms.
170
Tho. nuclear radius is known to be /?n = rn/l ,/** where
r, « 1.25 X 10"13 cm. Therefore
U (MeV)* 14.0/1 - -£ £ S - 0 . 5 8 4 + •••
•
where 5 is the nucleus's surface area. The coefficient of
surfaco tension is
0==|^.|« ^L2. MeV/cm*« 1.1 X 10IT J/m*.
8.23. The ultimoto volocity v0 is determined from tho con­
dition
= sin i « 3 x 10~3,
yth
where ut|, =
Neutrons for which v < u0 will be
trapped within the cavity. Their relative number is
3
o
___
3 / n \*Hh/
8.24. The lifetime is
x —
th
— -—
nav
2 X 10‘
»
.
1;7xl0-4 s,
where n is the hydrogen concentration and utb the neutron's
thermal velocity (see the solution of Problem 8.23).
8.25. The total number of spin stales for a system of two
particles with spins I and i — 1/2 (in h units) is
(2f + 1) (2/ + 1) or 2 (2/ + 1).
The number of states with the total spin /g = / -f 1/2 is
2/g + 1 = 2 (/ + 1), and the number of states with /g =
I — 1/2 is 21. Thus the scattering probability with paral­
lel orientations of I and i is
and that with the
antiparallel one is ^TfT ~ T'
8.26. From the laws of conservation of energy and momen12•
171
This pagination is wrong, it should be 172-73
turn,
EV= TN+ Tn- Q , - ^ = Pn + P„ .
where
Q = (mcl),L) — [(me*)«L1 f (me3),,],
we find Ihe required quantities
Th = pfc/2wN « 1 MeV and Tn = p\!2mn « 6 MeV.
8.27. Since the binding energy of nucleons is low, it is pos­
sible when estimating to assume that the longitudinal
momentum of the nucleons is p x « -jL. The transverse mo­
mentum of the nucleons insido the nucleus is due to the
Fermi-motion of the nucleons. The Fermi-momentum will
be estimated from the uncertainty relation
p ¥ ~ Ap ~
~ 100 MeV.
The estimate for the unglo of scattering of the nucleons
will be obtained from the relation
cp « — « 0.2 rad.
Pi
8.28. At the entry to the neutronguide the neutrons for which
v < ib will have a Maxwellian distribution of velocities, while
those for which v > n, will bo absent. Owing to the elastic­
ity of the scattering on the walls and ncglocling the flux
onto the detector, the neutrons will have an equilibrium
distribution of
dN{v, z) oc exp
vVdV
given that
E=-— - + m g z < E l,
El = mvll2.
Thus there will be no neutrons with E > El. Here z is the
hoighl up the neutronguide. For ultracold neutrons, the
exponent can bo changed for unity and tho neutron flux is
oc v dN oc ir* dv oc E dE.
The total flux at a height z is
O (z) = j
oc J E dE oc El~(m gz)2.
m gt
This flux vanishes at h = zmax
h = EDnig = ED2g w 5 in.
174
APPENDICES
I. Constants and Tables
1. Fundamental physical constants
SI
Velocity of light in vacuum,
c « 3x10® m/s
Elementary charge,
e=1.6xl0-1# C
Maas of a hydrogen atom,
mH= 1.67xl0-« kg
Boltzmann's constant,
&= 1.38x10-** J/K
Planck's constant,
h = h/2n = t .05Xi0“
34 J-s
Universal gas constant,
if=8.31 J/(K*mol)
Gravitational constant,
Y= 6.67x 10“u N.m*/kg*
Avogadro’
s number,
Loschmidt’
s number,
cos
3xl010 cm/s
4.8X10"10 CGSE
1.67X10g
1.38X10-“ erg/K
1.05X10-*7 e rg s
8.31x10’erg/(K-m)
6.67X10-8 dyn*cm*/g*
1V x= 6.02x10** mdl'1
n0= 2.69xlOl# cm-3
2. Electric, atomic, and nuclear constants
SI
Electric constant,
e0= 8.85xlO~1* F/m
Magnetic constant,
ji0= 1.26x10“
* H/m
Stefan-Boltzmann constant,
0 = 5.67x10-* W/(m*-K4)
CGS
5.67x10-* org/(s-cm**K4)
175
2. (Continued)
CGS
SI
a = e * f h c «1/137
meef = 0.511 MeV
mpc* = 938 MeV
m& c%= 1876 MeV
mae * = 3726 MeV
F in e s t r u c t u r e con sta n t,
E l e c t r o n r e s t m a ss,
P r o to n re st m a ss,
D e u tero n r e st m a ss,
o - p a r t ic le a t r e st m ass,
Rydberg's constant for an infinitely
heavy mass, 7?*=10 973 731 n r 1
Ionization potential of hydrogen atom,
I
| 109737,31 cm"1
7= 13.6 eV
B o h r r a d iu s
rB=0.53xl0 -10 m
0.53x10-* cm
B o h r m a gn eton ,
|iB=
=9.27X10-“ I/T
0.927X10-*® erg/G
= 5 °SX1»-" J/T
5.05x10"“ erg/G
N u cle a r m a gn eton ,
(*N= -2—
C la s s ic a l r a d iu s o f th e e le ctr o n ,
r0= -— 5- =2.82xl0~1* m
A to m ic m a ss u n it,
10-“ k g
2.82X10-1* cm
10-“ g
C o m p to n w a v e le n g th o f th e ele ctro n ,
Ae= -JL. = 3.86xl0-‘
»m
TUf/C
3.86X10-11 cm
C o m p to n w a v e le n g th o f th e p io n
A„ =
=1.4xl0-« m
1.4X10"1* cm
C o m p to n w a v e le n g th o f t h e n u cle on ,
ft
An= ■
m^e-=0.2XlO-» m
0.2x10-1* cm
Nuclear time tn = Nn /c «0.5x10“
“ s, where 7?n is the range of
the nuclear force
The temperature 293.4 K = 20.4 °C corresponds to
0.0253 eV
At an energy o
f
0.0253 eV a neutron has a velocity
p0= 2200 m/s
Wavelength of a neutron, Xa=0.287/%i/* (g in eV, Xp in
176
cm)-
3. Some astronomical constants
Sun's mass
Jfs = 1.99x10” g
Energy radiated by the Sun per second (luminous omittance)
Ls =3.86x10” W = 3.86x10” erg/s
Sun’
s radius
J?s=6.96x10” cm
Angular radius of the Sun at the average distance from the Earth
o r =4.65xl0"3 rad
Sun's temperature near its surface
f s =5.5X103 K
Earth's mass
AfE= 5.98x10” kg
Earth's radius (at the equator)
JlE=6.38x10’cm
Earth's temperature (average)
r E=300 K
Average orbital velocity of the Earth
pe = 3 x !0* cm/s
II. Some Non-System Units
SI
Units of length:
light year
parsec (pa).
angstrom (A)
fermi
COS
9.5x10” m
3.1X10” m
10~” m
10-” m
9.5X10” cm
3.1x10” cm
10"’cm
10-” cm
Unit of area:
barn
10-” m*
10-” cm*
Unit of time:
year
3.16X107 a
Unit of energy:
electron-volt, 1 eV = 1.6xl0-1® J = 1.6xl0-ls erg
177
III. Units of Some Physical Quantities in the SI and CGS Systems
and Their Relationships
of coulomb
uantity
electric fty*
electric charge
a
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IV. Elementary Particle Table
&
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* An antiparticle has the same rallies of mass, lifetime, spin, and laoapln T as the haslo particle hut Its values of elec;, letpon, and baryon charges, Its lsospln prelection Tz , and strangeness B are all opposite In sign.
** The symbols of corresponding antlpartlcles are Indicated on the right.________________________________________________
IV. (C on tin u ed >
V. Prefixes and Powers for Decimal
Multiple and Fractional Units
Name
Notation
Multiplier | Name
tera
T
10“
giga
mega
kilo
G
10*
M
10*
103
hecto
deca
deci
g
da
d
k
10*
101
io -i
1 centi
1 m illi
I micro
| nano
| Pico
| femto
atto
Notation
Multiplier
c
10->
m
n
10-»
10-«
10-®
P
IQ-™
f
a
10-ia
io -i*
Periodic Table
1Series 1
s
I
1
*
H
1
1.00787
hydrogen
2
2
LI
3 Be
4
6.830
8.0122
lithium
beryllium
2
Groups of
11
III
B
10.811
boron
IV
5
V
C
6 N
12.01115
14.0067
carbon
nitrogen
7
13 Si
11 Mg
12 A1
14 P
15
S 3 Na22.8888
24.31
26.9815
28.086
30.9738
silicon
magnesium
sodium
aluminium
phosphorus
4
4
20
K
10 Ca
40.08
38.102
potassium calcium
21
So
44.966
scandium
22
Tl
47.90
titanium
23
V
50.042
vanadium
Zn
65.37
zinc
Ga
31
60.72
gallium
Ge
32 As
72.50
74.92
germani­
arsenic
um
6
38
Rb
37 Sr
87.62
85.47
rubidium strontium
39
Y
88.905
yttrium
40
7
47
Ag
107.868
stiver
In
114.82
Indium
Sn
50 Sb
51
118.60
121.75
tin
antimony
8
Cs
55 Ba
132.905 137.34
barium
caesium
9
78
Au
196.967
gold
80
Fr
Ra
(226)
radium
5
20
Cu
63.546
copper
30
Zr
01.22
zirconium
41
33
Nb
92.008
niobium
5
6
7
10
87
(223)
francium
48
Gd
112.40
cadmium
56
Hg
200.69
mercury
88
40
57
|ZI La
138.91
lanthanum
72
Tl
81
204.37
thallium
Pb
82 Bi
83
207.10
208.98
lead
bismuth
89 \*L\ Ac
(227)
actinium
Ku
104
(280)
kurcbatovlum
Hf
178.40
hafnium
73
Ta
180.05
tantalum
* Lanthanides
58
Ce
140.12
cerium
69
Pr
140.01
praseody­
mium
60 Nd
144.24
neody­
mium
61 Pm 62 Sm 63 Eu
151.06
150.35
««>
prometh­ samarium europium
ium
64 Gd
157.26
gadoli­
nium
Er 60 Tm 70 Yb 71 Lu
66 Dv 67 Ho 68
174.97
173.04
168.03
167.26
164.03
162.50
thulium ytterbium lutetlum
erbium
bolmium
dyspro­
sium
The bracketed Integer Indicates the mass number of the most stable radio
65
Tb
158.02
terbium
of Chemical Elements
elements
VI
V III
V II
0
He
2
4.0026
helium
0
Ne
10
20.183
neon
S
16
82.064
sulpbur
Cl
17
35.453
chlorine
Ar
18
39.048
argon
24
25
Se
34
78.96
selenium
Br
35
70.00
bromine
42
Mo
05.04
molybdenum
43
Te
52
127.60
tellurium
I
53
126.904
Iodine
74
W
183.85
tungsten
75
Re
186.2
rhenium
Po
84
(210)
polonium
At
85
0
15.9004
oxygen
8
F
18.0984
fluorine
Mn
54.038
manganese
Cr
51.006
chromium
26 Fe
55.847
iron
27
Co
58.933
cobalt
28
N1
58.71
nickel
Kr
36
83.80
krypton
To 44 Ru 45 Rh
101.07 102.005
technetium ruthenium rhodium
46
Pd
106.4
palladium
Xe
54
131.30
xenon
76
Os
100.2
osmium
77
Ir
102.2
iridium
78
Pt
195.09
platinum
Rn 86
(222)
radon
w tallne
** Actinides
90 Th
232.038
thorium
01 Pa
(231)
protacti­
nium
92
U
238.03
uranium
93
97 Bk
(247)
berkellum
98
90
100 Fnt
(253)
fertnium
isotope.
cr
(249)
califor­
nium
Ks
(254)
einstei­
nium
Np 04 Pu 05 Am 06 Cm
(237)
(247)
(242)
(243)
neptu­
plutonium americium curium
nium
101 Md
103 Lr
(256)
(257)
mendele- (nobellum) lawrenylum
clum
TO
M ir
P u b lish e r s
TH E
w o u ld
REA D ER
be
g ra te fu l
for you r
com ­
m e n t s o n th e c o n t e n t , t r a n s la t io n a n d d e s ig n o f th is
book.
We
w o u ld a ls o
g e s tio n s
you
b e p l e a s e d t o r e c e iv e a n y o t h e r s u g ­
m ay
w is h
to
m ake.
O u r a d d ress is
M tr
2
P u b lish e r s
P ervy
R iz h s k y
P e r e u lo k
1-110, M o s c o w , G S P , 1 2 9 8 2 0
U SSR
ALSO FROM MIR PUBLISHERS
Elementary physics: problems and solutions
I. Gurskii
The book presents the entire course of elementary
physics in a consistent and concise form. The main
emphasis is made on solving typical problems. Most of
the problems are supplied with hints. A large number
of problems for independent solution is also included.
Intended for secondary school students, lecturers and
students of preparatory departments of technical col­
leges and especially for those who study physics in­
dependently.
Collected Problems in Physics
S. Kozel, D.Sc. (Phys.-Math.),
E. Rashba, D.Sc. (Phys.-Math.), and
S. Slavatinskii, D.Sc. (Phys.-Math.)
This collection contains more than 300
examination problems that were set over
the last few years to students
on the general physics course
at the Moscow Physico-Technical Institute.
Most of the problems are very complex
and difficult. They are akin to the
problems encountered in practice and those
that emerge from physics experiments.
Many of the problems require estimation
and their solution facilitates the development
of clear physical thinking and the formation
of concepts about the scales of physical phenomena
and quantities. Solutions or detailed hints
are given. Intended for physics students
at universities and institutes.
MiR PubiishERS M o sco w