UNIVERSITY OF LIMPOPO
FACULTY O F SCIENCE AND
AGRICULTURE
SCHOOL OF PHYSICAL AND MINERAL
SCIENCES
DEPARTMENT OF PHYSICS
MODERN PHYSICS
STUDY GUIDE
1
TABLE OF CONTENTS
1. Special Theory of Relativity
3
1.1. Special Relativity
1.2. Michelson-Morley Experiment
1.3. Time Dilation
1.4. Doppler Effect
1.5. Length Contraction
1.6. The Twin paradox
1.7. Relativistic Momentum
1.8. Mass and Energy
1.9. Energy and Momentum
1.10. The Lorentz transformation
1.11. Simultaneity
2. Wave- Particles Duality
28
2.1. Electromagnetic Waves
2.2. Black Body Radiation
2.3. Photo-Electric Effect
2.4. X-ray Production
2.5. Compton Effect
2.6. Pair Production & Pair annihilation
2.7. Photon Absorption
2.8. The Wave Properties of Particles
2.9. The Uncertainty Principle
3. Atomic Physics
53
3.1. Early models of the atom
3.2. Electron obits
3.3. Atomic Spectra
3.4. The Bohr Atom
3.5. Energy Levels and Spectra
3.6. Laser
4. Nuclear Physics
65
4.1. The nucleus of the atoms
4.2. The binding energy
4.3. The Nuclear Forces
4.4. Nuclear Models
5. Nuclear Transformation
77
5.1. Nuclear decays
5.2. Half- life and rate of decay
5.3. Nuclear Reactions
5.4. Nuclear Fission
5.5. Nuclear Fusion
2
1. Special Theory of Relativity
1.1 The Special Relativity.
All motion is relative; the speed of light in free space is the same for all observers
Measurements of time and space are affected by motion between observer and the observed.
Relativity connects space and time, matter and energy, electricity and magnetism.
All motion is relative; the speed of light in free space is the same for all observers.
A frame of reference is part of the description of the motion. To say that something is moving
always implies a specific frame of reference.
An inertial frame of reference is one in which Newton’s first law of motion holds
Any frame of reference that moves at constant velocity relative to an inertial frame is itself
an inertial frame. All inertial frames are equally valid- there is no universal frame of reference
that can be used everywhere and no such thing as “absolute motion.”
The theory of relativity deals with consequences of lack of universal frame of reference.
On June 30, 1905 Einstein formulated the two postulates of special
relativity:
1. The Principle of Relativity
The laws of physics are the same in all inertial frames of reference.
2. The Constancy of Speed of Light in Vacuum
The speed of light in vacuum has the same value c in all inertial frames
of reference.
Special relativity, which is what Einstein published in 1905, treats problems that involve
inertial frames of reference.
The two postulates of relativity are:
(i)
The principle of relativity
The laws of physics are the same in all inertial frames of reference.
(ii)
The constancy of the speed of light.
The speed of light in free space has the same value in all inertial frames of reference.
Measurement of space and time are not absolute, but depends on the relative motion between
the observer and what is being observed.
3
Relativistic and Newtonian mechanics agree for relative speeds much lower than the speed
of light. At higher speeds Newtonian mechanics fails and must be replaced by the relativistic
version.
1.2 The Michelson – Morley Experiment.
The negative results experiment
Michelson-Morley experiment was designed to measure the motion of the earth through the
“ether,” a hypothetical medium pervading the universe in which light waves were supposed
to occur.
To look for the earth’s motion through the ether, Michelson and Morley used a pair of light
beams formed by a half-silvered mirror as below
One light beam is directed to a mirror along a path perpendicular to the ether current, and the
other goes to a mirror along a path parallel to the ether current
Both beams end up at the same viewing screen.
The compensating plate ensures that both beams pass through the same thickness of air and
glass.
If the transit times of the two beams are the same, they will arrive at the screen in phase and
will interfere constructively.
An ether current due to the earth’s motion parallel to one of the beams, however, would cause
the beams to have different transit times and the result would be destructive interference at
the screen. This is the essence of the experiment.
Although the experiment was sensitive enough to detect the expected ether drift, to
everyone’s surprise none was found.
4
The negative results had two consequences.
(i)
First, it showed that the ether does not exist and so there is no such thing as
“absolute motion relative to the ether. All motion is relative to a specified frame
of reference, not a universal one.
(ii)
Second, the result showed that the speed of light is the same for all observers,
which is not true of waves that need a material medium in which to occur (such as
sound and water waves).
The Michelson-Morley experiment set the stage for Einstein’s 1905 special theory of
relativity, a theory that Michelson himself was reluctant to accept.
1.3 Time dilation.
A moving clock ticks more slowly than a clock at rest.
Measurements of time intervals are affected by relative motion between observer and what is
observed.
As a result, a clock that moves with respect to an observe ticks more slowly than it does
without such motion, and all processes (including those of life) occur more slowly to an
observer when they take place in a different inertial frame.
If someone in a moving spacecraft finds that the time interval between two events in the
spacecraft is t 0 , we on the ground would find the same interval has the longer duration, 𝑡. The
quantity, t 0 , which is determined by events that occur at the same place in the observer’s
frame of reference, is called the proper time of the interval between the events.
When witnessed from the ground, the events that mark the beginning and end of the time
interval occur at different places, and in consequence the duration of the interval appear
longer than the proper time. This effect is called time dilation (to dilate is to become larger).
Time does not run backward to any observer, for instance a sequence of events that occur at
some particular point at t1 , t 2 , t 3 ,... will appear in the same order to all observers everywhere,
though not necessarily with the same time interval t 2  t1 , t 3  t 2 ,... between each pair of
events.
Similarly, no distant observer, regardless of his state of motion, can see an event before it
happens – more precisely, before a nearby observer sees it-since the speed of light is finite
and a signal require a minimum period of tme L to travel a distance 𝐿.
c
5
Example 1.1
A spacecraft is moving relative to the earth. An observer on earth finds that, between 1 P.M.
and 2 P.M. according to her clock, 3601 s elapse on the spacecraft’s clock. What is the
spacecraft’s speed relative to the earth?
Solution
Here t 0  3600 s is the proper time interval on the earth and t  3601 s is the time interval
in the moving frame as measured from the earth. We proceed as follows:
t
t0
1 v2 c2
v2  t 
1 2   0 
c
t 
2
2


t 
 3600 s 
v  c 1   0   2.998  10 8 m / s 1  

 3601s 
t 
 7.1  10 6 m/s
6
2
1.4 Doppler Effect.
Why is the universe believed to be expanding?
The Doppler Effect is a wave phenomenon where a detected frequency (wavelength) is
different from the source frequency (wavelength) due to relative (radial) motion between the
source and the observer. The main characteristics of the Doppler Effect are:




When the source and/or observer move towards one another, the detected
frequency (wavelength) if higher (shorter) than that of the source, so it has been
“shifted” to a higher frequency (shorter wavelength).
When the source and/or observer move away from one another, the detected
frequency (wavelength) if lower (longer) than that of the source, so it has been
“shifted” to a lower frequency (longer wavelength).
The greater the shift, the greater the relative speed between the source and the
observer.
This effect does not depend on the distance between the source and the observer.
Light that has been shifted to higher frequency due to approach is called “blue shifted light”,
a light shifted to lower frequency is called “redshifted light”. To date, all distance galaxies
show a redshift in the light they emit, meaning they are all moving away from the Milky Way
and the earth.
The speeds of recession are observed to be proportional to distance, which suggests that the
entire universe is expanding. The proportionality is called Hubble’s law.
The relationship between the source frequency f 0 and the observed frequency f is
Doppler effect in sound
 1 v 
c
f  f 0 

V
1
c

7
(1.4.1)
Where 𝑐 = speed of sound
𝑣 = speed of observer (+ for motion towards the source, − for motion away
from it)
𝑉 = speed of the source (+ for motion towards the observer, − for motion
away from him)
If the observer is stationary 𝑣 = 0, and if the source is stationary 𝑉 = 0.
The Doppler effect in sound varies depending on whether the source, or the observer or both
are moving.
Doppler effect in light: we consider the three situations
1. Observer moving perpendicular to a line between him and the light source.
The proper time between ticks is t 0  1 f 0 , so between one tick and the next the time
t  t0
1  v 2 c 2 elapse in the reference frame of the observer. The frequency he finds is
1 v2 c2
1
accordingly f transverse  
t
t0
Transverse Doppler effect in light f  f 0 1  v 2 c 2
(1.4.2)
The observed frequency 𝑓 is always lower than the source frequency f 0
2. Observer receding from the light source.
Now the observer travels the distance vt away from the source between ticks, which means
that the light wave from a given tick takes vt c longer to reach him than the previous one.
Hence the total time between the arrival of successive wave is
T t
1 v c 1 v c
1 v c
1 v c
vt
 t0
 t0
 t0
2
2
c
1 v c
1 v c 1 v c
1 v c
And the observed frequency is
f (receding) 
1 v c
1 1 1 v c

 f0
T t0 1  v c
1 v c
(1.4.3)
The observed frequency 𝑓 is lower than the source frequency f 0 . Unlike the case of sound
waves, which propagate relative to a material medium it makes no difference whether the
observer is moving away from the source or the source is moving away from the observer.
8
3. Observer approaching the light source
The observer here travels the distance vt towards the source between ticks, so each light
wave takes vt c less time to arrive than the previous one. In this case T  t  vt c and the
result is
f (approaching )  f 0
1 v c
1 v c
(1.4.4)
The observed frequency is higher than the source frequency. Again, the same formula holds
for motion of the source toward the observer.
Equations (1.4.3) and (1.4.4) can be combined in a single formula
f  f0
Longitudinal Doppler effect in light
1 v c
1 v c
(1.4.5)
By adopting the convention that v is + for source and observer approaching each other and
– for source and observer receding from each other.
Example 1.2
A driver is caught going through a red light. The driver claims to the judge that the colour
she actually saw was green ( f  5.60  1014 Hz) and not red ( f 0  4.80  1014 Hz) because of
the Doppler effect. The judge accepts this explanation and instead fines her for speeding at
the rate R1-00 for each km/h she exceeded the speed limit of 80 km/h. what was the fine?
Solution
Solving Eq. (1.4.5) for 𝑣 gives
 5.60 2  4.80 2 
 f 2  f 02 
8

v  c 2

3
.
00

10
m
/
s

2 
2
2 
 f  f0 
 5.60   4.80  


 4.59  10 7 m/s  1.65  10 8 km/h
Since 1 m/s = 3.6 km/h.
the fine is therefore R(1.65 × 108 – 80) = R164 999 920-00
Example 1.3
A distance galaxy in the constellation Hydra is receding from the earth at 6.12 × 107 m/s. By
how is a green spectral line of wavelength 500 nm (1 nm =10 -9 m) emitted by this galaxy
shifted towards the red end of the spectrum?
Solution
Since   c f and 0  c f 0 from Eq.(1.4.5) we have
9
  0
1 v c
1 v c
Here 𝑣 = 0.204 𝑐 and 0  500 nm, so
  500nm
1  0.204
 615nm
1  0.204
Which is in the orange part of the spectrum. The shift is   0  115 nm. The galaxy is
believed to be 2.9 billion light-years away.
1 light year is the distance light can travel in vacuum in one year’s time. This
distance is equivalent to roughly 9,461,000,000,000 km.
1.5. Length Contraction.
Faster means shorter
Measurements of lengths as well as of time interval are affected by relative motion. The
length L of an object in motion with respect to and observer always appears to the observer
to be shorter than its length L0 when it is at rest with respect to him. This contraction occurs
only in the direction of relative motion. The length L0 of an object in its rest frame is called
its proper length.
Length contraction
L  L0 1  v 2 c 2
(1.5.1)
Like time dilation, length contraction is a reciprocal effect. To a person in a spacecraft,
objects on the earth appear shorter than they did when he was on the ground by the same
factor of 1  v 2 c 2 that the spacecraft appear shorter to somebody at rest. The proper length
L0 found in the rest frame is the maximum length any observer will measure. Only lengths
in the direction of motion undergo contraction. Thus to an outside observer a spacecraft is
shorter in flight than on the ground, but it is not narrower or thinner.
10
1.6 The Twin Paradox.
A long life, but it will not seem longer
This paradox involves two identical clocks, one of which remains on the earth while the other
is taken on a voyage into space at the speed v and eventually is brought back. It is customary
to replace the clocks with a pair of twins John and Jane, a substitution that is perfectly
acceptable because the processes of life – heartbeats, respiration, and so on – constitute
biological clocks of reasonable regularity.
John is 20 y old when he takes off on a space voyage at a speed of 0.8c to a star 20 lightyears away. To Jane, who stays behind, the pace of John’s life is slower than hers by a factor
of
1  v 2 c 2  1  0.80c  c 2  0.60  60%
2
To Jane, John’s heart beats only 3 times for every 5 beats of hers; John takes only 3 breaths
for every 5 of hers. Finally, John returns after 50 years have gone by according to Jane’s
calendar, but to John the trip has taken only 30 y. John is therefore 50 y old whereas Jane,
the twin who stayed home, is 70 y old.
Where is the paradox? If we consider the situation from the point of view of John in the
spacecraft, Jane on the earth is in motion relative to him at a speed of 0.80𝑐. Should not Jane
then be 50 y old when the spacecraft returns, while John is then 70- the precise opposite of
what was concluded above?
But the two situations are not equivalent. John changed inertial frames of references. Jane
remained in the same inertial frame during John’s whole voyage. The time dilation formula
applies to Jane’s observation of John, but not to John’s observation of Jane.
To look at John’s voyage from his perspective, we must take into account that the distance 𝐿
he covers is shortened to
L  L0 1  v 2 c 2  20light  years 1  0.80c  c 2  12 light-years
2
To John, time goes by at the usual rate, but his voyage to the star has taken L v  15 y and his
return another 15 y, for a total of 30 y. of course, John’s life span has not been extended to
him, because regardless of Jane’s 50-y wait, he has spent only 30 y on the roundtrip.
The nonsymmetric ageing of the twins has been verified by experiments in which accurate
clocks were taken on an airplane trip around the world and then compared with identical
clocks that had been left behind. An observer who departs from an inertial system and then
returns after moving relative to that system will always find his clocks slow compared with
clocks that stayed in the system.
Example 1.4
John and Jane each send out a radio signal once a year while John is away. How many signals
does John receive? How many does Jane receive?
11
Solution
On the outward trip, John and Jane are being separated at the rate 0.80c. From Doppler effect,
we find that each twin receives signals
T1  t 0
1 v c
1  0.80
 1y 
 3 y apart
1 v c
1  0.80
On the return trip, John and Jane are getting closer together at the same rate and each receives
signals more frequently, namely
T2  t 0
1 v c
1  0.80 1
 1y 
 y
1 v c
1  0.80 3
apart
To John, the trip to the star takes 15 y, and he receives 15 3  5 signals from Jane. During 15
y of the return trip, John receives 15 1 3  45 signals from Jane. For a total of 50 signals.
John therefore concludes that Jane has aged by 50 y in his absence. Both Dick and Jane agree
that Jane is 70 y old at the end of the voyage.
To Jane. John needs L0 v  25 y for the outward trip. Because the star is 20 light-years away.
Jane on the earth continues to receive John’s signals at the original rate of one every 3 y for
20 y after John has arrived at the star. Hence Jane receives signals every 3 y for 25 y + 20 y
= 45 y to give a total of 45 3  15 signals. (These are the 15 signals John sent out on the
outward trip.) Then, for the remaining 5 y of what is to Jane a 50-y voyage, signals arrive
from John at the shorter interval of 1 3 for an additional 5 1 3  15 signals. Jane thus
receives 30 signals in all and concludes that John has aged by 30 y during the time he was
away- which agrees with John’s own figure. John is indeed 20 y younger than his twin Jane
on his return.
Light-year 1 light year is the distance light can travel in vacuum in one year’s time. This
distance is equivalent to roughly 9,461,000,000,000 km
12
1.7 Relativistic Momentum.
Redefining an important quantity
Relativistic mass
m
m0
(1.7.1)
1 v2 c2
m0 is the proper mass (or rest mass) of an object, its mass when measured at rest relative to
an observer.
Relativistic momentum
p
m0 v
(1.7.2)
1 v2 c2
When v  c Eq. (1.7.2) becomes just p  m0 v , the classical momentum.
Relativistic momentum
Where
 
p  m0 v
(1.7.3)
1
(1.7.4)
1 v2 c2
Relativistic Second law
In relativity Newton’s second law of motion is given by
Relativistic Second law
F
dP d
 m0 v 
dt dt
(1.7.5)
This is more complicated than the classical formula F  m0 a because  is a function of v .
When v  c.  is very nearly equal to 1, and 𝐹 is very nearly equal to m0 v , as it should be.
Example 1.5
Find the acceleration of a particle of mass 𝑚 and velocity 𝒗 when it is acted upon by the
constant force F, where F is parallel to 𝒗.
Solution
From Eq. (1.7.5), since a  dv dt ,
F


d
v

mv   m d 
2
2
dt
dt 1  v c 



 dv
v2 c2
1

 m

32
 1  v 2 c 2 1  v 2 c 2  dt
ma

1  v 2 c 2 3 2

13

We note that F is equal to  3 ma , not ma . Merely replacing m by m in classical formulas
does not always give a relativistically correct results.
The acceleration of the particle is therefore


32
F
1 v2 c2
m
a
Even though the force is constant, the acceleration of the particle decreases as its velocity
increases. As v  c, a  0 , so the particle can never reach the speed of light, a conclusion
we expect.
Example 1.6
What is the momentum of a proton moving at the speed of v  0.86c ?
Solution
Using Equation 1.7.2, we obtain
p

m0 v
1 v2 c2
1.67  10
27


kg 0.86  3.00  10 8 m / s

1  0.86 
2
 8.44  10 19 kg.m/s
The units of kg.m/s are generally not convenient in solving problems of this type. Instead, we
manipulate equation 1.7.2 to obtain
pc 
mcv
1 v c
2
2

mc 2 v c 
1 v c
2
2

938MeV 0.86  1580 MeV
2
1  0.86 
Here we have used the proton’s rest energy. The momentum is obtained from this result by
dividing by the symbol 𝑐 (not its numerical value), which gives
p  1580 MeV/c
The units of MeV/c for momentum are often used in relativistic calculations because, as we
show later, the quantity 𝑝𝑐 often appears in these calculations. You should be able to convert
MeV/c to kg.m/s and show that the two results obtained for 𝑝 are equivalent.
1.8 Mass and Energy.
Where E0  mc comes from
2
As we recall from elementary physics, the work W done on an object by a constant force of
magnitude F that acts through the distance, s , where F is in the same direction as s, is given
14
by W  Fs . If no other forces act on the object and the objects starts from rest, all the work
done on it becomes kinetic energy, KE , so KE  Fs.
In the general case where 𝐹 need not be constant, the formula for kinetic energy is the integral
s
KE   Fds
0
In nonrelativistic physics, the kinetic energy of an object of mass 𝑚 and speed 𝑣 is
1
KE  mv 2 . To find the correct relativistic formula for KE we start from the relativistic form
2
of the second law of motion, which gives.
mv
v


d mv 
mv

KE  
ds   vd mv    vd 
2
2 

dt
0
0
0
 1 v c 
s
Integrating by parts (  xdy  xy   ydx )
KE



Kinetic energy
v
nv 2
1 v c
2
2
mv 2
1 v2 c2
mc 2
1 v c
2
2
 m
0
vdv
1 v2 c2


v
 mc 2 1  v 2 c 2 0
 mc 2
KE  mc 2  mc 2    1mc 2
(1.8.1)
This results states that the kinetic energy of an object is equal to the difference between mc 2
and mc 2 . Equation (1.8.1) may be written
Total energy
E  mc 2  mc 2  KE
(1.8.2)
If we interpret mc 2 as the total energy 𝐸 of the object, we see that when it is at rest and
𝐾𝐸 = 0, it nevertheless possesses the energy mc 2 . Accordingly mc 2 is called the rest energy
E0 of something whose mass is 𝑚. we therefore have
E  E0  KE
Where
Rest energy
E0  mc 2
(1.8.3)
If the object is moving, its total energy is
15
Total energy
mc 2
E  mc 2 
(1.8.4)
1 v2 c2
Example 1.7
A stationary body explodes into two fragments each of mass 1.0 kg that move apart at speeds
of 0.6c relative to the original body. Find the mass of the original body.
Solution.
The rest energy of the original body must equal to the sum of the total energies of the
fragments. Hence
m1c 2
E0  mc  m1c  m2 c 
2
And
m
2
2
1  v12 c 2

m2 c 2
1  v 22 c 2
E0
21.0kg   2.5 kg

2
2
c
1  0.60 
Mass can be created or destroyed, but when this happens, an equivalent amount of energy
simultaneously vanishes or comes into being, and vice versa. Mass and energy are different
aspects of the same thing.
Kinetic Energy at Low Speeds
When the relative speed v is small compared with 𝑐, the formula for kinetic energy must
1
reduce to the familiar mv 2 , which has been verified by experiment at such speeds. Let us
2
see if it is true. The relativistic formula for kinetic energy is
KE  mc 2  mc 2 
Kinetic energy
mc 2
1 v c
2
2
 mc 2
(1.8.5)
Since v 2 c 2 ≪ 1, we can use the binomial approximation (1 + 𝑥 )𝑛 ≈ 1 + 𝑛𝑥, valid for
|𝑥| ≪ 1, to obtain
1
1 v2 c2
 1
1 v2
,
2 c2
𝑣≪1
Thus we have the results
 1 v2  2
1
mc  mc 2  mv 2
KE  1 
2 
2
 2c 
𝑣≪1
At low speeds the relativistic expression for kinetic energy of a moving object does indeed
reduce to the classical one.
16
1.9 Energy and Momentum.
How they fit together in relativity
Total energy and momentum are conserved in an isolated system, and the rest energy of a
particle is invariant.
Let us look into how the total energy, rest energy, and momentum of a particle are related
E
Total energy
E2 
And square it to give
mc 2
(1.9.1)
1 v2 c2
m2c 4
1 v2 c2
mv
Momentum
p
We find that
p 2c 2 
(1.9.2)
1 v2 c2
m2v 2c 2
1 v2 c2
Now we subtract p 2 c 2 from E 2 :

 
2
m2c 4  m2v 2c 2 m2c 4 1  v 2 c 2
E p c 

 mc 2
2
2
2
2
1 v c
1 v c
2
2
2

Hence
Energy and Momentum
E 2  m2c 4  p 2c 2
(1.9.3)
Which is the formula we want. We note that, because mc 2 is invariant, so is E 2  p 2 c 2 : this
quantity for a particle has the same value in all frames of reference.
When a particle travels at a speed close to the speed of light (say v  0.99c ), which often
occurs in high-energy particle accelerators, the particle’s kinetic energy is much greater than
17
its rest energy; that is KE  E0 . In this case Eq. (1.9.3) can be written to a very good
approximation
E  pc
This is called the extreme relativistic approximation and is often useful for simplifying
calculations.
Massless Particles
Can massless particles exist? To be more precise, can a particle exist which has no rest mass
but which can nevertheless exhibit such particle-like properties as energy and momentum?
In classical mechanics, a particle must have rest mass in order to have energy and momentum,
but in relativistic mechanics this requirement does not hold.
From (1.9.1) and (1.9.2), when m  0 and 𝑣 ≪ 𝑐. It is clear that E  p  0. A massless
particle with a speed less than that of light can have neither energy nor momentum. However,
when 𝑚 = 0 and 𝑣 = 𝑐 , E  0 0 and p  0 0 , which are indeterminate; E and p can have
any values. Thus (1.9.1) and (1.9.2) are consistent with existence of massless particles that
possess energy and momentum provided that they travel with the speed of light.
𝐸  pc
𝑚= 0
Massless particle,
(1.9.4)
The conclusion is not that massless particles necessarily occur, only that the laws of physics
do not exclude the possibility as long as 𝑣 = 𝑐 and 𝐸 = 𝑝𝑐 for them. In fact, a massless
particle – the photon – indeed exists.
Electronvolts
The SI unit of energy is the joule (J), which is too large to be of convenience for atomic and
nuclear physics. A much more convenient unit is the electron-volt. In atomic physics the
usual unit of energy is the electronvolt (eV). One eV is the energy gained or lost by an electron
accelerated through a potential difference of 1 volt. Since 𝑊 = 𝑄𝑉


1 eV 1.602  1019 1.000V   1.602  1019 J
Higher energies in the atomic realm are expressed in kiloelectronvolt (keV),
megaelectronvolt (MeV) and gigaelectronvolt (GeV).
1 Kev = 10 3 eV
1 MeV  10 6 eV
1 GeV  10 9 eV
An example of a quantity expressed in MeV is the energy liberated when the nucleus of a
certain type of uranium atom splits into two parts. Each such fission event releases about 200
MeV; this is the process that power nuclear reactors and weapons.
The rest energies of elementary particles are often expressed in MeV/c 2 and GeV/c2. The
advantage is that the rest energy equivalent to a rest mass of, say, 0.938 GeV or 938 MeV
(the rest mass of the proton) is just E0  mc 2  0.938 GeV. If the proton’s kinetic energy is
5.000 GeV, finding its total energy is simple:
18
E  E0  KE  0.938  5.000GeV  5.938GeV
In a similar way the MeV/c and GeV/c are sometimes convenient units of linear momentum.
Suppose we want to calculate the momentum of a proton whose speed is 0.800c. From Eq.
(1.7.2) we have
p

mv
1 v2 c2

0.938GeV / c 0.800c 
2
1  0.800c  c 2
2
0.750GeV / c
 1.25 GeV/c
0.600
Example 1.8
An electron (m = 0.511 MeV/c2) and a photon (m = 0) both have momenta of 2.000 MeV/c.
calculate the total energy of each.
Solution
(a) From Eq. (1.9.3) the electron’s total energy is
E  m2c 4  p 2c 2 
E
0.511MeV c  c  2.00 MeV c  c
2 2
4
2
2
0.511MeV 2  2.000MeV 2  2.064MeV
(b) From Eq. (1.9.4) the photon energy is
E  pc  2.000 MeV c c  2.000 MeV
Example 1.9
What are the kinetic and relativistic total energies of a proton (E0 = 938 MeV) moving at a
speed of 𝑣 = 0.86𝑐?
Solution
Earlier we found the momentum of this particle to be 𝑝 = 1580 MeV/c. The energy can be
found from Eq. (1.9.3)
E
 pc2  mc 2 2  1580MeV 2  938MeV 2  1837 MeV
The kinetic energy follows from Eq. (1.8.1)
KE  E  E0  1837MeV  938MeV  899MeV
We could also have solved this problem by finding the kinetic energy directly from Eq. (1.8.5)
Example 1.10
Find the velocity and momentum of an electron (E 0 = 0.511 MeV) with a kinetic energy of
10.0 MeV.
19
Solution
The total energy is E = KE + E0 = 10.0 MeV + 0.511 MeV = 10.51 MeV
We can then find the momentum from Eq. (1.9.3)
p


2
1
1
E 2  mc 2 
c
c
10.51MeV 2  0.511MeV 2  10.5 MeV
Note that in this problem we could have used the extreme relativistic approximation, p  E c
. The error we would make in this case would be only 0.1 percent. The velocity can be found
by solving Eq.(1.8.4) for 𝑣
2
 mc 2 
v
 0.511MeV 
  1  
 1  
  0.9988
c
 10.51MeV 
 E 
2
Example 1.11
Calculate the momentum of a 1-MeV electron.
Solution
When we speak of a 1-MeV electron, we refer to the kinetic energy of the
particle. The rest energy of an electron being 0.511 MeV. We can therefore write the total
energy as
E = E0 + KE
= 0.511 MeV + 1.0 MeV
= 1.511 MeV
To calculate the momentum, we simply use Eq. (1.9.3)
p 2 c 2  E 2  E02
 1.511MeV   0.511MeV   2.022MeV 
2
2
2
 pc  1.422 MeV
 p  1.422 MeV
c
In relativistic calculation it is usually convenient to retain such designations rather than make
the cumbersome conversion to SI units.
1.10 The Lorentz transformation
The mathematical basis for comparing two descriptions is called a transformation.
Suppose we are in an inertial frame of reference S and find the coordinates of some event that
occurs at the time 𝑡 are 𝑥, 𝑦, 𝑧. An observer located in a different inertial frame S’ which is
moving with respect to S at the constant velocity v will find the same event occurs at the time
𝑡’ and has the coordinates 𝑥’, 𝑦’, 𝑧’. How are the measurements 𝑥, 𝑦, 𝑧, 𝑡 related to
𝑥’, 𝑦’, 𝑧’, 𝑡’?
20
Galilean Transformation.
Before special relativity, transforming measurements from one inertial system to another
seemed obvious. If clocks in both system are started when the origins of S and S’ coincide,
measurements in the 𝑥 direction made by S will be greater than those made by S’ by amount
𝑣𝑡, which is the distance moved in the 𝑥 direction. That is,
x '  x  vt
There is no relative motion in the y and z directions, and so
y'  y
z'  z
t'  t
Assume that
(1.10.1)
These set of equations is known as the Galilean transformation.
To convert velocity components measured in the S frame to their equivalents in the S’ frame,
we simply differentiate 𝑥’, 𝑦’, and 𝑧’ with respect to time:
dx '
v  '  vx  v
dt
'
x
v 'y 
dy '
 vy
dt '
v z' 
dz '
 vz
dt ,
(1.10.2)
Although the Galilean transformation and the corresponding velocity transformation seem
straightforward enough, they violate both of the postulates of special relativity. The first
postulate calls for the same equations of physics in both the S and S’ inertial frames, but the
equations of electricity and magnetism become very different when the Galilean
transformation is used to convert quantities measured in one frame into their equivalents in
the other. The second postulate calls for the same value of the speed of light c whether
determined in S or S’. If we measure the speed of light in the 𝑥 direction in the S system to
be 𝑐, however, in the S’ system it will be
c'  c  v
Clearly a different transformation is required if the postulates of special relativity are to be
satisfied. We would expect both time dilation and length contraction to follow naturally from
this new transformation.
21
Lorentz Transformation.
Lorentz transformation
x' 
x  vt
1 v2 c2
y'  y
z'  z
vx
c2
t' 
1 v2 c2
t
(1.10.3)
These equations comprise the Lorentz transformation. It is obvious that the Lorentz
transformation reduces to the Galilean transformation when the relative velocity 𝑣 is small
compared with the velocity of light 𝑐.
Example 1.12
Derive the relativistic length contraction using the Lorentz transformation.
Solution
Let us consider a rod lying along the 𝑥’ axis in the moving frame 𝑆’. an observer in this frame
determines the coordinates of its ends to be x1' and x 2' , and so the proper length of the rod is
L0  x2'  x1'
In order to find L  x 2  x1 , the length of the rod as measured in the stationery frame S at the
time t, we make use of Eq. (1.10.3) to give
x1' 
Hence
x1  vt
1 v c
2

x 2' 
2

x2  vt
1 v2 c2
L  x 2  x1  x 2'  x1' 1  v 2 c 2  L0 1  v 2 c 2
This is the same as Eq. (1.5.1)
22
Inverse Lorentz Transformation
The inverse Lorentz transformation converts measurements in the moving frame S’ to their
equivalents in S. To obtain the inverse transformation, primed and unprimed quantities in
Eqs.(1.10.3) are exchanged, and 𝑣 is replaced by – 𝑣:
x
Inverse Lorentz transformation
x '  vt '
1 v2 c2
y  y'
z'  z
vx '
c2
t
1 v2 c2
t' 
(1.10.4)
Example 1.13
Derive the formula for time dilation using inverse Lorentz transformation.
Solution
Let us consider a clock at the point 𝑥’ in the moving frame 𝑆’. When an observer in 𝑆’ finds
that the time is t1' , an observer in 𝑆 will find it to be t1 , such that
vx '
t  2
c
t1 
2
1 v c2
'
1
After a time interval of t 0 (to him), the observer in the moving system finds that the time is
now t 2' according to his clock. That is
t 0  t 2'  t1'
The observer in 𝑆, however, measures the end of the same time interval to be
vx '
t 2'  2
c
t2 
2
1 v c2
So to her the duration of the interval 𝑡 is
t  t 2  t1 
t 2'  t1'
1  v2 c

2
This is what we found earlier with the help of a light-pulse clock.
23
t0
1 v2 c2
Velocity Addition
We now turn to the Lorentz transformation for the correct scheme of velocity addition.
Suppose something is moving relative to both S and S’. An observer in S measures its three
velocity components to be
Vx 
dx
dt
Vy 
dy
dt
Vz 
dz
dt
V y' 
dy '
dt '
V z' 
dz '
dt '
While to an observer in S’ they are
V x' 
dx '
dt '
By differentiating the inverse Lorentz transformation equation for 𝑥, 𝑦, 𝑧, and 𝑡, we obtain
dx 
dx  vdt
'
1 v c
2
'
2
vdz '
c2
dt 
1 v2 c2
dt ' 
dy  dy '
dz  dz '
And also
dx '
v
dx dx '  vdt '
dt '
Vx 


dt
vdx '
v dx '
'
dt  2
1 2 '
c
c dt
Relativistic velocity transformation
V x'  v
Vx 
vV '
1  2x
c
Vy 
Similarly
Vz 
V y' 1  v 2 c 2
vVx'
1 2
c
V z' 1  v 2 c 2
vV '
1  2x
c
(1.10.5)
(1.10.6)
(1.10.7)
If Vx'  c, that is, if light is emitted in the moving frame S’ in its direction of motion relative
to S, an observer in frame S will measure the speed
Vx 
V x'  v
c  v cc  v) 


c
'
vc
cv
vVx
1 2
1 2
c
c
Thus observers in the car and on the road both find the same value for the speed of light, as
they must.
24
Example 1.14
Spacecraft Alpha is moving at 0.90𝑐 with respect to the earth. If spacecraft Beta is to pass
Alpha at a relative speed of 0.50𝑐 in the same direction, what speed must Beta have with
respect to the earth?
Solution
According to Galilean transformation, Beta would need a speed relative to the
earth of 0.90𝑐 + 0.50𝑐 = 1.40𝑐, which is impossible. According to Eq. (1.10.5). however,
with Vx'  0.50c and v  0.90c , the required speed is
V x'  v
0.50c  0.90c
Vx 

 0.97c
'

0.90c 0.50c 
vVx
1
1 2
c2
c
which is less than c.
Exercise
A spaceship moving away from the earth at a speed of 0.80c fires a missile
parallel to its direction of motion. The missile moves at a speed of 0.60c relative to the ship.
What is the speed of the missile as measured by observer on the earth?
(0.95c) Kr33
1.11 Simultaneity
Events that seem to take place simultaneously to one observer may not be simultaneous to
another observer in relative motion, and vice versa.
Let us examine two events-the off of a pair of flares, say- that occur at the same time t 0 to
somebody on the earth but at a different locations x1 and x 2 . What does the pilot of a
spacecraft in flight see? To him, the flare at x1 and t 0 appears at the time
t 
'
1
t 0  vx1 c 2
1 v2 c2
While the flare at x 2 and t 0 appears at the time
t  vx2 c 2
t 2'  0
1 v2 c2
Hence two events that occur simultaneously to one observer are separated by a time interval
of
t 2'  t1' 
Or
t ' 
vx1  x 2  c 2
1 v2 c2
v L c2
1 v2 c2
25
(1.11.1)
To an observer moving at the speed v relative to the other observer: who is right? The question
is, of course, meaningless: both observers are “right” since each simply measures what he
sees.
EXERCISES
1. How fast must a spacecraft travel relative to the earth for each day on the spacecraft
3
c
2
A spacecraft receding from the earth at 0.97c transmit data at the rate 1.00  10 4
pulses/s. At what rate are they received?
The frequencies of the spectral lines in light from a distant galaxy are found to be
two-thirds as great as those of the same lines in light from nearby stars. Find the
recession speed of the distant galaxy.
A distant galaxy is moving away from the Earth at such high speed that the blue
hydrogen line at a wavelength of 434 nm is recorded at 600 nm, in the red range of
the spectrum. What is the speed of the galaxy relative to the earth? (0.31c)Krane35
If the angle between the direction of motion of a light source of frequency f 0 and the
to correspond to 2 d on the earth?
2.
3.
4.
5.
direction from it to an observer is θ, the frequency f the observer finds is given by
f  f0
1 v2 c2
1  v c  cos
Where v is the relative speed of the source. Show that this formula includes Eqs.
(1.4.2) to (1.4.4) as special cases.
6. (a) show that when v ≪ 𝑐, the formulas for the Doppler effect both in light and in
sound for an observer approaching a source, and vice versa, all reduce to
f  f 0 1  v c  , so that f f  v c . [Hint for x ≪ 1, 1 1  x   1  x.] (b) What do
the formulas for an observer receding from a source, and vice versa, reduce to when
v≪ c?
7. A physics professor claims in court that the reason he went through the red light
  650nm  was that, due to his motion, the red colour was Doppler shifted to green
  550nm  . How fast was he going?
5.0 10 m / s 
7
8. An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a
spacecraft moving at 0.90c relative to the earth. What is the height as measured by an
observer in the same spacecraft? By an observer on the earth?
9. An astronaut is standing in a spacecraft parallel to its direction of motion. An observer
on the earth finds that the spacecraft speeds is 0.60c and the astronaut is 1.3 m tall.
What is the astronaut height as measured in the spacecraft?
10. How much time does a meter stick moving at 0.100c relative to an observer take to
pass the observer? The meter stick is parallel to its direction of motion.
11. A meter stick moving with respect to an observer appears only 500 mm long to her.
What is its relative speed? How long does it take to pass her? The meter stick is
parallel to its direction of motion.
26
12. A spacecraft antenna is at an angle of 10o relative to the axis of the spacecraft. If the
spacecraft moves away from the earth at a speed of 0.70c, what is the angle of the
antenna as seen from the earth?
13. Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays
on the earth. Each twin sends the other a signal once a year by his own reckoning. (a)
How many signals does A send during the trip? How many does B send? (b) How
many signals does A receive? How many does B receive?
14. A woman leaves the earth in a spacecraft that makes a round trip to the nearest star, 4
light-years distant, at a speed of 0.9c. How much younger is she upon her return than
her twin sister who remained behind?
(5.0 y)
15. (a) An electron’s speed is doubled from 0.2c to 0.4c. By what ratio does its momentum
increase? (b) what happens to the momentum ratio when the electron’s speed is
doubled again from 0.4to 0.8c?


16. At what speed does the kinetic energy of a particle equal its rest energy? ( 3 / 2 c )
17. An electron has a kinetic energy of 0.100 MeV. Find its speed according to classical
and relativistic mechanics.
18. A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle
in terms of c.
(0.9989c)
19. (a) The speed of a proton is increased from 0.20c to 0.40c. By what factor does its
kinetic energy increase? (b) The proton speed is again doubled, this time to 0.80c. By
what factor does its kinetic energy increase now?
20. Find the SI equivalent of the mass unit MeV/c2 and momentum unit MeV/c.
21. What is the energy of a photon whose momentum is the same as that of a proton
whose kinetic energy is 10.0 MeV?
22. Find the momentum (in MeV/c) of an electron whose speed is 0.600c.
(0.383)
23. Find the total energy and kinetic energy (in GeV) and the momentum (in GeV/c) of a
proton whose speed is 0.900c. The mass of the proton is 0.938 GeV/c 2.
24. Find the momentum of an electron whose kinetic energy equals its rest energy of 511
keV.
(885 keV/c)
25. Verify that v c  pc E .
26. Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.500
GeV.
(0.963c; 3.372 GeV/c)
2
27. Find the total energy of a neutron (m = 0.940 GeV/c ) whose momentum is 1.200
GeV/c
28. A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its
mass (in MeV/c2) and speed (as a fraction of c)
(874MeV/c2; 0.37c)
29. An observer detects two explosions, one that occurs near her at a certain time and
another that occurs 2.00 ms later 100km away. Another observer finds that the two
explosions occur at the same place. What time interval separates the explosions to the
second observer?
(1.97 ms)
30. An observer detects two explosions that occur at the same time, one near her and the
other 100 km away. Another observer finds that the two explosions occur 160 km
apart. What time interval separates the explosions to the second observer?
31. Observer O sees a red flash of light at the origin at 𝑡 = 0 and a blue flash of light at
𝑥 = 3.26 km at a time 𝑡 = 7.73 μs. What are the distance and the time interval
27
between the flashes according to observer O ' , who moves relative to O in the
direction of increasing x with a speed of 0.625𝑐? Assume that the origins of the two
coordinate systems line up at t  t '  0 .
( x '  2.3 km, t '  1.07 μs)
32. A body moving at 0.500c with respect to an observer disintegrates into two fragments
that move in opposite directions relative to their centre of mass along the same line
of motion as the original body. One fragment has a velocity of 0.600c in the backward
direction relative to the centre of mass and the other has a velocity of 0.500c in the
forward direction. What velocities will the observer find?
33. A man on the moon sees two spacecrafts, A and B, coming toward him from opposite
directions at the relative speeds of 0.800c and 0.900c. (a) What does a man on A
measure for the speed with which he is approaching the moon? For the speed with
which he is approaching B? (b) What does a man on B measure for the speed with
which he is approaching the moon? For the speed with which he is approaching A?
(a) 0.800c; 0.988c
(b) 0.900c; 0.988c
34. Rocket A leaves a space station with a speed of 0.826c. Later, rocket B leaves in the
same direction with a speed of 0.635c. What is the speed of rocket A as observed from
rocket B?
(0.402c)
28
2. Wave – Particle Duality
In classical physics particles and waves are described separately.
That is: 1. we have the mechanics of particles.
2. we also have optics of waves.
The two are described by different experiments and principles.
In Modern Physics the waves and particles have the quality or character of being twofold
(i.e. duality). That duality can be seen in electromagnetic waves.
Under certain conditions they exhibit diffraction, interference, and polarization – which
are wave properties.
Under certain conditions they exhibit the property of streams of particles.
Wave-particle duality and special relativity are the basis of understanding Modern
Physics.
2.1 Electromagnetic Waves
Coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave
behaviour
Electromagnetic waves have time-varying electric and magnetic fields that propagate
through
space.
Electromagnetic waves have energy and momentum property.
Electromagnetic waves have the property of interference, reflection, diffraction and
polarization.
Electromagnetic waves are represented by electric and magnetic fields which vary
sinusoidally with time.
29
Electromagnetic wave covers a broad range of spectrum, or the spectrum of
electromagnetic waves covers a broad range of frequency and wavelength.
The Electromagnetic Spectrum:
The speed of electromagnetic waves is free space (vacuum) is given by
c
1
 0 0
 2.998  10 8 m.s-1
 0 is the electric permittivity of free space = 8.852 ×10-1 F.m-1.
 0 is the magnetic permittivity of free space = 4𝜋 ×10-7 H.m-1.
So light consists of electromagnetic waves.
The general equation 𝑐 = 𝑣𝜆 is true for all electromagnetic (e/m) waves.
From low frequency radio waves to high frequency x-rays and  -rays.
30
Electromagnetic waves obey the superposition principle:
When two of more waves of the same nature travel past a point at the same time, the
instantaneous amplitude there is the sum of the instantaneous amplitude of the individual
waves.


In constructive interference, the in phase waves reinforce one another.
In destructive interference, the out of phase waves cancel each other.
Note
1. e/m are transverse, i.e. both E and B are perpendicular to the direction
for propagation of the wave and to each other.
2. Magnitudes of E and B have their relation 𝑬 = 𝒄𝑩
3. The waves travel in vacuum with a constant speed, 𝑐.
2.2 Black Body Radiation
Only the quantum theory of light can explain its origin
An object at any temperature emits radiation called thermal radiation. This radiation
depends on the temperature and properties of a given object. At low temperatures, the
wavelengths of thermal radiation are in the infrared region.
As the temperature increases, thermal radiation consists of a continuous distribution from the
infra-red part of the spectrum through the visible spectrum to the ultra-violet part of the
spectrum.
31
The diagram show how the peak wavelength and total radiation intensity vary with
temperature. (although this plot shows relatively high temperatures, the same relation
remains true down to absolute zero temperature.)
5500 K curve corresponds to the temperature of the surface of the sun.
For low temperatures, the radiation drops and the peaks occurs at high wavelengths.
The total intensity I (Power per unit area) emitted from the surface of an ideal radiator at
absolute temperature T is given by the Stefan-Boltzmann law 𝐼 = 𝜎 T4
Where σ is the fundamental Stefan-Boltzmann constant
𝜎 = 5.6705 × 10-8 W/m2.K4
The wavelength at the peak of the spectrum is related to the temperature T by
𝜆maxT= 2.898 × 10-3 m.K
This is known as the Wien’s displacement law.
For the sun’s temperature we have
𝜆𝑚𝑎𝑥 =
2.898  103
 500 nm
5500
Which is the visible region of the spectrum?
Blackbody
A Blackbody is an assumed body that completely absorbs all wavelengths of thermal
radiation coming to it. Such bodies do not reflect light, and appear black.
Conversely all blackbodies heated emit thermal radiation with same spectrum in accordance
with classical physics.
32
Both the distribution of blackbody radiation as a function of wavelength, known as Plank
law cannot be explained using classical physics. This fact has led to the development of
quantum mechanics.
Blackbody Radiation – the thermal radiation emitted by a blackbody heated to a given
temperature.
Blackbody Temperature – the effective temperature at which a blackbody emits
blackbody radiation.
Thermal Radiation – the radiation emitted by a body as a result of its temperature (i.e.
thermal motion of its electrons)
2.3 Photo-electric Effect
The energies of electrons liberated by light depend on the frequency of the light
Photo-electric effect is the emission of electrons from the surface of a conductor when light
strikes its surface.
The emitted electrons (called photo-electrons) absorb energy from the incident radiation.
This enables them to overcome the potential barrier which confines them in the material.
Energy of a photon
E  hf 
hc

The minimum energy, an electron requires to overcome the potential barrier is called the
work function (W) or (φ)
Work function
  hf 0
33
(2.3.1)
Photo-electric effect can be used to measure maximum kinetic energy (KEmax) of the emitted
electrons. This can be achieved by using a variable voltage. This voltage is increased until a
zero current across the circuit is reached. This means no emitted electrons have sufficient
energy to reach the cathode. This is called the stopping potential (Vo). From this
measurement, KEmax can be determined i.e.
1
𝐾𝐸𝑚𝑎𝑥 = 𝑒𝑉𝑜 = mv2 max
2
Stopping potential
KE max  eV0
(2.3.2
Photoelectric effect
hf  KE max  
(2.3.3)
Wave theory of light
For monochromatic light
(a) If the intensity is increased, the number of electrons is increased, their 𝐾𝐸 is
increased.
(b) 𝐾𝐸 is independent of the frequency of light.
Particle theory (Photon theory)
Light is concentrated in small packets called photons.
Each photon of light of frequency 𝑣 has the energy 𝒉𝒗.
Increasing the intensity of light increased the number of photons.
When a photon collides with a metal an electron is ejected.
All the photon energy is absorbed or transferred to an electron.
If 𝒉𝒗 > 𝛷𝒐 electrons are ejected from the metal i.e. 𝒉𝒗 = 𝑲𝑬𝒎𝒂𝒙 + 𝜱𝒐
34
From the photon theory we have:
(a) The greater the intensity of light, the more photons are incident and the more
electrons are ejected.
(b) If the frequency in increased, the maximum KE of electrons is increased linearly
i.e. 𝑲𝑬𝒎𝒂𝒙 = 𝒉𝒗 − 𝜱𝒐
(c) If the frequency f is less than the cut off frequency v where fo no electrons will be
ejected from the metal, e.g.
” blue light caused sodium (Na) to emit electrons but red light does not.”
Phenomenon
Can be explained in terms of
waves.
Can be explained in terms of
particles.
Reflection
Refraction
Interference
Diffraction
Polarization
Photoelectric
effect
Most commonly observed phenomena with light can be explained by waves. But the
photoelectric effect suggested a particle nature for light. Then electrons too were found to
exhibit dual natures.
Kinetic energy of electrons
KE max  hf  
35
The excess energy appears as the kinetic energy of the electrons.
Example 2.1
(a) What are the energy and momentum of a photon of red light of wavelength 650 nm?
(b) What is the wavelength of a photon of energy 2.40 eV?
Solution


Converting to electron-volt
E
E
hc


6.63  10
34
J .s 3.00  108 m / s
 3.06  1019 J
650  109 m
3.06  1019 J
 1.91 eV
1.60  1019 J / eV
Problems of this sort are simplified if we express the combination ℎ𝑐 in units of eV.nm
hc 1240eV .nm
E

 191 eV

650nm
The momentum is found in a similar way
p
h


1 hc 1 1240eV .nm

 191 eV/c
c  c 650nm
The momentum could also be found directly from energy:
p
E 191eV

 1.91 eV/c
c
c
(b) Solving for 𝜆, we find
hc 1240eV .nm


 517 nm
E
2.20eV
Example 2.2
The work function for tungsten metal is 4.52 eV. (a) What is the cut-off wavelength c for
tungsten? (b) What is the maximum kinetic energy of the electrons when radiation of
wavelength 198 nm is used? (c) What is the stopping potential in this case?
Solution
(a) c 
hc


1240eV .nm
 274 nm
4.52eV
(b) At the shorter wavelength,
in the ultraviolet region
K max  hf   
hc


1240eV .nm
 4.52eV  174 eV
198nm
(c) The stopping potential is the voltage corresponding to Kmax:
K
174eV
Vs  max 
 174 V
e
e

36
2.4.
X-ray Production (Bremsstrahlung)
x-rays consist of high-energy photons
X-rays were discovered in 1895 by W.Rontgen, Deutshland (German) Physicist.
The kinetic energy of electrons is transformed into electromagnetic energy.
X-ray machines provide suitable sufficient intensity for electron flow.
X-ray machines have three principal segments:
1. Ca control panel.
2. High voltage power supply.
3. The x-ray tube.
X-ray production can be explained as the inverse/reverse of the photo-electric effect.
37
X-rays are e/m waves of very short wavelength, with high energies.
Electrons loose most or all of their kinetic energy during a collision with atoms of a metal
target. This energy resurfaces as x-rays.
X-ray spectra
The x-ray spectrum can also be defined as the energy distribution of the radiation produced
during an x-ray exposure.
The x-ray spectrum wavelengths have two parts:
1. A continuous spectrum with a cut-off at some 𝜆min
2. A series of peak superimposed on the continuous spectrum.
The cut-off wavelength, 𝜆min becomes shorter as the supplied voltage is increased. But the
peak and remain in the same wavelength. But these peaks occupy different positions for
different materials.
So peaks are a characteristics of a material used.
A continuous spectrum is produced by electrons slowing down after collision with atoms,
thus converting most of all of their kinetic energy to 𝑥-rays.
The cut-off wavelength is the same for all target materials at a certain supplied voltage.
This corresponds to the maximum kinetic energy an electron can acquire:
 𝐸 = ℎ𝑣 = ℎ 𝜆
𝑐
 i.e. as 𝐸 increases, 𝜆 decreases.
The higher the accelerating voltage, the higher the electron kinetic energy and the shorter
the cut off wavelength.
38

i.e.
E  hf max  eV

hc
min
 eV
 min 
X-ray production
hc
eV
1.24  10 6
V.m
min 
V
(1.4.1)
The line spectrum gives the characteristics of the material used.
This originates from rearrangement of electron structure of the target atoms after being
bombarded by electrons.
The lines (characteristics) are produced by electron transition between energy levels in the
target atoms. The innermost, lowest energy levels are involved.
39
Example: 2.3
Find the shortest wavelength and its corresponding frequency present in the radiation from
an X-ray machine whose accelerating potential is 50 000 𝑉.
Solution:
From Eq. (1.4.1) we have
min 
1.24  106 V .m
 2.28  1011 m  0.0248 nm
5.00  104 V
This wavelength corresponds to the frequency
3.00  10 8 m / s
f max 

 1.21  1019 Hz
11
min
2.48  10 m
c
1.5. Compton Effect
Further confirmation of the photon model
In his explanation of the Compton scattering experiment, Arthur Compton treated the x-ray
photons as particles and applied conservation of energy and conservation of momentum to
the collision of a photon with a stationary electron.




Compton effect can be explained as the collision of a photon with an electron.
Quantum theory, regards photons as particles with no rest mass.
A collision between a photon and an electron is demonstrated below:
Compton scattering diagram representation:
An x-ray photon strikes a stationery electron, and is scattered at an angle θ from its initial
direction and electron recoils at an angle Φ from the photons initial direction.
So in this collision a photon loses an amount of energy equal to that gained by the electron
(i.e. 𝐾𝐸 of the electron).
The incident photon has the frequency v, whereas the scattered photon has the lower
frequency ν’, so that
40
𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑎𝑛 𝑥 − 𝑟𝑎𝑦 𝑝ℎ𝑜𝑡𝑜𝑛 = 𝐾𝐸 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
𝒉𝝂 − 𝒉𝝂′ = 𝑲𝑬 ……………………………
2.5.1
also, for massless particles we have
𝑬 = 𝒑𝒄
2.5.2
So the x-ray photon’s momentum is given by
Photon momentum
v
pE h
c
c
2.5.3
Since momentum is a vector quantity it must be conserved in both x and y- directions.
The photon has h
v
v'
and h momenta respectively before and after collision.
c
c
The electron has 0 and p momenta before and after collision respectively.
𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
So
i.e 𝑥:
𝒚:
v
v'
h  0  h cos   p cos 
c
c
2.5.4
v'
sin   p sin 
c
2.5.5
0h
We now want to relate the difference in wavelength of the initial photon and the scattered
photon and the angle θ between the two.
These are both measurable quantities.
Firstly, we multiply 2.5.4 and 2.5.5 by c and rearrange to get
hv  hv' cos   pc cos 
41
hv' sin   pc sin 
2.5.6
Square both equations 5.6 then add them
hv2  2hvhv'cos  hv'2  p 2 c 2
2.5.7
Next consider the following two equations for energy of a particle
E  KE  mo c 2
E
2.5.8
p 2 c 2  mo c 4
2
2.5.9
Equating 5.8 and 5.9 we obtain:
KE  m c   p c  m c
2 2
2
2
2
o
4
o
KE 2  2mo c 2 KE  p 2c 2
2.5.10
Substituting hv  hv'  KE in 2.5.10 gives
hv 2  2hv hv'  hv'2  2mo c 2 hv  hv'  p 2 c 2
2.5.11
Substituting 2.5.11 in 2.5.7 gives
hv 2  2hv hv'  hv'2  2mo c 2 hv  hv'  hv 2  2hv hv'cos  hv'2
2hvhv'1  cos   2mo c 2 hv  hv'
2.5.12
Dividing 2.5.12 by 2h 2 c 2 we obtain,
v v'
1  cos   mo c  v  v' 
cc
h c c 
But
2.5.13
v 1
v' 1
 and 
such that,
c 
c '
 ' 
c 
h
1  cos 
mo c
2.5.14
h
is the Compton wavelength so that 2.5.14 becomes
mo c
 '  c 1  cos 
Compton effect
'   
h
1  cos  
mc
42
2.5.15
2.5.16
Example 2.4
X-rays of wavelength 10.0 pm are scattered from a target. (a) Find the wavelength of the xray scattered through 45o. (b) Find the maximum wavelength present in the scattered x-rays.
(c) Find the maximum kinetic energy of the recoil electrons.
Solution
(a) From  '  c 1  cos  , and so
 '    c 1  cos 45o 
= 10.0 𝑝𝑚 + 0.293𝜆𝑐
(b)
= 10.7 𝑝𝑚
 ' is maximum when 1  cos    2 , in which case
 '    2c  10.0 pm  4.9 pm  14.9 pm
(c) The maximum recoil kinetic energy is equal to the difference between the energies
of the incident and scattered photons, so
1 1 
KE max  h f  f '  hc  ' 
  


Where ' is given in (b). hence
KE max 
6.626  10


J .s 3.00  10 8 m / s  1
1 



12
10 m / pm
 10.0 pm 14.9 pm 
34
 6.54  10 15 J
Which is equal to 40.8 keV.
2.6
Pair Production & Pair Annihilation
Energy into matter
This is the case where a photon materializes as an electron (e-) and a positron e+ (positive
electron).
So in pair production electromagnetic energy in converted into matter.
Electron-positron pair occurs near an atomic nucleus so as to preserve the conservation laws
of momentum.
43
Also the sum of the charges (e- + e+) is zero like that of a photon.
The total energy, including the rest energies of positron and electro are equal to that of
photon energy.






Linear momentum is also conserved with the help of the nucleus
Pair production is produced from gamma rays (𝛾)
The reverse of pair production is pair annihilation
This occur when an electron and a positron are next to each other
The two come together due to opposite charges.
Both particles vanish simultaneously, and their rest mass becomes the energy in the
form of two gamma rays: e - + e+ → 𝛾 + 𝛾
Example 2.5
Show that pair production cannot occur in empty space.
Solution
From conservation of energy
Cons E:
hf  2mc 2
44
The angles  are equal in order that momentum be conserved in the transverse direction. In
the direction of motion of the photon, for momentum to be conserved it must be true that
hf
 2 p cos 
c
Cons p:
hf  2 pc cos 
Since p  mv for the electron and positron,
v
hf  2mc 2   cos 
c
Because v c  1 and cos  1 ,
hf  2mc 2
But conservation of energy requires that hf  2mc 2 . Hence it is impossible for pair
production to conserve both energy and momentum unless some other object is involved in
the process to carry away part of the initial photon momentum.
2.7
Photon Absorption
Photons of light, 𝑥 −rays, and 𝛾-rays interact with matter through:
(a)
(b)
(c)
Photoelectric effect.
Compton effect.
Pair production.
At low photon energies, the photoelectric effect is the chief mechanism of energy loss.
The importance of the photoelectric effect decreases with increasing energy,
to be succeeded by Compton scattering. The greater the atomic number of the
absorber, the higher the energy at which the photoelectric effect remains
significant, at intermediate energies (i.e. few tens of keV) Compton effect is the dominant. In
lighter elements, Compton scattering becomes dominant at photon energies of a few tens of
keV, whereas in the heavier ones this does not happen until photon energies of nearly 1MeV
is reached.
Pair production is more likely at photon energy which exceeds the threshold of 1.02 MeV.
The greater the atomic number of the absorber, the lower the energy at which pair production
takes over as the principal mechanism of energy loss by gamma rays.
the intensity, I, of photons is equal to the rate at which they transport energy per unit crosssectional area.
the fractional energy,  dI , lost by the beam through a thickness dx is proportional to dx
I
45
i.e. -𝑑𝐼⁄𝐼 = 𝜇𝑑𝑥 where 𝜇 is the linear attenuation coefficient.
the value of 𝜇 depends on the energy of the photon and the nature of the absorbing material.
→ upon integration of radiation intensity is
I  I 0 e  x
(2.7.1)
The intensity of radiation decreases exponentially with the thickness 𝑥.
The relation between the thickness 𝑥 and 𝜇 are 𝐼⁄𝐼𝑜 = 𝑒 −𝜇𝑥
x =
And the absorber thickness is
ln(𝐼𝑜/𝐼)
𝜇
(2.7.2)
Example 2.6
The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1. (a) Find
the relative intensity of a beam of 2.0-MeV gamma rays after it has passed through 10 cm
of water. (b) How far must such a beam travel in water before its intensity is reduced to 1
percent of its original value?
Solution
(a)


Here x  4.9m 1 0.10m  0.49 and so, from Eq. (2.7.1)
I
 e  x  e 0.49  0.61
I0
The intensity of the beam is reduced to 61 percent of its original value after passing through
10 cm of water
(b) Since
I0
I
 100, Eq. (2.7.2) yields
x
ln I 0 I 


ln 100
 0.94m
4.9m 1
46
2.8 The Wave Properties of Particles
A moving body behaves in certain ways as though it has a wave nature
De Broglie postulate:
Because photons have wave and particle characteristics, perhaps all form of matter have
wave as well as particles properties.
According to relativity, a photon with zero rest mass has P  E
E  hf  hc
hf
hc h
 pE 



c
c
c 
Photon wavelength

c
and also
so that
h
p
(2.8.1)
But particles of mass 𝑚, and speed 𝑣 has p  mv
Then the de Broglie’s wavelength of the particle is  
For matter waves, the frequency is
h
h

p mv
𝑓 = 𝐸⁄ℎ
So the dual nature of matter can be observed in these two equations
𝑚𝑣 and 𝐸 demonstrate the particle nature
𝜆 and 𝑓 demonstrate the wave nature.
m
Where 𝑚 is the relativistic mass
m0
1
47
v2
c2
Example 2.7
Find the de Broglie wavelengths of (a) a 46-g golf ball with a velocity of 30 m/s, and (b) an
electron with a velocity of 107 m/s.
Solution
(a) Since 𝑣 ≪ 𝑐, we can let  = 1. Hence

h
6.63  10 34 J .s

 4.8  10 34 m
mv 0.046kg 30m / s 
The wavelength of the golf ball is so small compared with its dimensions that we would not
expect to find any wave aspects in its behaviour.
(b) Again 𝑣 ≪ 𝑐, so with m  9.1  10 31 kg, we have
h
6.63  10 34 J .s


 7.3  10 11 m
31
7
mv 9.1  10 kg 10 m / s



The dimensions of atoms are comparable with this figure- the radius of the hydrogen atom,
for instance, is 5.3  10 11 m. it is therefore not surprising that the wave character of moving
electron is the key to understanding atomic structure and behaviour.
Example 2.8
Find the kinetic energy of a proton whose de Broglie wavelength is 1.000 fm =1.000×10-15
m, which is roughly the proton diameter.
Solution
A relativistic calculation is needed unless 𝑝𝑐 for the proton is much smaller than the proton
rest energy of E0 = 0.938 GeV
pc  mvc 
hc

4.136  10 eV .s 2.99810 m / s   1.240 10 eV  1.240 GeV

15
8
9
1.000  10
15
m
Since pc  E0 a relativistic calculation is required, the total energy of the proton is
E  E02  p 2 c 2 
0.939GeV 2  1.240GeV 2  1.555 GeV
The corresponding kinetic energy is
KE  E  E0  1.555  0.938GeV  0.617 MeV
Exc. An electron has de Broglie wavelength of 2.00 pm = 2.00 × 10-12 m. Find its kinetic
energy (answer 292 keV)
48
2.9
The Uncertainty Principle
We cannot know the future because we cannot know the present
by classical mechanics it is possible to measure the position and speed of a
particle with very small uncertainty.
but by quantum theory it is impossible to make such a measurement with good accuracy.
If a measurement of position is made with precision ∆𝑥 and a simultaneous
measurement of momentum, px, is made with precision, p x , then the product of the
two uncertainty can never be smaller than ℏ⁄2
ℎ
i.e xp x ≥ ℏ⁄2, where ℏ = 2𝜋
it is impossible to know both the exact position and exact momentum of an object
at the same time
*
if ∆𝑥 is very small, then ∆𝑝 is large, and conversely if ∆𝑥 is very large, then ∆𝑝 is small
*
these uncertainties arise from quantum structure of matter.
*
an incoming photon of momentum ℎ⁄𝜆, transfers it to an electron, during collision.
*
so after collision the electron’s momentum is ℎ⁄𝜆,
*
the position of this electron is within one wavelength, such that 𝛥𝑥 = 𝜆
ℎ
𝛥𝑝𝛥𝑥 = 𝜆 𝜆 = ℎ
49
this represent the minimum in the product of uncertainties, so that ΔpΔx ≥ ℏ .
xp 
Uncertainty principle

2
(2.9.1)
It is not possible to make a simultaneous determination of the position and the
momentum of a particle with unlimited precision.
Another uncertainty is when measuring the energy of the system 𝛥𝐸, at time interval
𝛥𝑡 which is
Et 
Uncertainty in energy and time

2
(2.9.2)
It is not possible to make a simultaneous determination of the energy and the time
coordinate of a particle with unlimited precision.
Heisenberg at road block
Example 2.9
An electron moves in the 𝑥 direction with a speed of 3.6  10 6 m / s . We can measure its
speed to a precision of 1%. (a) With what precision can we simultaneously measure its
position? (b) What can we say about its motion in the 𝑦 direction?
Solution
(a) The electron momentum is
p x  mvx  9.11 1031 kg 3.6  106 m / s  3.3  1024 kg.m/s



The uncertainty p x is 1% of this value, or 3.3  10 26 kg.m/s. The uncertainty in
position is then
x 

1.05  10 34 J .s

 3.2 nm
p x 3.3  10  26 kg.m / s
Which is roughly 10 atomic diameters.
(b) If the electron is moving in the 𝑥 direction, then we know its speed in the 𝑦 direction
precisely; thus p y  0 . The uncertainty relationship yp y ~  then requires that
50
y be infinite. We can therefore know nothing at all about the electron’s 𝑦
coordinate.
Example 2.10
Estimate the minimum velocity of a billiard ball (𝑚 ~ 100 g) confined to a billiard table of
dimension 1 m.
Solution
For x ~ 1 m, we have
p x ~
So
 1.05  1034 J .s

 1  1034 kg.m/s
x
1m
p x 1  10 34 kg.m / s
v x 

 1  10 33 m/s
m
0.1kg
Thus quantum effects suggest motion of the billiard ball with a speed of the order of
1
of the diameter of an
1  10 33 m/s. At this speed, the ball would move a distance of
100
atomic nucleus in a time equal to the age of the universe! Once again, we see that quantum
effects are not observable with macroscopic objects.
EXERCISES
1. Calculate the energy of a 700-nm photon.
(1.77 eV)
2. Calculate the wavelength and frequency of a 100-MeV photon
3. Calculate the wavelength radiated by the hot plate at 237 oC and the wavelength
radiated by the human skin at 37 oC.
4. Find the shortest wavelengths and corresponding frequencies present in the
radiation from an x-ray machine whose accelerating potential is 50 000 V and 20
kV respectively.
5. What voltage must be applied to an X-ray tube for it to emit X-rays with a
minimum wavelength of 30 pm?
6. X-rays of wavelength 10.0 pm are scattered from a target. (a) Calculate the
wavelength of the x-ray scattered through 45o. (b) Calculate the maximum
wavelength present in the scattered x-rays. (c) Calculate the maximum kinetic
energy of the recoil electrons.
7. A beam of x-ray is scattered by a target. At 45o from the beam direction, the
scattered x-rays have wavelength of 2.2 pm. Calculate the wavelength of the x-ray
in the direct beam.
(1.5 pm)
51
8. An x-ray of the initial frequency 3.0 x1019 Hz collides with an electron and is
scattered through 90o. Find the new frequency.
(2.4 x 1019 Hz)
9. Find the energy of an x-ray photon which can impart a maximum energy of 50 keV
to an electron.
10. At what scattering angle will incident 100-keV x-ray leave a target with energy of
90 keV?
(64o)
11. Find the change in wavelength of 80-pm x-ray that are scattered 120o by a target. (b)
Find the angle between the directions of the recoil electron and the incident photon.
(c) Find the energy of the recoil electron.
12. A positron collides head on with an electron and both are annihilated. Each particle
had a kinetic energy of 1.00-MeV. Calculate the wavelength of the resulting
photons.
(0.821 pm)
13. A positron with a kinetic energy of 2.000 MeV collides with an electron at rest and
the two particles annihilated. Two photons are produced; one moves in the same
direction as the incoming positron and the other moves in the opposite direction.
Calculate the energies of the photons.
14. Calculate the frequency of an x-ray photon whose momentum is 1.1 x 10-23 kg.m/s2
15. How much energy must a photon have if it is to have the momentum of 10-MeV
proton?
16. The threshold wavelength for photoelectric emission in tungsten is 230 nm. What
wavelength of light must be used in order for electrons with maximum kinetic
energy of 1.5 eV to be ejected?
(180 nm)
17. An incident light has wavelength 3.3 x 10-7 m, the maximum external energy of the
emitted photoelectrons is 5.6 x 10-19 J. (i) Calculate the metal work function in eV
and (ii) the threshold wavelength for this metal.
18. The work function for tungsten metal is 4.52 eV. (a) Calculate the cut-off
wavelength for tungsten? (b) Calculate the maximum kinetic energy for electrons
when radiation of wavelength 198 nm is used? Calculate the stopping potential in
this case?
(274 nm , 1.74 V)
(p76 Krane)
19. Show that the thickness x1/2 of an absorber required to reduce the intensity of a
beam of radiation by factor of 2 is given by (i) x1/2 = 0.693/μ. And (ii) calculate the
absorber thickness needed to produce an intensity reduction of a factor of 10.
20. Show that the intensity of radiation absorbed in a thickness x of an absorber is given
by Ioμx when μx << 1. (b) if μx = 0.100, what is the percentage error in using this
formula instead of
I = Io e-μx
21. The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m-1. Find the
thickness of lead required to reduce by half the intensity of beam of such gamma
rays.
(8.9 mm)
52
22. The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1 and
52 m-1 in lead. What thickness of water would give the same shielding for gamma
rays as 10 mm of lead?
(11 cm)
23. The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1. (a)
Find the relative intensity of a beam of 2.0-MeV gamma rays after it has passed
through 10 cm of water. (b) How far must such a beam travel before its intensity is
reduced to a one percent of its original value?
24. Calculate the de Broglie wavelength of a 10 g bullet travelling at 5000 m/s.
25. Calculate the de Broglie wavelength of an electron with a velocity of 2.2 x 10 6 m/s.
26. Find the de Broglie wavelength of a 1.00-MeV proton. Is the relativistic calculation
needed?
27. Compare the uncertainties in the velocities of an electron and a proton confined in a
1.00-nm box.
28. The position and momentum of a 1.00-keV electron are simultaneously determined.
If its position is located to within 0.100 nm. What is the percentage of uncertainty in
its momentum?
(3.1 percent)
29. (a) How much time is needed to measure the kinetic energy of an electron whose
speed is 10.0 m/s with uncertainty of no more than 0.100 percent? How far will the
electron have travelled in this period of time? (b) Make the same calculations for a
1.00-g insect whose speed is the same. What do these sets of figures indicate?
53
3. Atomic Physics
3.1
Early Models of an atom
An atom is largely empty space
All of matter is made up of atoms
Atoms also have a structure with electrons being part of its constituent.
Thomson Model
JJ Thomson discovered an electron in 1897. An atom is a positively charged homogeneous
sphere with negatively charged electrons embedded in it. This was modelled as the plumpudding model of an atom.
Rutherford’s Model
The negatively charged electrons orbit around a positively charged nucleus, the same way
planets revolve around the sun.
Nucleus must have a radius 10-15 to 10-14m, electron orbit being about 10-10m.
About 99.9% of the atoms mass is in the nucleus, atom volume is dominated by empty
space
54
The big black dot represents the nucleus and the small red dots are the electrons in their
orbits around the nucleus.
Nuclear Dimension
Rutherford scattering gives us a way to find an upper limit to nuclear dimensions. An alpha
particle will have its smallest R when it approaches a nucleus head on, which will be
followed by 180o scattering. At instant of closest approach, the initial kinetic energy KE of
the particle is entirely converted to the potential energy, and so at that instant
2 Ze 2
KE initial  PE 
4 0 R
1
Since the charge of the alpha particle is 2𝑒 and that of the nucleus is 𝑍𝑒 . Hence
Distance of closest approach
R
2Z e 2
4 0 KE
(3.1.1)
Example 3.1
Find the distance of closet approach of 8.0-MeV alpha particle incident on a gold foil.
Solution
zZe 2
1
d
 279 1.44eV .nm 
 28  10 6 nm  2.8  10 14 m
6
4 0 KE
8.0  10 eV
Bohr’ Model
This was added in the Rutherford’s model.
This theory combined Plank’s quantum theory, and Einstein’s photon theory of light.
Electrons move around the nucleus in circular orbits under Coulomb’s force of attraction.
Certain electron orbits are stable, so no radiation is emitted by these orbits.
55
Radiation is emitted by an atom when an electron moves from an orbit of higher energy to
an orbit of lower energy.

𝒄
Ei – Ef = 𝒉 𝝀
with Ei > Ef
3.2 Electron orbits
A hydrogen atom has one electron.
mv 2
For a circular orbit, the centripetal force Fc =
is provided by the electrostatic force
r
between an electron and a nucleus
Fe 
Thus
e2
4 o r 2
1
For the orbit to be stable
Fe  Fc
e 2 mv 2
=
4 o r 2
r
1
56
v
e
(3.2.1)
4 o mr
Which is the electron velocity
Also an electron in hydrogen atom has
KE 
1 2
mv
2
PE  
e2
4 o r
The total energy is
ETOT  KE  PE 
1 2
e2
mv 
2
4 o r
Substituting for 𝑣 we get
ETOT  
e2
8 o r
(3.2.2)
The total energy of the electron is negative. This reflects the fact that the electron is bound
to the nucleus. If E were greater than zero, an electron would not follow a closed orbit
around the nucleus.
Example 3.2
Experiments indicate that 13.6 eV is required to separate a hydrogen atom into a proton and
an electron; that is, its total energy is E = -13.6 eV. Find the orbital radius and velocity of
the electron in hydrogen atom.
Solution
Since 13.6 eV = 2.2 × 10-18 J, from Eq. (3.2.2)
57
1.6  10 C 
r

8 E
8 8.85  10 F / m 2.2  10 J 
19
e2
12
2
18
0
 5.3  10 11 m
An atomic radius of this magnitude agrees with estimates made in other ways. The electron
velocity can be found from Eq. (3.2.1):
v
e
4 o mr

1.6  10 19 C
4 8.85  10 12 F / m9.1 10 31 kg 5.3  10 11 m
 2.2  10 6 m/s
Since 𝑣 ≪ 𝑐, we can ignore special relativity when considering the hydrogen atom.
3.3 Atomic Spectra
Each element has a characteristic line spectrum
Light emitted by excited atoms is seen as a line spectrum not continuous spectrum.
Wavelengths observed are called emission line spectrum.
58
Atoms at ground state can also absorb light so that electrons get to a higher energy level.
This gives rise to absorption line spectrum.
These spectra serve as characteristics for identifying elements i.e they are the finger
prints of the elements
Line spectra
These spectra consist of discrete wavelength.
They are produced by atoms in gas (vapour) at moderate pressures
The frequency of the emitted photon (quantum) is given by 𝛥𝐸 = ℎ𝑓
Band spectra
This is produced by molecules and gases at high pressures. Many energy levels close to
one another are involved. These levels arise from vibrational and rotational states.
Transitions from these levels give rise to a series of bands, each having many tiny lines.
Absorption spectra
Each element has a characteristic line spectrum. Spectroscopy is a useful tool for
analyzing the composition of an unknown compound. The absorption line spectrum
consists of s bright background with dark lines corresponding to the missing
wavelengths. The emission spectrum consists of dark background with bright lines.
59
Wavelengths in the spectrum of an element are found to fall into sets called spectral
series.
Each and every element has a spectral series consisting of
1. Lyman series,
1 1 
 R 2  2 ,

n 
1
n = 2, 3, 4, …
2. Balmer series,
1 
 1
 R 2  2 ,

n 
2
n = 3, 4, 5, …
3. Paschen series,
1 
1
 R 2  2 ,

n 
3
n = 4, 5, 6, …
4. Bracket series,
1 
 1
 R 2  2 ,

n 
4
n = 5, 6, 7, …
5. Pfund series,
1 
 1
 R 2  2 ,

n 
5
n = 6, 7, 8, …
1
1
1
1
1
The quantity R, is known as the Rydberg constant R = 1.097 × 107 m-1
In the Balmer series for hydrogen atom, Hα corresponds to 𝑛 = 3, Hβ to 𝑛 = 4, H𝛾 to
𝑛 = 5 and H𝜎 to 𝑛 = 6. The series limit corresponds to 𝑛 = ∞ which occurs at 𝜆 =
4⁄ Balmer series contains wavelength in the visible part of the spectrum. Lyman
𝑅
series contain wavelengths in the ultraviolet part of the spectrum. Paschen, Brackett, and
Pfund series contain wavelength in the infrared part of the spectrum in chronological
order.
3.3 The Bohr Atom
Electron waves in the atom
For an electron around the nucleus in an orbit, de Broglie wavelength is

h
mv
With the electron velocity, 𝜈, given by
v
Orbital electron wavelength

e
4 o mr
h 4 o r
e
m
(3.4.1)
Substituting the radius r = 5.3× 10-11 m of an electron orbit in λ we obtain
  33 × 10-11 m.
This is the same as the circumference of the electron orbit
60
2𝜋𝑟 = 2𝜋(5.3× 10-11 ) = 33 × 10-11 m.
So the electron orbit of the hydrogen atom corresponds to one complete electron wave
joined to itself.
This generally corresponds to: an electron can circle a nucleus only if its orbit contains
an integral number of de Broglie wavelengths.
This combines particle and wave nature of the electron as the electron wavelength
depends on the orbital velocity. So the condition for orbit stability is
n  2rn ,
Condition for orbit stability
𝑛 = 1, 2, 3, …
(3.4.2)
rn , the radius of the orbit containing 𝑛 wavelengths. The integer 𝑛, is called the
quantum number of the orbit. Substituting  for the electron wavelength equation
2rn 
nh 4 o r
.
e
m
So that the possible orbits are those with
n 2 h 2 o
,
rn 
me2
𝑛 = 1, 2, 3, …
(3.4.3)
The innermost orbit radius is often called the Bohr radius. It is denoted by the symbol a0
as
rn  n 2 a0
(3.4.4)
3.4 Energy Levels and Spectra
A photon is emitted when an electron jumps from one energy level to a lower level
Various orbits have different electron energy levels. given in terms of orbital radius rn
61
Electron energy levels are
En  
e2
8 0 rn

e2
e2
 k
4 0 2rn
2rn
1
Substituting rn from the previous section we obtain:
me 4  1  E1
,
En  
 
8 0 h 2  n 2  n 2
𝑛 = 1, 2, 3, …
(3.5.1)
Now E1  2.18  1018 J  13.6eV
The energies given by the above equation are called energy levels. The lowest energy
level, E1, is called the ground state of the atom. The higher energy levels, E2, E3, E4, …
are called excited states. As the quantum number 𝑛 increases, the energy En approaches
zero (0). So that in the limit n  , E   0 , and the electron is not bounded by the
nucleus, to form an atom. The work or energy needed to remove an electron from an
atom is called its ionization energy.
Example 3.3
An electron collides with a hydrogen atom in its ground state and excites it to a state of
𝑛 = 3. How much energy was given to the hydrogen atom in this inelastic (KE not
conserved) collision?
Solution
From Eq.(3.5.1) the energy change of hydrogen atom that goes from an initial state of
quantum number ni to a final state of quantum number n f is
E  E f  Ei 
 1
E1 E1
1 



E

1
 n2 n2 
n 2f ni2
i 
 f
Here ni  1, n f  3, and E1  13.6 eV, so
 1 1
E  13.6 E1  2  2 eV  12.1 eV
3 1 
Example 3.4
Hydrogen atoms in states of high quantum numbers have been created in the laboratory
and observed in space. They are called Rydberg atoms. Find the quantum number of the
Bohr orbit in a hydrogen atom whose radius is 0.0100 mm. (b) what is the energy of a
hydrogen atom in this state?
Solution
(a) From Eq.(3.4.4) with rn  1.00  105 m
62
n
rn
1.00  10 5 m

 435
a0
5.29  10 11 m
(b) From Eq. (3.5.1)
En 
E1  13.6eV

 7.19  10 5 eV
2
2
n
435
Example 3.5
Find the longest wavelength present in the Balmer series of hydrogen, corresponding to
Hα line.
Solution
In the Balmer series the quantum number of the final state is n f  2 . The longest
wavelength in this series corresponds to the smallest energy difference between energy
levels. Hence the initial state must be ni  3 and
 1
1 
1 
 1
 R 2  2   R 2  2   0.139 R



3 
2
 n f ni 
1

1
1

 6.56  10 7 m  656 nm
0.139 R 0.139 1.097  10 7 m 1


This wavelength is near the red end of the visible spectrum.
3.5 Lasers
How to produce light waves all in step
A laser is a beam of highly coherent and very monochromatic light. This results from cooperative emission from many atoms. Laser is an acronym for: Light Amplification by
Stimulated Emission of Radiation (LASER)
The laser is a device that produce a light beam with these remarkable properties:
(a) The light is coherent with all waves exactly in phase.
(b) The light is very nearly monochromatic.
(c) A laser beam travels in straight parallel lines (doesn’t diverge) i.e collimated
(d) The laser beam is extremely intense than light from any other source.
63
When lasers were invented in 1960 they were called “a solution looking for a problem”.
Lasers have found utilities in varied applications like science, information technology,
medicine, industry, law, entertainment, and military.
For example, when they are used as supermarket barcode scanner, in compact disc
devices, and lasers printers.
Laser Applications
Medical applications
Welding and Cutting
Surveying
Garment industry
Laser nuclear fusion
Communication
Laser printing
CDs and optical discs
Spectroscopy
Heat treatment
Barcode scanners
Laser cooling
64
EXERCISES
1. Determine the distance of closed approach of 1.00-MeV protons incident on gold
(1.14 x 10-13 m)
nuclei.
2. How much kinetic energy must alpha particle have before its distance of closest
approach to gold nucleus is equal to the nuclear radius of 7.0 x 10 -15 m? (33 MeV)
3. What is the shortest wavelength present in the Bracket series of the spectral lines?
(1.46 μm)
4. What is the shortest wavelength present in the Paschen series of the spectral lines?
5. Calculate the three longest wavelengths in the Paschen series.
(1875 nm, 1281 nm, 1094 nm) (p188 Krane)
6. Find the wavelength of the spectral line that corresponds to a transition in hydrogen
from the n = 10 state to the ground state. In what part of the spectrum is this?
(92.1nm; ultraviolet)
7. Find the wavelength of the spectral line that corresponds to a transition in hydrogen
from the n = 6 state to n = 3 state. In what part of the spectrum is this?
8. Find the wavelengths of the transitions from n1 = 3 to n2 = 2 and from n1 =4 to to n2
= 2. In what part of the spectrum is this?
(656.1 nm, 486.0 nm)
(p192 Krane)
9. Calculate the two longest wavelengths of the Balmer series of triply ionized
beryllium (Z = 4).
(41.0 nm, 30.4 nm)
(p194 Krane)
10. How many different wavelengths would appear in the spectrum of hydrogen atoms
initially in the n = 5 state?
11. How much energy is required to remove an electron in the n = 2 state from a
hydrogen atom?
(3.4 eV)
12. An electron in the n = 5 state of hydrogen. To what states can the electron make
transitions, and what are the energies of the emitted radiations?
65
4. Nuclear Physics
Nuclear structure involves forces different from gravitational and electromagnetic forces.
4.1 The Nucleus of an Atom
Atoms are smaller than the wavelength of visible light → no hope of seeing an atom. Atoms
they can only be described about how mass and electric charge are distributed in their
volumes.
The nucleus has a positively charged protons and neutral neutrons. Nuclei are made up of
protons and neutrons bound together by the strong force. Both protons and neutrons are
referred to as nucleons. The number of protons is called the atomic number and determines
the chemical element. Nuclei of a given element (same atomic number) may have different
numbers of neutrons and are then said to be different isotopes of the element.
Mass of proton ≈ mass of neutron ≈ 1840 times the mass of electrons.
Protons and neutrons are held together by strong nuclear forces.
This nuclear force is a short-range force
The atomic nucleus is described by:
(a) atomic number, Z, - the number of protons in the nucleus, (this is equal to the
number of electrons in the atom).
(b) Neutron number, N, - the number of neutrons in the nucleus.
(c) Mass number, A, - the number of nucleons in the nucleus, i.e A = N + Z
The number of protons and neutrons in nuclei are demonstrated using symbols.
i.e
A
Z
X , 𝑋 is chemical symbol of the element.
66
e.g
70
31
Ga , Gallium has a mass number 70 and atomic number 31.
So it contains 31 protons and 39 neutrons.
Isotopes are nuclei that contain the same number of protons but different number of
neutrons.
28
29
30
e.g 116C , 126 C , 136 C and 146 C or 14
Si , 14
Si and 14
Si
Atomic masses are always expressed in mass units (u)
where 1 u = 1.66054× 10-27 kg
Atomic masses are measured in terms of atomic mass units with the carbon-12 atom
defined as having a mass of exactly 12 amu. It is also common practice to quote the rest
mass energy E=m0c2 as if it were the mass. The conversion to amu is:
1 u = 1.66054 x 10-27 kg = 931.494 MeV/c2
the energy equivalent of the atomic mass unit is
E = mc2 = (1.66054 × 10-27)(2.99792 × 108)2 = 931.494 MeV
Mass of proton, neutron and electron
Particle
Kg
U
Mev/c2
Proton
1.673 x 10-27
1.007 276
938.3
Neutron
1.675 x 10-27
1.008 665
939.6
Electron
9.109 x 10-31
0.000 549
0.5110
Nuclear Size and Density
Various types of scattering experiments suggest that nuclei are roughly spherical and appear
to have essentially the same density. The data are summarized in the expression called the
Fermi model:
Nuclei radii
R  R0 A1 3
The value of R0 is
R0  1.2  1015 m  1.2 fm
(4.1.1)
From Eq.(4.1.1) we find that the radius of 126 C nucleus is
67
R  1.212 fm  2.7 fm
13
238
Similarly, the radius of the 107
47 Ag nucleus is 5.7 fm and that of the 92 U nucleus is 7.4 fm.
where r is the radius of the
nucleus of mass number A. The
assumption of constant density
leads to a nuclear density
4.2. The Binding Energy
The mis
sing energy that keeps a nucleus together
Usually the total mass of a nucleus is less than the sum of the mass of protons and
neutrons. This means that the total energy of the nucleus is less than the energy of
the separated nucleons. The difference in energy is called the binding energy.
68
The total binding energy is the energy required to break the nucleus into its constituent
protons and neutrons.
with mP – mass of proton, mN – mass of neutron, and mA – mass of the nucleus.
Δm = ZmP + NmN – mA is the mass deficit.
Eb = Δmc2
Binding energy
(4.2.1)
in electronvolts this become Eb = Δm × 931.5 MeV / u
Example 4.1
The binding energy of the neon isotope 1020 Ne is 60.647 MeV. Find its atomic mass.
Solution
Here Z  10 and N  10 , from Eq.(4.2.1)
    
 931.49EMeV / u
m ZA X  Zm 11 H  Nm n  

b

m 1020 Ne  101.007825u   101.008665u  
160.647 MeV
 19.992 u
931.49 MeV / u
Binding Energy per Nucleon
The binding energy per nucleon for a given nucleus is an average found by dividing its total
binding energy by the number of its nucleons it contains. Thus the binding energy per
nucleon for 12 H is (2.2 MeV)/2 = 1.1 MeV/nucleon, and for 209
83 Bi is (1640 MeV)/209 = 7.8
MeV/nucleon. The binding energy per nucleon plotted against the number of nucleons for
various atomic nuclei is shown below.
69
Example 4.2
Find the total binding energy B and also the binding energy per nucleon B/A for 2656 Fe30 and
238
92
U 146 .
Solution



B  Nm n  Zm p  m ZA X N c 2
For 2656 Fe30 with N = 30 and Z = 26
B  30  1.008665u  26  1.007825u  55.934939u 931.5MeV / u 
= 492.3 MeV
B
 492.3MeV  / 56  8.791 MeV per nucleon
A
For 238
92 U 146
B  146  1.008665u  92  1.007825u  238.050785u 931.5MeV / u 
= 1802 MeV
B
 1802 MeV  / 238  7.571 MeV per nucleon
A
70
Example 4.3
(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotope
42
20 Ca . Given table of atomic masses.
(b) Find the energy needed to remove a proton from this nucleus.
(c) Why are these energies different?
Solution
(a) Removing a neutron from 2042 Ca leaves 2041Ca . From table of atomic masses. The
mass of 2041Ca plus the mass of a free neutron is
40.962278u  1.008665u  41.970943u
The difference between this mass and the mass of 2042 Ca is 0.012321 u, so the
binding energy of the missing neutron is
0.012321u 931.49meV / u   11.48 MeV
(b) Removing a proton from 2042 Ca leave the potassium isotope 1941 K . A similar
calculation gives a binding energy of 10.27 MeV for the missing proton.
(c) The neutron was acted upon only by attractive forces whereas the proton was also
acted upon by repulsive electric forces that decrease its binding energy.
4.3 The Nuclear Forces
There are four forces of nature: strong nuclear forces, weak nuclear forces,
electromagnetic forces, and gravitational forces.
The strong nuclear forces are responsible for holding protons stable in the nuclei of atoms.
The weak nuclear force is responsible for certain types of radioactive decay.
Nuclear forces are very complex in nature. Some of the important characteristics of these
forces are:
1. Nuclear forces act between a pair of neutrons, a pair of protons and also between a
neutron, proton pair, with the same strength. This shows that nuclear forces are
independent of charge.
2. Nuclear forces are the strongest forces in nature. The magnitude of nuclear forces is
100 times that of electrostatic forces 1038 times that of gravitational forces between
nucleons. That is why nucleons are held together in a nucleus In spite of electrostatic
force repulsion between protons.
3. The nuclear forces are very short range forces. They are operative up to distances of
the order of a few fermi.
4. The variation of nuclear forces with the distance between nucleons is not known
71
exactly. However, we observe that,
(i) Nuclear forces are negligible, when distance between nucleons is more than 10 fermi.
(ii) When nucleons are brought closer, nuclear force of attraction develops which goes
on increasing rapidly with decreasing distance. However, nuclear forces do not obey
inverse square law.
(iii) When distance between nucleons becomes less than 0.5 fermi, the nuclear forces
become strongly repulsive.
5. The nuclear forces show saturation properties i.e. each nucleon interacts with its
immediate neighbours only, rather than with all the other nucleons in the nucleus.
6. The nuclear forces are dependent on spin or angular momentum of nuclei.
The nuclear stability
For light nuclei atoms, they are stable if Z = N.
For heavy nuclei atoms, they are stable if the number of neutrons exceeds the number of
protons.
Most stable nuclei atoms have even values of A = N + Z
Certain atoms have exceptionally stable nuclei.
Their N and Z values are called magic numbers.
They are Z or N = 2, 8, 20, 28, 50, 82, 126.
72
4.4. Nuclear Models
The two types of nuclear models shall be discussed.
(a) The liquid-drop model which account for the nuclear binding energy.
(b) The shell model which accounts for the existence of magic numbers
The Liquid-Drop Model
If we consider the sum of the following five types of energies, then the picture of a nucleus
as a drop of incompressible liquid roughly accounts for the observed variation of binding
energy of the nucleus:
Volume energy. When an assembly of nucleons of the same size is packed together into the
smallest volume, each interior nucleon has a certain number of other nucleons in contact
with it. So, this nuclear energy is proportional to the volume or A such that
EV  aV A
Surface energy. A nucleon at the surface of a nucleus interacts with fewer other nucleons
than one in the interior of the nucleus and hence its binding energy is less. This surface
energy term takes that into account and is therefore negative and is proportional to the
surface area so that.
2
E s  a s A 3
Coulomb Energy. The electric repulsion between each pair of protons in a nucleus
contributes toward decreasing its binding energy. Each proton repels every other proton in
2
the nucleus, like k e e
so the binding energy depends on the square of the atomics
r
number Z such that:
Ec  ac
Z2
A
1 3
or
Ec  ac
Z Z  1
A
1 3
Asymmetry energy (also called Pauli Energy). An energy associated with the Pauli
exclusion principle. Were it not for the Coulomb energy, the most stable form of nuclear
matter would have the same number of neutrons as protons, since unequal numbers of
neutrons and protons imply filling higher energy levels for one type of particle, while
73
leaving lower energy levels vacant for the other type. Asymmetry occurs to heavy nuclei
having excess neutrons compared to protons.
( A  2Z ) 2
E a  a a
A
Pairing energy. An energy which is a correction term that arises from the tendency of
proton pairs and neutron pairs to occur. An even number of particles is more stable than an
odd number.
E p    A, Z 
The semi-empirical mass formula states that the binding energy will take the following
form:
(4.4.1)
Volume
term
Surface
term
Asymmetry
term
Coulomb
term
pairing
term
Example 4.4
The atomic mass of the zinc isotope 3064 Zn is 63.929 u. Compare its binding energy with the
prediction of the Eq.(4.4.1)
Solution
The binding energy of 3064 Zn is from Eq.(4.2.1)
Eb  301.007825u   341.008665u   63.929u 931.49Mev / u   559.1 MeV
The semi-empirical binding energy formula, using the coefficients, gives
74

2
0.59530 29  19.016  33.5  MeV  561.7 MeV
Eb  14.164  13.0 64 3 
1
3 
64
64  3
64 4 

The plus sign is used for the last tern because 3064 Zn is an even-even nucleus. The difference
between the observed and calculated binding energies is less than 0.5%
Example 4.5
Isobars are nuclides that have the same mass number A. Derive a formula for the atomic
number of the most stable isobar of a given A and use it to find the most stable isobar of
𝐴 = 25.
Solution
To find the value of Z for which the binding energy Eb is a maximum, which corresponds
to maximum stability, we must solve
dE b
 0 for Z. from Eq.(4.4.1) we have
dZ
dE b
a
4a
  13 2Z  1  4  A  2Z   0
dZ
A
A 3
1
Z
1
a 3 A 3  4a 4
2a 3 A
1
3
 8a 4 A
1

0.595 A 3  76
1
1.19 A 3  152 A 1
For 𝐴 = 25 this formula gives 𝑍 = 11.7, from which we conclude that 𝑍 = 12 should
25
be the atomic number of the most stable isobar of 𝐴 = 25. This nuclide is 12
Mg , which is
25
in fact the only stable A = 25 isobar. The other isobars, 11
Na and 1325 Al , are both radioactive.
The Shell Model
This model assumes that nucleons move in well-defined orbits in the nucleus.
The quantized states occupied by nucleons can be explained by quantum numbers.
The allowed states can be explained by using the Pauli’s exclusion principle, where two
fermions cannot occupy the same quantum state.
The spin-orbit effect in nucleons is due to the nuclear forces. This model accounts for the
observed magic numbers. This explains the extra stability of nuclei having even number of
protons and neutrons.
In nuclear realm, closed shell effects are associated with the following magic nucleon
number: 2, 8, 20, 28, 50, 82, 126.
For example, 116Sn has a magic number of protons (50) and 54Fe has a magic number of
neutrons (28). Some nuclei, for example 40Ca and 208Pb, have magic numbers of both
protons and neutrons; these nuclei have exceptional stability and are called " doubly
magic."
75
EXERCISES
1. State the number of neutrons and protons in each of the following: 36 Li; 1022 Ne ;
94
40
Zr ; 180
72 Hf.
2. Ordinary boron is a mixture of the 105 B and 115 B isotopes and has a composite atomic
mass of 10.82 u. What percentage of each is present in ordinary boron?
3. Estimate the nuclear radius of 27
13𝑍14 .
4.
What limits the size of a stable nucleus?
(short range strong nuclear force)
5. What happens to the atomic and mass number of a nucleus when it (a) emits an
alpha particle, (b) emits an electron, (c) emits a positron, (d) captures an electron?
6. Both 148 O and 198 O undergo beta decay. Which would you expect to emit a positron
and which an electron? Why?
7. It is believed possibly on the basis of the shell model that nuclide of z= 110 and A
=294 may be exceptionally long-lived. Estimate its nuclear radius.
(7.9 fm)
8. Which nucleus would you expect to be more stable, 73 Li or 83 Li; 136 C or 156 C.
( 73 Li; 136 C)
9. Both 148 O and 198 O undergo beta decay. Which would you expect to emit a positron
and which an electron? Why?
56
10. Calculate the binding energy per nucleon in 20
10 Ne and in 26 Fe.
(8.03 MeV; 8.79 M eV)
197
11. Find the biding energy per nucleon in 79
35 Br and in 79 Au.
12. The binding energy of 24
12 Mg is 198.25 MeV. Find its atomic mass.
13. Find the energies needed to remove a neutron from 42 He, then to remove a proton,
finally to separate the remaining neutron and proton. Compare with the total binding
energy of 42 He.
(20.6 MeV; 5.5 MeV; 2.2 MeV; 28.3 MeV)
76
14. Use the semi-empirical binding-energy formula to calculate the binding energy of
40
20 Ca. What is the percentage discrepancy between this figure and the actual binding
energy?
(Calculated 347.95 MeV; actual 342.05 MeV, which is 1.7% less)
15. Which isobar of A = 75 does the liquid-drop model suggest is the most stable?
127
16. Use the liquid-drop model to establish which of the mirror isobars 127
52 Te and 53 I
decays into the other. What kind of decay occurs?
127
( 127
53 I is stable; 52 Te undergoes negative beta decay)
77
5. Nuclear Transformations
Despite the strength of the forces that hold nucleons together to form an atomic nucleus,
many nuclides are unstable and spontaneously change into other nuclides by radioactive
decay. And all nuclei can be transformed by reactions with nucleons or other nuclei that
collide with them.
5.1 Nuclear Decays
Alpha (α) decay
Alpha decay
Visual representation of alpha decay
78

Alpha decay, or α-decay, is a type of radioactive decay in which an atomic nucleus emits
an alpha particle and thereby transforms (or 'decays') into an atom with a mass number 4
less and atomic number 2 less. For example, uranium-238 decaying through α-particle
emission to form thorium-234 can be expressed as
Because an alpha particle is the same as the nucleus of a helium-4 atom - consisting of two
protons and two neutrons and thus having mass number 4 and atomic number 2 - this can
also be written as:
Notice how, on either side of the nuclear equation, both the mass number and the atomic
number are conserved: the mass number is 238 on the left side and (234 + 4) on the right
side, and the atomic number is 92 on the left side and (90 + 2) on the right side.
The alpha particle also has a charge +2, but the charge is usually not written in nuclear
equations.
When a nucleus emits an alpha particle, the remaining nucleus is different from the original
one. The changing of one element to another is transmutation.
Alpha decay is a form of a nuclear fission, where the parent atom split into two daughter
nuclei products.
Alpha decay is fundamentally a quantum tunnelling process.
Disintegration energy
Q  mi  m f  m x c 2
(5.1.1)
Where mi , m f and mx are masses of initial nucleus, final nucleus and particle respectively.
Alpha-particle energy
KE 
A4
Q
A
79
(5.1.2)
Example 5.1
Find the kinetic energy of the alpha particle emitted in the alpha decay of 226Ra.
Solution
The decay process is

226
88
 
Ra138  222
86 Rn136  
 

Q  m 226 Ra  m 222 Rn  m 4 He c 2  226.025403u  222.017571u  4.002603u 931.5MeV / u 
= 4.871 MeV
K 
A4
222
Q
.4.871MeV  4.785 MeV
A
226
Example 5.2
The nucleus 226Ra decays by alpha emission with a half-life of 1602 y. It also decays by
emitting 14C. Find the Q value for 14C emission and compare with that for alpha emission.
(see Example above)
Solution
If 226Ra emits 14C, which contains 6 protons and 8 neutrons, the resulting nucleus is 212Pb,
so the decay process is
226
Ra 212 Pb 14 C

 
  
Q  m 226 Ra  m 212 Pb  m 14C c 2
 226.025403u  211.991871u  14.003242u 931.5MeV / u 
 28.215 MeV
We used atomic masses because the electron masses cancel.
Beta (β) decay
In nuclear physics, beta decay (β decay) is a type of radioactive decay in which a proton is
transformed into a neutron, or vice versa, inside an atomic nucleus. This process allows the
atom to move closer to the optimal ratio of protons and neutrons. As a result of this
transformation, the nucleus emits a detectable beta particle, which is an electron or positron.
There are two types of beta decay, known as beta minus and beta plus. Beta minus (β−)
decay produces an electron and electron antineutrino, while beta plus (β+) decay produces a
positron and electron neutrino; β+ decay is thus also known as positron emission.
80
The following are the type of beta decays
n0 → p + e- + 𝝂’
……………………………..(beta minus (β-) decay)
p → n0 + e+ + 𝝂
……………………………..(beta plus (β+) decay)
p + e- → n0 + 𝝂
……………………………..(electron (e-) capture)
these decay processes may transmute one chemical element into another
for example:
137
55
Cs → 138
56 Ba + e + 𝒗’ ………………β decay
22
11
+
+
Na → 22
10 Ne + e + 𝒗 ………………β decay
26
13
Al + e- → 26
12 Mg + 𝒗 ………………β decay or K-capture
An electron emitted in β decay is not an orbital electron, the electron is created within the
nucleus itself. One of the neutrons changes to a proton and in order to conserve a charge it
throws off an electron.
But in electron capture it is the orbital electron that is swallowed by the proton, preferably
the electron in the innermost orbital, the K-orbital hence the name “K-capture”.
Example 5.3
23
Ne decays to 23Na by negative beta emission. What is the maximum kinetic energy of the
emitted electrons?
Solution

23
Na12  e   
This decay process is 1023 Ne13 11
And the Q value is

 

Q  m 23 Ne  m 23 Na c 2
Using atomic masses:
 22.994465u  22.989767 u 931.5MeV / u 
 4.375MeV
Except for a small correction for the kinetic energy of the recoiling nucleus, the maximum
kinetic energy of the electrons is equal to this value. (This occurs when the neutrino has a
81
negligible energy. Similarly, the maximum neutrino energy occurs when the electron has a
negligibly small kinetic energy.)
Example 5.4
40
K is an unusual isotope, in that it decays by negative beta emission, positive beta
emission, and electron capture. Find the Q values for these decays.
Solution
40
19
The process for negative beta decay is
And Q value is
  

K 21  2040 Ca 20  e   

Q   m 40 K  m 40 Ca c 2
 39.963999u  39.962591u 931.5MeV / u 
 1.312 MeV
The process for positive beta emission is
And the Q value is
  
40
19
K 21 1840 Ar22  e   


Q   m 40 K  m 40 Ar  2me c 2
 39.963999u  39.962384u  2  0.000549u 931.5MeV / u 
 0.482 MeV
For electron capture,
40
19
K 21  e  1840 Ar22  
  
and

Qec  m 40 K  m 40 Ar c 2
 39.963999u  39.962384u 931.5MeV / u 
 1.504 MeV
Gamma (𝜸) decay
Gamma rays are often produced alongside other form of radiation like alpha or beta.
When a nucleus emits an α or β particle, the daughter nucleus is sometimes left in an
excited state.
A general equation for gamma decay is ZA X* → ZA X + 𝛾
82
And specifically it is
60
Ni* → 60 Ni + 𝛾
Like an atom a nucleus can be in an excited state. When it jumps down to a lower energy
state, or ground state it emits a photon.
Example 5.5
The helium isotope 26 He is unstable. What kind of decay would you expect it to undergo?
Solution
For light nuclei atoms, they are stable if Z = N. For heavy nuclei atoms, they are stable if
the number of neutrons exceeds the number of protons. The instability of 26 He must be due
to an excess of neutrons. This suggest that 26 He undergoes negative beta decay to become
the lithium isotope 36 Li whose neutron/proton ratio is more consistent with stability:
6
2
He36 Li  e 
This is, in fact, the manner in which 26 He decays.
5.2
Half-life and Rate of Decay
Radioactive Half-Life
The radioactive half-life for a given radioisotope is a measure of the tendency of the
nucleus to "decay" or "disintegrate" and as such is based purely upon that probability.
The predictions of decay can be stated in terms of the half-life, the decay constant, or the
average lifetime. The relationship between these quantities is as follows.
Nuclear Decay Probability
Radioactive decay is a statistical process which depends upon the instability of the
particular radioisotope, but which for any given nucleus in a sample is completely
unpredictable. The decay process and the observed half-life dependence of radioactivity can
83
be predicted by assuming that individual nuclear decays are purely random events. If there
are N radioactive nuclei at some time t, then the number ΔN which would decay in any
given time interval Δt would be proportional to N:
where 𝝀 is a constant of proportionality (decay constant).
The minus sign shows that N keeps on decreasing.
For the limit t  0, the above equation becomes:
dN
 N
dt
Upon rewriting we have
dN
 dt
N
Integrating:
N
t
dN
N N   0 dt
0
 N 
  t
ln
N
 0
N  N 0 e t
Thus
(5.2.1)
Where N0 is the number of radioactive nuclei present at 𝑡 = 0. N is the remaining nuclei
after time t. Thus the number of parent nuclei decrease exponentially with time. The decay
dN
rate, R  N 
is called the activity of the sample.
t dt
R
dN
 N 0 e t  R0 e t
dt
where R0  N 0 the decay rate at 𝑡 = 0.
So at any other time 𝑡, R  N .
Activity law
R  R0 e  t
(5.2.2)
84
This also implies that the decay rate and amount of emitted radiation also follow the same
type of relationship: R  R0 e  t
Radioactive Decay Constant
The rate of radioactive decay is typically expressed in terms of either the radioctive halflife, or the radioactive decay constant. They are related as follows:
The decay constant is also sometimes called the disintegration constant. The half-life and
the decay constant give the same information, so either may be used to characterize decay.
Another useful concept in radioactive decay is the average lifetime. The average lifetime is
the reciprocal of the decay constant as defined here.
For example, free neutrons decay with a half-life of about 10.3 minutes. This corresponds to
a decay constant of .067/min and an average lifetime of 14.8 minutes or 890 seconds.
The half-life of a radioactive substance is the time it takes for half of a given number of
radioactive nuclei to decay.
Letting N 
N0
2
and t  T 1 in N  N 0 e t gives
2
N0
 N 0 e  t
2
2  e  t
 T1 
2
ln 2


0.693

A mostly used unit of activity is curie (𝐶𝑖), named after a polish-French physicist Marie
Curie for pioneering research in radioactivity.
1Ci  3.7  1010 decays/s
The SI unit of activity is the Becquerel (Bq)
1Bq = 1 decays/s
85
 1Ci  3.7  1010 decays/s  3.7  1010 Bq
The mean lifetime is related to the half-life by
 mean 
Mean lifetime
1


T1
2
ln 2

T1
2
0.693
 1.44T1 2 .
Example 5.6
222
86
Rn alpha decays to 218
84 Po with half-life of 3.82 days. How long does it take 60.0 percent
of a sample of radon to decay?
Solution
N
N
N
 e t  t  ln
 t  ln 0
N0
N0
N
From Eq. (5.2.1)
1
t

ln
N0
N
Here   0.693 / T1 2  0.693 / 3.82d and N  1  0.600N 0  0.400N 0 , so that
t
3.82d
1
ln
 5.05 d
0.693 0.400
Example 5.7
Find the activity of 1.00 mg of radon, 222 Rn , whose atomic mass is 222 u.
Solution
The decay constant of radon is

0.693
0.693

 2.11  10 6 s 1
3.8d 86400 s / d 
T1 2
The number of atoms in 1.00 mg of 222 Rn is
1.00  10 6 kg
N
 2.71  1018 atoms
 27
222u  1.66  10 kg / u

Hence




R  N  2.11 106 s 1 2.71 1018 nuclei
 5.72  1012 decays/s = 5.72 TBq = 155 Ci
86
Example 5.8
What will the activity of the above radon sample be exactly one week later?
Solution
The activity of the sample decays according to Eq. (5.2.2). since R0  155 Ci
t  2.11 106 s 1 7.00d 86400s / d   1.28
R  R0 e  t  155Ci e 1.28  43 Ci
Therefore
Example 5.9
The half-life of 198Au is 2.70 days.
(a) What is the decay constant of 198Au?
(b) What is the probability that any198 Au nucleus will decay in one second?
(c) Suppose we had a 1.00-μg sample of 198Au. What is its activity?
(d) How many decays per second occur when the sample is one week old?
Solution
(a)  
0.693 0.693 1d
1h

.
.
 2.97  10 6 s 1
t1 2
2.70d 24h 3600 s
(b) The decay probability per second is just the decay constant, so the probability of any
198
Au nucleus decaying in one second is 2.97  10 6 .
(c) The number of atoms in the sample is determined from the Avogadro constant N A
and the molar mass M:
N



mN A 1.00  10 6 g 6.02  10 23 atoms / mole

 3.04  1015 atoms
M
198 g / mole



R  N  2.97  106 s 1 3.04  1015  9.03  109 decays per second = 0.244 Ci
(d) The activity decays according to R  R0 e  t


R  9.03  109 decays / s e 0.693 / 2.70d 7 d   1.50  109 decays/s
Radiometric Dating
Radioactivity makes it possible to establish the ages of many geological and biological
specimens. Because the decay of any particular radio-nucluide is independent of its
environment, the ratio between the amounts of that nuclide and its stable daughter in a
87
specimen depends on the latter’s age. The greater the proportion of the daughter nuclide,
the older the specimen.
t
Geological dating
1

ln
N0
N
(5.2.3)
Where N is the number of atoms of parent nuclide remaining and N 0 is the number of parent
nuclide in the beginning.
Example 5.10
A piece of wood from the ruins of an ancient dwelling was found to have a 14C activity of
13 disintegrations per minute per gram of its carbon content. The 14C activity of living
wood is 16 disintegrations per minute per gram. How long ago did the tree die from which
the wood sample came?
Solution
If the activity of a certain mass of carbon from a plant or animal that was recently alive is
R0 and the activity of the same mass of carbon from the sample to be dated is R, then from
Eq. (5.2.2)
R  R0 e  t
To solve for the age 𝑡 we proceed as follows:
e t 
And
R0
R
1 R
 t  ln 0  t  ln 0
R
R
 R
  0.693T
y and
 0.693
5760
12
t
1

ln
R0
R
 16
13
and so
R0 5760 y 16

ln
 1.7  10 3 y
R
0.693 13
Example 5.11
The half-life of 235U is 7.04  10 8 y. A sample of rock, which solidified with the Earth
4.55  10 9 years ago, contains N atoms of 235U. How many 235U atoms did the same
rock have at the time it solidified?
Solution
The age of the rock corresponds to
4.55  10 9 y
 6.46 half-lives
7.04  10 8 y
88
Since each half-life reduces N by a factor of 2, the overall reduction in N has been
2 6.46  88.2
The original rock therefore contained 88.2N atoms of 235U
5.3 Nuclear Reactions
The Q value of the nuclear reaction
A B C  D
Is defined as the difference between the rest energies of A and B and the rest energies of C
and D:
Q value of nuclear reaction Q  m A  mB  mC  mD c 2
(5.3.1)
If Q is a positive quantity, energy is given off by the reaction. If Q is a negative quantity,
enough kinetic energy KEcm in the center-of-mass system must be provided by the reacting
particle so that KE cm  Q  0.
Exc
Calculate the threshold kinetic energy for the reaction
p13H 2 12 H 1  12H1
(-4.033 MeV)
Example 5.12
Find the minimum kinetic energy in the laboratory system needed by an alpha particle to
cause the reaction 14 N  , p 17O . The masses of 14N, 4He, 1H, and 17O are respectively
14.00307 u, 4.00260 u, 1.00783 u, and 16.99913 u.
Solution
Since the masses are given in atomic mass units, it is easiest to proceed by finding the mass
difference between reactants and products in the same units and then multiplying by 931.5
MeV/u. thus we have
Q = (14.00307 u + 4.00260 u–1.00783 u – 16.99913 u)(931.5 MeV/u) = -1.20 MeV
89
The minimum kinetic energy KE cm in the center-of-mass system must therefore be 1.20
MeV in order for the reaction to occur
Nuclear Fission
Nuclear fission is a decay process in which unstable nuceus splits into two fragments of
comparable mass.
Fission occurs more readily in unstable heavy nuclei, (e.g. 235
92 U).
Fission is initiated by capture of thermal neutrons by heavy nucleus.
In nuclear fission energy of about 200 MeV per event is released.
i.e
n + 235
U → 236
U → N1 + N2 + neutrons
92
92
for example
236
141
92
n + 235
92 U → 92 U → 56 Ba + 36 Kr + 3n
90
Nuclear Fusion
In nuclear fusion, two light nuclei fuse together to form one larger nucleus.
The mass of the final nucleus is less than the combined masses of the original nuclei. So
there is a loss of mass accompanied by the release of energy.
For example
1
1
H + 11 H → 21 H + e+ + 𝝂
2
1
H + 11 H → 23 He + n + 3.3 MeV
3
1
H + 21 H → 42 He + n + 17.6 MeV
2
1
H + 21 H → 31 H + 11 H + 4.0 MeV
EXERCISES
1. Tritium ( 31 H) has a half-life of 12.5 y against beta decay. What fraction of a sample
will remain undecayed after 25 y?
(¼)
2. Find the probability that a particular nucleus of 38Cl will undergo beta decay in any
1.00-s period. The half-life of 38Cl is 37.2 min.
(3.10  10-4)
3. The activity of a certain radionuclide decreases to 15 percent of its original value in
10 d. Calculate its half-life.
4. One g of 226Ra has an activity of nearly 1 Ci. Determine the half-life of 226Ra.
(1.6  103 y)
9
5. The half-life of 238
92 U against alpha decay is 4.5  10 y. Find the activity of 1.0g of
238
(1.23  104 Bq)
U.
91
6. The half-life of the alpha-emitter 210Po is 138 d. what mass of 210Po is needed for a
(2.22  10-9 kg)
10-mCi source?
7. The half-life of 24Na is 15.0 h. how long does it take 80 and 75 percent of a sample
of this nuclide to decay?
(34.8 h, 30.0 h)
8. Tritium, the hydrogen isotope of mass 3, has a half-life of 12.3 y. what fraction of
the tritium atoms remains in the sample after 50.0 y?
9.
(0.0598)
222
86𝑅𝑛 has a half-life of 3.82 days.
(i)
How long does it take for 87.5 percent of this sample of radon to decay?
(ii)
How long does it take 12.5 % of the sample to remain undecayed?
(iii)
How long does it take 50% of the sample to decay?
10. The activity of a sample of an unknown radionuclide is measured at daily intervals.
The results in MBq, are 32.1, 27.2, 23.0, 19.5, and 16.5. Find the half-life of the
radionuclide.
11. The relative radiocarbon activity in a piece of charcoal from the remains of an
ancient campfire is 0.18 that of a contemporary specimen. How long ago did the fire
(1.4  104 y)
occur?
12. Find the minimum kinetic energy in the laboratory system needed by an alpha
particle to cause the reaction
14
17
14
4
1
17
7𝑁 (𝛼, 𝑝) 8𝑂 The masses of 7𝑁 , 2𝐻𝑒 , 1𝐻 , and 8𝑂
are respectively 14.00307 u, 4.00260 u, 1.00783 u, and 16.9913 u.
13. Calculate the disintegration energy when 232
92 U (mass = 232.037131 u) decays to
Th (228.028716 u) with emission of α (4.002602 u).
228
90
14. The radionuclide 232U alpha-decays into 228 Th. (a) Find the energy released in the
decay. (b) Is it possible for 232U to decay into 231U by emitting a neutron? (c) Is it
possible for 232U to decay into 231Pa by emitting a proton? The atomic masses of
231
U and 231Pa are respectively 231.036270 u and 231.035880 u.
15. By how much must the atomic mass of a parent exceed the atomic mass of a
daughter when (a) an electron is emitted, (b) a positron is emitted, and (c) an
electron is captured?
16. Show that it is energetically possible for 64Cu to undergo beta decay by electron
emission, positron emission, and electron capture and find the energy released in
each case.
17. Carry out the calculations of Exercise 16 for 80Br.
( 2.01 MeV; 0.85 MeV; 1.87 MeV)
18. Calculate the maximum energy of the electrons emitted in the beta decay of 12B.
92
19. Find the minimum antineutrino energy needed to produce the inverse beta-decay

reaction.
p   n  e  .
(1.80 MeV)
20. Complete these nuclear reactions:
6
3
Li  ? 47 Be 01n
35
17
9
4
Cl  ?1632 S  24He
Be 24He 324 He  ?
79
35
Br  12H  ?2 01 n
 H ; H ; n; Kr 
2
1
1
1
1
1
79
36
21. Find the minimum energy in the laboratory system that a neutron must have in order
to initiate the reaction 01 n168 O  2.20 MeV 136 C  24He
22. Find the minimum energy in the laboratory system that a proton must have in order
to initiate the reaction p  d  2.22 MeV  p  p  n
(3.33 MeV)
23. Use the semi-empirical binding –energy formula to calculate the energy that would
be released if 238U nucleus were to split into two identical fragments.
24. In their old age, heavy stars obtain part of their energy by the reaction
4
2
He + 126 C 168 O
How much energy does each such event give off?
25. The initial reaction in the carbon cycle from which stars hotter than the sun obtain
their energy is
1
1
H 126 C 137 N  
Find the minimum energy the proton must have to come in contact with the 126 C
nucleus.
(2.37 MeV)
26. Calculate the total energy released if 1.00 kg 235U undergoes fission, taking the
disintegration energy per event to be Q = 208 MeV.
93