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Module G485.3 Nuclear
Physics
student answer booklet
Lesson 33 questions – The Atom
1.
Describe briefly the two conflicting theories of the structure of the atom.
…………………………………………………………………………………………………………….
The English scientist Thomson suggested that the atom, which is a neutral particle, was made
of positive charge with ‘lumps’ of negative charge inset in it - rather like the plums in a
pudding. For this reason it was known as the Plum Pudding theory of the atom.
Rutherford explained it this way. He knew that the alpha particles carried a positive
charge so he said that the positive charge of the atom was concentrated in one place
that he called the nucleus, and that the negatively charged particles, the electrons,
were in orbit around the nucleus. Most of the mass was in the nucleus
………………………………………………………………………………………………………. (4)
2.
Why was the nuclear model of Rutherford accepted as correct?
…………………………………………………………………………………………………………….
Rutherford’s prediction using the idea of Coulomb law repulsion was verified by experiment.
It also enables experimental values of nuclear charge to be obtained, ie atomic number.
………………………………………………………………………………………………………. (2)
3.
What would have happened if neutrons had been used in Rutherford’s experiment?
Explain your answer.
…………………………………………………………………………………………………………….
…………………………………………………………………………………………………………….
They would not have been repelled
so it is unlikely that any would ‘bounce back’. Some could be absorbed by the nucleus.
………………………………………………………………………………………………………. (2)
4.
What would have happened if aluminium had been used instead of gold in the alpha
scattering experiment? Explain your answer.
…………………………………………………………………………………………………………….
The charge on the nucleus is much smaller
so deflection would be smaller.
………………………………………………………………………………………………………. (2)
5.
What three properties of the nucleus can be deduced from the Rutherford scattering
experiment? Explain your answer.
…………………………………………………………………………………………………………….
Small, massive and positive.
………………………………………………………………………………………………………. (3)
6
Describe briefly one scattering experiment to investigate the size of the
nucleus of the atom.
Include a description of the properties of the incident radiation which
makes it suitable for this experiment.
In your answer, you should make clear how evidence for the size of the
nucleus follows from your description.
Any seven from:
α - particle scattering
suitable diagram with source, foil, moveable detector
2 or more trajectories shown
vacuum
most particles have little if any deflection
large deflection of very few
reference to Coulomb’s law /elastic scattering
alphas repelled by nucleus (positive charges)
monoenergetic
OR electron scattering
High energy diagram with source sample, moveable detector / film
Vacuum
Electron accelerator or other detail
Most have zero deflection
Characteristic angular distribution with minimum
Minimum not zero
De Broglie wavelength
Wavelength comparable to nuclear size hence high energy B1 × 7
Clearly shows how evidence for the size of the nucleus follows from
what is described. (1)
[Total 8 marks]
Practical advice
These questions are to help your students to think about the Rutherford ideas.
Answers and worked solutions
1
The English scientist Thomson suggested that the atom, which is a neutral particle,
was made of positive charge with ‘lumps’ of negative charge inset in it - rather like the
plums in a pudding. For this reason it was known as the Plum Pudding theory of the
atom.
Rutherford explained it this way. He knew that the alpha particles carried a positive
charge so he said that the positive charge of the atom was concentrated in one place
that he called the nucleus, and that the negatively charged particles, the electrons,
were in orbit around the nucleus. Most of the mass was in the nucleus
2
Rutherford’s prediction using the idea of Coulomb law repulsion was verified by
experiment. It also enables experimental values of nuclear charge to be obtained, ie
atomic number.
3
They would not have been repelled so it is unlikely that any would ‘bounce back’.
Some could be absorbed by the nucleus.
4
The charge on the nucleus is much smaller so deflection would be smaller.
See the equation
TAP 521-7: Rutherford scattering data
5
Small, massive and positive.
External reference
This activity is taken from Resourceful Physics
Lesson 34 questions – Nuclear Forces
1.
Fig. 1 shows two protons A and B in contact and at equilibrium inside a
nucleus.
A
B
Fig. 1
Proton A exerts three forces on proton B. These are an electrostatic force
FE, a gravitational force FG and a strong force FS.
(a)
On Fig. 1, mark and label the three forces acting on proton B.
Assume that every force acts at the centre of the proton.
forces FS and FG acting inwards, force FE acting outwards - all through centre of proton;
3 forces 2/2, 2 forces 1/2, marked and labelled (2)
[2]
(b)
Write an equation relating FE, FG and FS.
FE = FS + FG;
accept FE + FS + FG = 0 allow ecf from (a)
[1]
(c)
The radius of a proton is 1.40 × 10–15 m.
Calculate the values of
(i)
FE
FE = Q2 / (4π ε0 r2) (1)
= (1.6 × 10–19)2 / [4π × 8.85 × 10–12 (1.4 × 10–15)2] = 117.4 N (1)
use of r = 1.4 × 10–15 m (–1) once only
FE = ............... 117.4...................... N
[2]
(ii)
FG
FG = m2 G / r2 (1)
= (1.67 × 10–27)2 × 6.67 × 10–11 / (1.4 × 10–15)2 = 9.5 × 10–35 N (1)
FG = ............. 9.5 × 10–35........................ N
[2]
(iii)
FS.
FS = 117.4 N / same as FE allow ecf (1)
FS = ............ 117.4........................ N
[1]
(d)
Comment on the relative magnitudes of FE and FG.
......................... FE >> FG so FG negligible / insignificant / can be
ignored or AW
............................................................................................................
............
[1]
(e)
Fig. 2 shows two neutrons in contact and at equilibrium inside a
nucleus.
Fig. 2
Without further calculation, state the values of FE, FG and FS for
these neutrons.
(i)
FE = ...............0................................................................... N
[1]
(ii)
FG = ............... 9.5× 10–
35
................................................................... N
[1]
(iii)
FS = ................. 9.5 × 10–
35
................................................................ N
[1]
[Total 12 marks]
2.
This question is about the strong and electrostatic forces inside a
nucleus.
The figure below shows how the strong force (strong interaction) and the
electrostatic force between two protons vary with distance between the
centres of the protons.
strong
force
force
repulsive
electrostatic
force
0
0
N
P
distance
between centres
attractive
(a)
Label on the figure the regions of the force axis which represent
attraction and repulsion respectively.
[1]
(b)
(i)
On the figure above, mark a point which represents the
distance between the centres of two adjacent neutrons in a
nucleus. Label this point N.
Explain why you chose point N.
correct point N - where strong line crosses distance axis;
at N (resultant) force is zero; (1)
so neutrons must be at equilibrium; (1)
any 1
not just ‘forces equal’
2
..................................................................................................
.............
[2]
(ii)
On the figure, mark a point P which represents the distance
between two adjacent protons in a nucleus.
Explain why you chose point P.
..................... correct point P; (1)
at P electrostatic and strong forces balance (or AW);
(1)...........
..................................................................................................
.............
..................................................................................................
.............
..................................................................................................
.............
[2]
(c) On the figure, sketch a line to show how the resultant force between
two protons varies with the distance between their centres. Pay
particular attention to the points at which this line crosses any other
line.
crosses axis at P; allow P on either curve if forces equal (1)
crosses e/s force line at point vertically above N; (1)
generally correct shape, entirely above strong line; (1)
[3]
(d)
(i)
Write an expression for the electrostatic force between two
point charges Q which are situated at a distance x apart.
(F =) Q2/[4πε0(x)2] allow (F =) Q1 Q2 /[4πε0(x)2] (1)
[1]
(ii)
The electrostatic force between two protons in contact in a
nucleus is 25 N.
Calculate the distance between the centres of the two
protons.
25 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 [d]2) subs. (1)
d = 3.0(3) × 10–15 m allow 3 × 10–15 m (1)
distance = .......... 3.0(3) × 10–15............................ m
[2]
[Total 11 marks]
Lesson 35 Answers - Nuclear Properties
(/14)
1
4
2
He
(a)
an alpha particle
(b)
a proton
(c)
a hydrogen nucleus
(d)
a neutron
(e)
a beta particle
(f)
a positron.
1
1
H or 11 p
1
0
1
1
H
n
0
1
0
1
e
e
2
13
6
(a)
carbon-13
(b)
nitrogen-14
(c)
neon-22
(d)
tin-118
118
50
(e)
iron-54
54
26
3
22
10
C
14
7
N
Ne
Sn
Fe
acts only on nearest neighbour / when nuclei are 1 diameter apart; (1)
either
so force holding nucleons/ neutrons together independent of size of
nucleus (1)
or
reference to b so distance apart (of nucleons) must be constant;
so density of nucleus is independent of size; (1)
3
[3]
Lesson 37 Answers- Quarks
1.
(i)
up down down / udd;
(ii)
Q
u (+)2/3
d –1/3
(iii)
B
(+)1/3
(+)1/3
1
S
0
0
u values (1)
d values (1)
2
so for neutron Q = 0
B=1
S=0
1
[4]
2.
(i)
leptons;
(ii)
neutrino / muon / tau(on);
1
1
[2]
3.
neutron is udd / proton is uud; (1)
quarks are: up down strange top bottom charm; (1)
either up / u has Q = (+)2/3, B = (+)1/3;
or down / d has Q = –1/3, B = (+)1/3; (1)
quarks are fundamental particles; (1)
for every quark there is an antiquark; (1)
antiquarks have opposite values of Q, B and S (compared to quark) (1)
quarks are held together by strong force / gluons (1)
Q, B and S are conserved in (quark) reactions (1) any 2
5
[5]
4.
baryon: two examples
neutron; (1)
3 particles quoted, including one wrong gets 1/2 only
quark composition: proton
neutron
uud; (1)
udd; (1)
proton; (1)
(aware consists of 3 quarks, unspecified, gets 1/2)
stability: proton stable inside (stable) nucleus; (1)
proton possible decay / half life = 1032 years when free; (1)
allow any half life > 1030 years
neutron stable inside (stable) nucleus; (1)
neutron half life = 10/15 minutes when free; (1)
any 6
[6]
5.
lepton: two examples:
positron; (1)
neutrino; (1) any 2 (2)
(allow muon, tauon)
3 particles including one wrong gets 1 only
electron; (1)
composition: fundamental (- no quark components); (1)
forces: weak force / interaction; (1)
electron / positron - (also) electromagnetic / electrostatic force; (1)
where found:electron - in atom, outside nucleus or in β– decay; (1)
positron (rarely) emerging from (high mass) radioisotopes /
in β + decay / accelerating-colliding machines; (1)
neutrino - travelling in space eg from Sun
or emitted (with electron / positron) in beta decay; (1)
allow ONCE ‘resulting from high energy particle collisions’
any 6
[6]
Lesson 38 answers – beta decay
1.
(i)
leptons;
1
(ii)
neutrino / muon / tau(on);
1
[2]
2.
(i)
weak (force / interaction); (1)
(ii)
3
1H
(iii)
d → u + e + v ; (2)
1
→ 32He + 0–1e + v ; (1)
1
d → u gets 1/2
u
d → u + e / β + v is not in simplest form, so gets ½
d
baryon reaction 10n → 11p + 0–1e + v gets ½
2
[4]
3.
(a)
hadron
baryon
lepton
neutron
proton
electron
neutrino
(b)
4 lines correct 2/2: 3 lines correct 1/2: 2 or 1 line correct 0/2 (2)
2
(i)
10–15 minutes - any value within range (1)
1
(ii)
weak force / interaction (1)
1
(iii)
d → u + e– + ν (-bar)
(u) (u)
2
omits e– or ν loses 1 each (2)
(d) (d)
charge: –1/3 (+ 2/3 –1/3) → 2/3 (+2/3 –1/3) –1 (+0) (1)
baryon number: 1/3 (+1/3 + 1/3) → 1/3 (+1/3 + 1/3) + 0 (+0) (1)
nuclear values:
charge 0 = 1 – 1 (+ 0) and baryon no. 1 = 1 + 0 gets 1/2
2
(i)
arrowed line plus ‘resultant’ / pr label
1
(ii)
anti- (1) neutrino (1) is emitted
carried away some momentum (1)
shows neutrino momentum vector (1)
(iv)
(c)
any 3
3
[12]
Lesson 39 Answers – Radioactive Properties
1.
(a)
(b)
He nucleus, a few cm / 3 to 10 cm
About 1 m / 0.3 to 2 m / several m, 1 to 10 mm Al / 1 mm Pb
(high energy) e-m radiation, 1 to 10 cm of Pb / several m of concrete
only 2 correct 1 mark, only 4 correct 2 marks
B3
Source, absorbers placed in front of detector on diagram
Explanation of how results identify the source
(2 marks possible)
Allowance for background (max 2)
(allow for distance expt to a max 2)
B1
B2
[6]
2.
mass change/charge change/range/speed of emission/monoenergetic v
range of speed/alpha emitted from only high mass nuclei/number of
particles in the decay/other sensible suggestion or further detail
any three
3
[3]
3.
α helium nucleus β electron γ photon/e-m radiation/energy (1)
α charge +(2e) mass 4mp/4u β charge –(e) mass me γ charge 0 mass 0
(2)
α emission energy 3 – 7 MeV β emission energy 1 – 2 MeV γ
emission energy about 1 – 2 MeV or all of the same order of
magnitude/AW (1)
α monoenergetic from given nuclide β range of emission energies from
given nuclide from zero to a maximum γ monoenergetic from given
nuclide or comparison in terms of velocities (1)
α range 3 – 7 cm of air β range 1 – 2 m of air γ range inverse square
law in air/ order of kms (1)
α absorbed by paper β absorbed by thin/ 1 mm Al sheet γ up to cm of Pb
sheet (1)
α strongly ionising β weakly ionising γ hardly ionising at all (1)
any other sensible comparison (1)
6
max 6 marks
Quality of written communication
2
[8]
Lesson 40 answers – Radioactive Decay and Half Life
1.
A: the number of (undecayed) nuclei which decay per second/rate of
decay of nuclei
λ: the probability of a given nucleus decaying in the next second or in unit
time/the (decay) constant relates the activity to the number of undecayed
nuclei
N: the number of undecayed nuclei/nuclei of the original nuclide
(remaining)
1
1
1
[3]
2.
(a)
29; 34
(b)
λ = 0.693/T = 0.693/(120 × 3.2 × 107) = (1.8 × 10–10 s–1) accept ln 2
(c)
(i)
(ii)
(iii)
(iv)
2
1
Q = CV = 1.2 × 10–12 × 90; evidence of calculation (= 1.1 ×
10–10 C)
2
n = Q/e = 1.1 × 10–10/1.6 × 10–19; = 6.9 × 108 allow sig. fig.
variations
2
A = λN; N = 6.9 × 108/1.8 × 10–10; = 3.8 × 1018 using 7.0 gives
3.9
3
1 y is less than 1% of 120 y so expect to be within 1%/
using e–λt gives exactly 1% fall/ problem of random emission
or other relevant statement
1
[11]
3.
(a)
(b)
number of decayed U-238 nuclei = ½ × number of undecayed U-238
nuclei; (1)
so 1/3 of U-238 has decayed and 2/3 of U-238 has not decayed; (1)
(so ratio = 2/3)
either λ = 0.693 / T½ = 0.693 / (4.47 × 109) (= 1.55 × 10–10 y–1)
subs. (1)
N = N0 e–λt so N / N0 = e–λt and ln (N / N0) = –λ t
ln (0.667) = –1.55 × 10–10 t
alg. / arith. (1)
9
so t = 2.61 × 10 y
ans. (1)
or
N / N0 = (½)x so 0.667 = (½)x and ln (0.667) = x ln(0.5)
2
3
and
(c)
either N0 = (5.00 / 238) × 6.02 × 1023
= 1.26 × 1022 atoms
subs. (1)
ans. (1)
2
exponential decay graph for U: starts from N0 and approaches t
axis; (1)
exponential growth of Pb from zero: approaches a constant value of
N0 ; (1)
lines sensibly ‘mirror images’; (1)
3
or
(d)
x = 0.584 then t = x T½ = 0.584 × 4.47 × 109 = 2.61 × 109 y
N0 = (5.00 × 10–3) / (1.67 × 10–27 × 238) (1)
= 1.26 × 1022 atoms (1)
[10]
4.
(i)
(ii)
(take lns of both sides) appreciate ln e–λt = –λt; and ln C/Co = ln C – ln Co
or when multiplying logs add
2
gradient = 0.056 h–1 allow ± 0.002 h–; T = ln2/λ = ln2/gradient =
ln2/0.056 h;
T = 12.4 h allow ± 0.4 h
3
[5]
5.
(a)
(i)
(ii)
(iii)
radioactive implies the emission of ionising radiation (1)
OR emits alpha, beta and gamma radiation (1)
1
nuclide refers to a particular nuclear structure (with a stated
number
of protons and neutrons) (1)
1
half-life is the (average) time taken for the activity to fall to
half its
original value (1)
1
(b)
time / hour
activity of
activity of
activity of
material / Bq
(c)
nuclide X /Bq
nuclide Y /Bq
0
4600
4200
400
6
3713
3334
379
12
3002
2646
356
18
2436
2100
336
24
1984
1667
317
30
1619
1323
296
36
1333
1050
283
(i) and (ii) 2100 as first figure to be filled in for nuclide X (1)
1667 (1)
1050 (1)
idea of subtraction for nuclide Y (1)
correct values for the ones given in nuclide Y column
(1)
5
sensible graph plotted (1)
extrapolation done (1)
value 70 ± 5 hours (1)
3
OR
A = A0 e–λt (1)
ln A = ln A0 – λt
e.g. when A = 296, t = 30 h
5.6904 = 5.9915 – λ × 30 (1)
0.3011/30 = 0.01004 = λ
τ = ln 2/ λ = 69.0 h answers will vary slightly dependent on starting
and
finishing times (1)
(d)
separate the two nuclides (before starting the count) (1)
by chemical means (if possible) (1)
OR
using a centrifuge or diffusion (if isotopes)
OR
sensible idea about shielding against one of the emitted particles
(e)
decay constants or half lives are different (1)
half-life at the start is approximately that for X (1)
X decays more rapidly than Y so after a long time the half-life is that
for Y (1)
in between it has a value intermediate between the two (which
varies) (1)
MAXIMUM 3
OR
dealt with mathematically, along the lines of
2
3
two separate exponential decays (1)
when added together do not give an exponential graph (1)
with back up maths (1)
[16]
Lesson 41 answers - Change in energy: Change in mass
Calculating using Erest = mc2
These questions show you how to calculate changes in energy from changes in mass, using
2
Einstein’s relation Erest = mc linking the rest energy of a particle to its mass.
Transmutation of chemical elements
The dream of the ancients was alchemy: turning base metals into gold. Although this is
chemically impossible, at the end of the nineteenth century radioactivity was discovered by
Henri Becquerel. When alpha and beta radiation are emitted atomic nuclei are ‘transmuted’
from one element to another. For example:
238
92uranium
 24 alpha 234
90 thorium.
In 1932 using protons (hydrogen nuclei) accelerated through a potential difference of 800 000
V, two English physicists, Cockcroft and Walton, carried out the first artificial transmutation: by
bombarding lithium with the protons they produced two helium nuclei:
1
7
4
1H  3Li  2
He  24He.
Change in mass
Notice that in both these reactions the mass number and charge (proton number) are
conserved. Energy, however, is only conserved if you take account of changes to the rest
energy – in effect of changes to the masses – of the particles.
In Cockcroft and Walton’s experiment, the masses of the particles are:

H: 1.0073 atomic mass units

Li: 7.0160 atomic mass units

He: 4.0015 atomic mass units.
An atomic mass unit, symbol u, is equal to 1.6605  10
1
–27
kg.
Show that the mass decreases in this reaction.
1
7
4
1H  3Li  2
He  24He.
Calculate m in atomic mass units and in kilograms.
Change in energy
2
The energy of the protons was 800 000 electron volts (800 keV). The lithium was in
solid form so the nuclei would only have been vibrating due to thermal energy, less
than an electron volt.
The reaction was captured in this photograph:
Two pairs of alpha particles, emerging in opposite directions, can be seen in the
photograph.
From the range of the tracks through the cloud chamber the energy of the alpha
particles was measured to be 8.5 MeV each.
Show that the total kinetic energy of the particles increases, and calculate E in MeV
and in joules.
3
If the increase in kinetic energy comes from the decrease in rest energy you should
2
expect E = mc . Calculate the ratio of the change in kinetic energy to the change in
–1
mass E/m in J kg .
4
Show that the value of the ratio E/m is approximately consistent with the
2
relationship E = mc .
16
–1
The large value of c 2 (9  10 J kg : use this value from now on in calculations) means that
a small change in mass represents a vast change in rest energy. This relationship between
2
mass and energy is why particle physicists measure masses in MeV / c ; any unit of energy
2
divided by c is a unit of mass.
Creating massive particles
Energy is ‘materialised’ in matter–antimatter production. A photon of electromagnetic radiation
can produce an electron and a positron. In this case, the energy of the photon vanishes and
the rest energy of the particles appears. (This reaction needs to take place near to the
nucleus of a heavy atom to conserve momentum but this is not going to affect your
calculations here.)
In this bubble chamber photograph a photon enters from the bottom. It is uncharged and so
produces no observable track. After some distance the photon disappears and produces the
electron–positron pair. These two charged particles ionise the liquid in the chamber and
bubbles form near the ions and are photographed.
In this case the chamber is filled with liquid hydrogen mixed with liquid neon. It is held under
pressure which is released just as the particles enter the chamber to encourage bubbles to
form and enlarge near the ions.
5
The bubble chamber is in a magnetic field, so charged particles bend due to the force
Bqv on a moving charge. How does the photograph show that the two particles have
opposite charges?
6
The mass of the electron is 5.5  10 u. What is the minimum energy photon that will
produce an electron–positron pair? From what part of the electromagnetic spectrum
–34
–1
is this? (Planck constant h = 6.63  10 J Hz .)
–4
Nuclear binding energy
If protons and neutrons (together known as nucleons) are bound together in a nucleus, the
bound nucleus must have less energy than the nucleons of which it is made. That is, the rest
energy of the nucleus must be less than the sum of the rest energies of its nucleons. In turn,
this means that the mass of the nucleus must be less than the sum of the masses of its
nucleons.
The simplest compound nucleus is the deuteron, the nucleus of hydrogen-2. It consists of a
proton and a neutron bound together by the strong nuclear force. The masses of these
particles are:

proton: 1.0073 u

neutron: 1.0087 u

deuteron: 2.0136 u.
7
Calculate the difference in mass between a deuteron and one proton and one neutron.
8
Calculate the binding energy of the deuteron in J and in MeV.
9
Calculate the binding energy per nucleon of the deuteron.
10
Express the difference in mass as a percentage of the sum of the masses of the
proton and neutron.
Mass change in nuclear fission
A possible reaction for the nuclear fission of uranium-235 is:
235
1
133
99
92 U  0 n 51 Sb 41 Nb
 410 n.
The masses of the particles are

U-235 = 235.0439 u

Sb-133 = 132.9152 u

Nb-99 = 98.9116 u

neutron (n) = 1.0087 u.
11
Show that the energy change per atom of uranium is about 200 MeV and calculate
m/m.
Summary
2
Einstein's famous equation Erest = mc reveals a Universe that is not as simple as it seems at
first sight. The mass of a particle is generally a very large part of its total energy. The
existence of rest energy was not suspected until after Einstein had predicted it, because the
change in mass is usually so small, because changes in energy are usually a small fraction of
the rest energy. Only in nuclear reactions where m/m ~ 0.1% or more are you able to see
the change in mass, accompanied by what appears to be a huge change in energy.
Hints
1
Compare masses of H plus Li with mass of two He nuclei.
2
Two 8.5 MeV alpha particles come out, but one 800 keV proton goes in.
3
Compare the answers to questions 1 and 2.
4
c2!
Don’t expect to get exactly the speed of light. Remember to take the square root of
5
What is the difference between forces F and - F?
6
Start with the mass of an electron in atomic mass units. Convert to kilograms. Write
down the mass of an electron–positron pair. Use Erest = mc2 to get the rest energy of
the pair in joules. Then use E = hf.
7
Do this one in the same way as question 1.
8
Erest = mc2 again. But now use the electron charge to get to electron volts and MeV.
9
How many nucleons in a deuteron?
10
Best to take the difference as a fraction of the mass before.
11
Add up before and after masses in atomic mass units first. Don’t forget there’s one
extra neutron to start with and four extra neutrons afterwards. Then convert mass
changes first to joules and then to MeV.
Answers and worked solutions
1.
Mass of H plus Li  1.0073 u  7.0160 u  8.0233 u
Mass of two He  2  4.0015 u  8.0030 u
Difference m  8.0030 u  8.0233 u  0.0203 u
So we can find the mass difference in kg:
m  0.0203 u 1.660510 27 kg  3.370810 29 kg
2.
Increase in energy:
E  2  8.5 MeV  0.8 MeV  16.2 MeV
In joules:
E  (16.2 106 eV)  (1.6 10 19 J eV 1 )  2.6 10 12 J
3.
2.60 10 12 J
E

 7.7 10 16 J kg 1 .
m 3.37 10  29 kg
2
2
16
–1
8
–1
4.
If E = mc , then c = 7.7  10 J kg , so c = 2.8  10 m s .
5.
The force on a moving charged particle is Bqv. If the charge q changes sign, the
direction of the force is reversed, so the curvature is opposite.
6.
The mass of an electron or positron is equal to:
(5.5  104 u)  (1.66  1027 kg)  9.1 1031 kg.
2
From Erest = mc , the rest energy of an electron–positron pair is:
Erest  2  9.110 31 kg  (3 108 m s 1 ) 2  1.6 10 13 J.
If this energy is supplied by a photon of energy E = hf, then:
f 
1.6 10 13 J
6.63 10 34 J Hz 1
 2.5 10 20 Hz.
This is the frequency of a gamma ray.
7.
The mass difference is:
2.0136 u  (1.0073 u  1.0087 u)  0.0024 u.
I
n kg the mass difference is:
 0.0024 u  (1.66  1027 kg)  3.98  1030 kg.
8.
Binding energy  3.98  10 30 kg  (3  108 m s 1)2
 –3.58  10 13 J
–
3.58  10 13 J
1.6  10 19 J eV 1
 –2.2 MeV .
 –2.2  10 6 eV
9.
The deuteron has two nucleons so the binding energy per nucleon is
–2.2 MeV / 2 = –1.1 MeV.
10.
As a percentage the mass difference is equal to:
0.0024 u
 1.2 10 3 100  0.1% (approximat ely )
1.0073 u  1.0087 u
11.
Mass after  132.9152 u  98.9116 u  (4 1.0087 u)  235 .8616 u
Mass difference  236.0526 u  235 .8616 u  0.191 u.
Change in rest energy 
0.191 u  (1.66 10 -27 kg )  (3 10 8 m s 1 ) 2
1.6 10
19
J eV
1
The ratio is given by:
m / m  0.191 u / 236 u  8.110 4 ~ 0.1%.
External reference
This activity is taken from Advancing Physics chapter 18, 200S
 1.78 10 8 eV  178 MeV.
Lesson 42 - Fusion answers
1
a) 221H = 32He + 10n
b)
mass change in fusion of 221H= 0.0035u = 5.8x10-30kg
Energy release = 5.8x10-30kg x 9x1016
= 5.2 x 10-13J per fusion of 4.0 u
4u=6.6 x10-27kg
Energy release per kg = 5.2 x 10-13J/6.6 x10-27kg
= 7.9x1014J/kg
enough for a town of 20000 for a lifetime!!
2a) X is a positron
b) Y is a 32He
c) Z is a proton
d) 611p = 42He + 211p + 20+1e
or 411p = 42He + 20+1e