Things that don’t change
Question 10S: Short Answer
1. The upper numbers (nucleon number) add up to the same on the left and right: 1 = 1 + 0 + 0.
2. The lower numbers (charge number) add up to the same on the left and right: 0 = 1 + (– 1) + 0.
3. The upper (nucleon) number is zero for both (even though an electron has some mass).
4. The reaction produces one lepton and one antilepton. The electron has lepton number 1. The
antineutrino has lepton number – 1. Together, they add to zero. The total lepton number is zero
before and after.
5. The upper numbers (nucleon number) add up to the same on the left and right: 0 + 1 = 1 + 0.
6. The lower numbers (charge number) add up to the same on the left and right: – 1 + 1 = 0 + 0.
7. An electron (a lepton) vanishes on the left. A neutrino (also a lepton) is created on the right. The
total lepton number remains equal to 1. If the neutrino were an antineutrino the lepton number
would go from + 1 to – 1, and not be conserved.
8. The total nucleon number 49 remains constant: 49 = 49 + 0 + 0.
9. The total nucleon number 62 remains constant: 62 = 62 + 0 + 0.
10. The electric charge on the nucleus is decreased, the charge being carried away by the positron:
26 = 25 + 1 + 0.
11. The positive electric charge on the nucleus increases, a negative charge being carried away by
the electron: 26 = 27 + (– 1) + 0.
12. A lepton–antilepton pair is produced, so the lepton number does not change.
13. The antineutrino ‘cancels’ the increase in lepton number due to producing an electron.
14. The charge on the nucleus changes. So the number of electrons around the nucleus changes,
altering the chemical properties of the atom.
Beta decay and conservation
Question 20S: Short Answer
1. Charge on left-hand side = 6e. Charge on right hand side = 7e – 1e = 6e.
2. Number of nucleons on left-hand side = 14 = 6 protons + 8 neutrons. Number of nucleons on
right-hand side = 14 = 7 protons + 7 neutrons.
3. The electron is created as a neutron changes into a proton in the nucleus.
4. m = 13.999948 u – [13.999231 u + 0.000549 u] = 0.000168 u
5.

E  0.000168 u  1.66  10 27 kg u 1  3  10 8 m s 1
6.
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
2
 2.5  10 14 J
E 
2.5  10 14 J
1.6  10
19
J eV
1
 1.56  10 5 eV  160 keV
7. If the same energy is released in each decay then all the electrons should leave with the same
energy. They don’t. If an electron leaves with less than 160 keV then either the law of
conservation of energy does not hold or something else has the energy.
8. The particle must be electrically uncharged, and difficult to detect. It must have very small mass,
since some electrons take away almost all of the total energy. The unseen particle must have
energy and momentum (and also spin, but that is not obvious from the information given in the
question).
9. This when the unseen particle has very nearly zero energy and the electron has almost all of the
available 160 keV. It cannot get more than this because this is the maximum energy released in
the decay.
10. The unseen particle carries away 160 eV – 100 eV = 60 eV.
Creation and annihilation
Question 30S: Short Answer
1.
E rest  mc 2

E rest  9 .11  10  31 kg  3.00  10 8 m s – 1

2
 8 .2  10 14 J.
2. Energy in eV is energy in joules divided by 1.60  10–19 J eV–1:
8.2  10 14 J
 5.1  10 5 eV  0.51 MeV
1.60  10 19 J eV 1
3. 0.51 MeV since electrons and positrons have identical mass.
4. 1.02 MeV. Two particles, each of rest energy 0.51 MeV, are annihilated.
5. 0.51 MeV each. The two photons share the total 1.02 MeV.
6. 1.02 MeV. The photon has to create the rest energy of two particles, each 0.51 MeV.
7. Approximately 104 pairs. Each pair requires approximately 1 MeV = 1  106 eV and 10 GeV = 10
9
 10 eV is available.
8. The 1.28 MeV photon has enough energy to create an electron–positron pair. The positron from
the decay can annihilate with any nearby electron.
9. Approximately 2 GeV. The rest energy of a proton–antiproton pair is 2000 times greater than the
roughly 1 MeV rest energy of an electron–positron pair.
10. Their 0.51 MeV energy is too small (minimum 1.02 MeV).
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Rutherford scattering:
Energy and closest approach
Question 70S: Short Answer
1. The units are:
CC
C 2 J 1 m 1  m
= J.
2. In the expression
electrical potential energy =
2e  Ze
4 0 r
dividing by e gives
2Ze
4 0r
for the energy in eV. Multiply by 10–6 to get the energy in MeV.
3. Substituting values gives
2  79  1.6  10 19 C
 10 6
EP =
12
2 1
1
14
4  8.85  10
C J m  1.0  10
m
 22.7 MeV.
4. Halves, because the potential energy is proportional to 1 / r .
5. About 4.6  10–14 m, where the graph reaches 5 MeV.
6. About 8.0  10–14 m.
7. About 12 MeV.
8. The charge on the nucleus is smaller, so the potential energy is smaller in the same ratio.
9. The ratio of the charges, 79 / 50 = 1.58.
10. About 3  10–14 m.
11. About 0.75  10–14 m.
12. The alpha particle approaches to about 2.5 times the radius of the nucleus. Attractive forces
between nucleons might begin to be important, and modify the scattering.
13. Inserting values:
82  82  1.6  10 19 C
EP =
 10 6
12
2 1
1
14
4  8.85  10
C J m  1.0  10
m
 967 MeV.
Rutherford scattering:
Directions of forces
Question 80S: Short Answer
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1. The repulsive force is along the line joining the alpha particle and the nucleus.
Path of alpha particle scattered by nucleus
C
B
FA
A
nucleus
2. The repulsive force is along the line joining the alpha particle and the nucleus.
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Path of alpha particle scattered by nucleus
C
FB
B
FA
A
nucleus
3. The force at B is four times as large as the force at A, because the distance is halved and the
force varies as 1 / r 2.
4. The force on the nucleus is equal and opposite to the force on the alpha particle. But because the
nucleus is much more massive, it recoils only slightly.
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Path of alpha particle scattered by nucleus
C
FB
B
FA
A
nucleus
FN
5. Particle moves slowest at B, because this is the distance of closest approach, the particle has
been decelerating due to repulsive force. After B it accelerates away.
6. Here the particle is furthest from the nucleus, the alpha particle has been accelerated between B
and C and is therefore going fastest at C.
7. The alpha particle path is symmetrical about the line from the nucleus to B. So the net change of
momentum of the nucleus is along this direction.
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Path of alpha particle scattered by nucleus
C
FB
B
FA
nucleus
A
recoil
8. Uphill, because the particle is approaching the nucleus but being pushed away from it.
9. Along contour, because the particle is travelling at right angles to the direction of the force on it.
10. Downhill, because the particle is travelling away from the nucleus and is being pushed away from
it.
11. Potential at B larger than potential at A by a factor of 2, because the distance is halved and the
potential varies as 1 / r .
Electrons measure the size of nuclei
Question 90S: Short Answer
1. sin 30 = ½ =  / d approximately.
2.
sin   1.22

d
1.22
4
 0.305.
The angle whose sin is 0.305 is 17.8.

3.
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Energy  100 MeV
 10 8 eV
 10 8 eV  (1.6  10 19 J eV 1 )
 1.6  10 11 J.
4.
p

E
c
1.6  1011 J
3.0  108 m s 1
 0.53  1019 kg m s 1.
5.


h
p
6.6  10 34 J s
0.53  10 19 J m 1 s
 12  10 15 m.
6. Ratio of wavelength to radius = 12  10–15 m / 1.2  10–15 m = 10; ratio of wavelength to diameter
= 5.
7. Substitute
E
p
c
in
h

p
gives
hc

.
E
8. From
hc

E
 is inversely proportional to E. Since  for 100 MeV is 12  10–15 m then  for 400 MeV is 3 
10–15 m.
9. Since sin  = 0.67, then 0.67 = 1.22  / d , and  / d = 0.55.
10. Since  for 400 MeV is 3  10–15 m, and  / d = 0.55, then d = 5.5  10–15 m and r = 2.7  10–15 m.
11. (2.7 / 1.2)3 = 11.4. It would be closer to 12 but for rounding errors.
12. Diameter of the uranium-238 nucleus = 15  10–15 m. Choosing = 2 d gives  = 30  10–15 m,
about five times the wavelength of 400 MeV electrons (3  10–15 m). So scale down energy by a
factor 5, to say 80 MeV.
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Scattering and scale
Question 100S: Short Answer
1.
Energy  1 GeV
 10 9 eV
 10 9 eV  (1.6  10 19 J eV 1 )
 1.6  10 10 J.
2.
p

E
c
1.6  10 10 J
3.0  10 8 m s 1
 5.3  10 19 J m 1 s.
3.


h
p
6.6  10 34 J s
5.3  10 19 J m 1 s
 1.2  10 15 m
or approximately 1 fm.
4. Units of h c / E are
(J s)  (m s 1 )
 m.
J
5. Down to about 1 fm.
6. Yes, because the nucleus is of the order 1 fm in size, and the wavelength is about 1 fm.
7. The energy is 100 times larger, so the scale is 100 times smaller, 0.01 fm.
8. The scale is 105 times smaller than a nucleus, so the energy is 105 times larger, or 105 GeV. This
is not practicable in accelerating machines. However, cosmic rays reach 1011 GeV.
9. A good answer would take several examples of big accelerators, and compare the energies they
use with the scales they probe. You might be able to find information about the cost of such
accelerators.
Putting quarks together
Question 110S: Short Answer
1. uuu has charge
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
2
3
e
2
3
e
2
3
e  2e.
2. uud has charge

2
3
e
2
3
e
1
3
e  1e.
3. udd has charge

2
3
e  31 e  31 e  0.
4. ddd has charge
 31 e  31 e  31 e  1e.
5. Proton has charge +1 e so could be uud.
6. Neutron has charge 0 so could be udd.
7. – has charge –1 e so could be ddd.
8. ++ has charge +2 e so could be uuu.
9. A down quark, charge –1/3 e changes to an up quark, charge +2/3 e, so the charge carried away
must be –1 e if charge is to be conserved.
10. u u has charge +2/3 e –2/3 e = 0.
11. d d has charge –1/3 e + 1/3 e = 0.
12. u d has charge +2/3 e + 1/3 e = 1 e.
13. d u has charge –1/3 e – 2/3 e = –1 e.
14. The + with charge + 1 e could be u d .
15. The – with charge – 1 e could be d u .
16. The neutral 0 could be either or both of u u and d d. In fact it is an equal mixture of the two.
Spectra and energy levels
Question 150S: Short Answer
1. The electron is always bound to the nucleus, so energy must be provided to remove it. If the
energy of a free electron is zero, the energy of a bound electron is less, therefore negative.
2.
7.86  10 19 J
1.6  10 19 J eV 1
 4.91 eV.
3.
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E  hf
hc



(6.63  10 34 J Hz 1 )  ( 3.00  10 8 m s 1 )
253  10 9 m
 7.86  10 19 J.
The jump is from level at 7.86  10–19 J to the lowest level.
4.
E  hf
 8.84 eV  6.70 eV
 2.14 eV
 3.42  10 19 J
(multiply by e).
f 
3 . 42  10  19 J
E

h
6 . 63  10  34 J Hz 1
 5 . 16  10 14 Hz .
Wavelength  
3 .00  10 8 m s 1
c

f
5 .16  10 14 Hz
 581 nm.
5. Longest wavelength corresponds to smallest photon energy.The smallest energy difference is
5.46 eV – 4.91 eV = 0.55 eV. The electron must jump up in energy if a photon is absorbed.
6.
Frequency f 
c 3.00  108 m s 1


591 109 m
 5.08  1014 Hz.
Energy E  hf
 (6.63  10 34 J Hz 1 )  (5.08  1014 Hz)
 3.36  10 19 J
 2.10 eV
(divide by e). The lower level = – 5.14 eV, so the upper level = – 5.14 eV + 2.10 eV = – 3.04 eV.
The energy is negative because in all these levels the electron is bound.
7. The spectral line is split into a closely spaced pair (doublet)
8. The best graph to plot is of 1 /  En against n. This spaces the points equally and should give a
straight line.
9. Longest wavelength is the smallest difference, from n = 2 to n = 1. The energy difference is:
 5.45  10 19 J  (21.8  10 19 J)  16.35  10 19 J.
The wavelength is:
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c
f
hc

E
(6.63  10 34 J Hz 1 )  (3.00  10 8 m s 1 )

16.35  10 19 J
 120 nm
which is less than the 400 nm which is roughly the boundary between visible and ultraviolet.

10. To shift from 120 nm to more than 400 nm needs a red-shift of at least 400 nm / 120 nm = 3.3.
11.
Frequency f 
c 3.00  10 8 m s 1


486  10  9 m
 6.17  10 14 Hz.
Energy E  hf
 (6.63  10 34 J Hz 1 )  (6.16  1014 Hz)
 4.09  10 19 J
 2.55 eV
(divide by e). This is equal to the gap between the n = 4 and n = 2 levels, i.e. – 0.85 eV – (– 3.40
eV) = 2.55 eV.
How small could a hydrogen atom be?
Question 160S: Short Answer
1.
EP 
e2
 e2
 9.0  10 9 J C 2 m 
r
40 r
 9.0  10 9 J C 2 m 
1.6  10
19
C

2
1.0  10 10 m
 23  10 19 J.
2. As for question 1 with a radius 10 times smaller, so the energy is 10 times larger: EP = – 230 
10–19 J.
3. The electrical binding is stronger at the smaller radius (more negative).
4. The waves have half a wavelength fitting into the diameter, so  = 4.0  10–10 m.
5. The radius is 10 times smaller so the wavelength is 10 times shorter:  = 0.4  10–10 m.
6. More half-wavelength loops have to be fitted into the same length, so the wavelengths will all be
smaller.
7.
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mv 

h

6.6  10 34 J s
4.0  10 10 m
 1.65  10 24 kg m s 1.
8. As for question 7 with the wavelength 10 times smaller, so the momentum is 10 times larger: m v
= 16.5  10–24 kg m s–1.
9. The smaller the space the shorter the wavelength. But the shorter the wavelength the greater the
momentum, since m v = h / .
2
1
10. E K  2 mv .
Since
mv 2 
then
EK 
11.
mv 2
m
1 mv 
.
2 m
2

1 1.65  10 24 kg m s 1
EK 
2
9.1 10 31 kg

2
 15  10 19 J.
12.
EK

1 16.5  10 24 kg m s 1

2
9.1  10 31 kg

2
 1500  10 19 J.
The momentum is 10 times larger, so the momentum squared is 100 times larger, making the
kinetic energy 100 times larger.
13. The total energy is:
EP  EK  23  10 19 J  15  10 19 J
 8  10 19 J.
If the total energy is negative, the electron is bound (energy must be supplied for it to escape).
14. The total energy is:
EP  EK  230  10 19 J  1500  10 19 J
 1270  10 19 J.
15. The total energy of the rocket is positive, if it can escape.
16. For the smaller atom, radius r = 0.1  10–10 m, the total energy is positive, so there is enough
kinetic energy for the electron to escape. The electron is not bound. The larger atom, r = 1.0 
10–10 m, has a negative total energy, so energy must be supplied to remove the electron, and the
electron is bound.
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Carrots and guitar strings
Question 170S: Short Answer
1.
p
h

and
n 
2L
n
so
pn 
2.
nh
.
2L
p  mv
so
p 2  m 2v 2 .
Dividing both sides by 2 m gives
1 p2

2 m
1
2
mv 2  EK .
3. From question 2:
1 p2
2 m
and from question 1:
nh
pn 
.
2L
Thus the energy En of the nth level is given by:
EK 
En 

1 p n2
2 m
1
2
n2
h2
4L2 m
.
4. Inserting values in the equation from question 3 gives:
E n  21 n 2
6.6  10 J s
4  10 m  9.1 10
34
9
 n  0.6  10
2
5.
19
2
J.
The energies are:
n  1 : E1  0.6  10 19 J
n  2 : E 2  2.4  10 19 J
n  3 : E 3  5.4  10 19 J
n  4 : E 4  9.6  10 19 J.
6.
For  = 400 nm:
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2
31
kg
E  hf
hc



(6.6  10 34 J s)  (3  10 8 m s 1 )
400  10 9 m
 5  10 19 J (approx.).
For  = 700 nm:
(6.6  10 34 J s)  (3  10 8 m s 1 )
E 
700  10 9 m
 3  10 19 J (approx.).
7. The transitions from 1–3 (4.8  10–19 J), 2–3 (3.0  10–19 J) and 3–4 (4.2  10–19 J) all fall in the
energy range of photons of visible light.
8. Yes, at least some transitions fall in the visible range, so according to this model carotene can
absorb visible light.
The hydrogen spectrum
Question 180S: Short Answer
1.
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0
E6 = –0.38 eV
E5 = –0.54 eV
E4 = –0.85 eV
E3 = –1.50 eV
E2 = –3.40 eV
E1 = –13.60 eV
2. E = E4 – E2 = –0.85 eV – (–3.4 eV) = 2.6 eV. In joules E = 2.6 eV  1.6  10–19 J eV–1 = 4.0 
10–19 J. Using E = hf , the frequency f = 4.0 10–19 J / 6.6  10–34 J s = 6.0  1014 Hz and  =
3.0 108 m s–1 / 6.0  1014 Hz = 5.0 10–7 m. The photon lies in the visible part of the spectrum.
3.
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0
E6 = –0.38 eV
E5 = –0.54 eV
E4 = –0.85 eV
E3 = –1.50 eV
E2 = –3.40 eV
E1 = –13.60 eV
4.
fn 
13.6 eV  1
1 
 2  2
h
n 
2
5. k = 13.6 eV / h = 13.6 eV  1.6 10–19 J eV–1 / 6.6  10–34 J s = 3.3  1015 Hz
6. Ultraviolet photons have a higher frequency and must therefore have a higher energy than visible
photons. The energy jumps that created them must be bigger. They correspond to jumps from the
nth level down to the ground state (n = 1)
fn 
13.6 eV 
1 
1  2 
h
 n 
7.
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0
E6 = –0.38 eV
E5 = –0.54 eV
E4 = –0.85 eV
E3 = –1.50 eV
E2 = –3.40 eV
E1 = –13.60 eV
Particles and interactions
Question 40M: Multiple Choice
1. Photons – the carriers of the electromagnetic interaction.
2. Bosons – carriers of all interactions are bosons with spin a multiple of 1, as opposed to fermions
with spin a multiple of ½.
3. Positrons – or antielectrons, with charge + e.
4. Leptons – the ‘light’ particles which do not feel the strong interaction.
5. Neutrinos – notice that ‘photons’ would be as good an answer if bosons were not excluded.
6. Fermions – more than one of these ‘matter-like’ particles cannot occupy any particular quantum
state.
7. Nucleons – a collective term is needed for protons and neutrons because they are alike as far as
the strong nuclear interaction is concerned.
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8. Nucleon, lepton – the number of protons plus neutrons is the same; the emission of a lepton is
accompanied by the emission of an antilepton (e.g. electron, antineutrino).
9. Neutrino, lepton – the positron is an antilepton, and the neutrino is a lepton, so the change in
lepton number is zero.
10. Photons – which interact with charged particles.
11. Antiparticle – it can be any such pair, even quark–antiquark.
12. Photons – the photons can then create almost anything for which there is enough energy.
13. Fermions – they all have spin ½ and obey the Pauli exclusion principle.
14. Photons – the charge on the electron can be interpreted as the quantum amplitude for interaction
of photons and electrons.
15. Proton, electron, neutrino – the proton conserves nucleon number. The electron and antineutrino
together conserve lepton number.
16. Antiparticles – antiparticles have charge opposite to that of their particle partners.
Keeping momentum and energy unchanged
Question 60M: Multiple choice
1. B, D, F, H. Initial momentum is zero in all pictures. Final photon momentum is non-zero in the
horizontal direction in A, C, E, F.
2. All. Either the particles are at rest (A, B, C, D) or have equal and opposite velocities (E, F, G, H).
3. B, D, F, H. The photon momenta are equal and opposite.
4. A, B, C, D. Where the particle is at rest.
5. A, C, E, G. The photons are in opposite vertical directions but the same horizontal direction.
6. C, G. The photons both travelled to the left while the initial momentum was zero.
7. None. There is always zero initial vertical momentum. After, the photons all have equal and
opposite vertical components of momentum.
8. B, D, F, H. Momentum is conserved. Energy is conserved if the photons have the right energy.
Charge is conserved.
9. B, D. The initial energy is all rest energy, so the energy of the photons will be equal to the rest
energy.
10. F, H. The initial energy exceeds the rest energy, by the kinetic energy of the particles. The
photons can acquire the total energy available.
11. All. A negative charge is annihilated by a positive charge.
Creation from annihilation
Question 50C: Comprehension
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1. Positrons: are antiparticles of electrons, they have the same mass but positive charge.
2. Antiparticles include the fact that antiparticles have all properties except mass opposite to those
of their particle counterparts.
3. Antimatter: is made of antiparticles. No bulk antimatter is known in the Universe, but antiparticles
can be produced in accelerators.
4. Mutually annihilate: the opposite properties of particle and antiparticle all cancel, and the rest
energy (mass) goes into kinetic energy of a photon (which has zero mass).
5. Erest = m c2 : you could show that the mass of an electron corresponds to 0.5 MeV approximately.
Or give an example from nuclear reactions.
6. It can return whence it came, into electron and positron: a photon (near a nucleus) can
create an electron–positron pair.
7. Emerge from the encounter: pairs of particles and antiparticles emerge.
8. The heat of stars: because the stars are hot, particles have a lot of random thermal energy per
particle. This may be enough to cause an event in a collision.
9. Temporarily simulate that heat: accelerated particles are collided, so that their kinetic energy is
available in the collision event.
10. Primordial Big Bang: the hypothetical origin of the Universe, when space and time were created
and a hot fireball of space, time and matter exploded.
11. By annihilating electrons and positrons in the laboratory: the best way is to make
accelerated particles collide head on, to get the maximum energy into the products of the
collision.
Finding parts of protons
Question 120C: Comprehension
1. Stanford Linear Accelerator (SLAC): A good answer will have a labelled sketch showing a
series of electrodes with an alternating accelerating p.d. across them in pairs. The electron gun in
a cathode ray tube is a linear accelerator – you should explain how it accelerates electrons.
2. Energies up to 20 GeV: 20 GeV = 2  1010 eV = 3.2  10–9 J.
3. The electron exchanges a virtual photon with a charged particle: Virtual photons ‘exist’ only
inside an interaction. They are not part of what goes in or comes out. The force between electron
and proton in an atom can also be thought of as the exchange of a virtual photon. From the
modern point of view, virtual photons are the carriers of the electromagnetic force.
4. The photon transfers energy and momentum: The equation momentum = m v does not apply
except approximately at low velocities. The momentum p of any particle is given by
2
2
E rest
 E total
 pc  .
For photons the mass and so the rest energy is zero. Thus E = p c and p = E / c.
2
5. Created in particle–antiparticle pairs: Electron–positron annihilation is used in PET brain
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scans.
6. Triplets of particles called quarks: proton = uud (charge = + 2/3 e + 2/3 e – 1/3 e = 1 e);
neutron = udd (charge = + 2/3 e – 1/3 e – 1/3 e = 0).
7. Energy equal to thousands of times their rest energy or mass: The rest energy Erest = m c 2
of an electron is about 0.5 MeV. 20 GeV is 40 000 times the rest energy.
8. Time dilation: In relativity, time as seen for an object moving relatively to you seems to run slow.
The time dilation factor in this case is 40 000 (the ratio total energy / rest energy). So the motion
of the quarks really is seriously slowed down!
9. Feynman had seen that it all boiled down to hitting just one of them at a time: Rutherford
scattering is the interaction of one alpha particle with one nucleus. To achieve this the target foil
must be very thin.
10. At least you don’t persuade yourself that you know something you don’t really know : The
term ‘electric current’ was invented when people didn’t know what flowed.
How Niels Bohr began quantum theory
Question 140C: Comprehension
1. The energy of the photon is E = h f. This must be the difference between energies E2 and E1.
Thus h f = E2 – E1, so that f = (E2 – E1) / h.
2. The difference in energy between two levels is h f. Thus Balmer’s equation can be written h fn = h
 constant (1 / 22 – 1 / n2). This can be seen as the difference between two states of energy En =
– h  constant / n2 and energy E2 = h  constant / 22.
3. The electrical potential energy is
e2
.
4 0 r
It is negative because energy must be supplied to take the electron further away from the proton,
to a large distance where the potential energy is taken to be zero.
EP  
4.
Force 
e2
4 0 r 2
and
mv 2
.
r
Putting these equal gives
force 
mv 2
e2

.
r
40 r 2
Multiply both sides by r to get
mv 2 
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e2
.
4 0 r
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5. The kinetic energy is ½ m v2. Since
e2
4 0 r
the kinetic energy
mv 2 
1 e2
  21 EP .
2 4 0 r
Thus adding kinetic and potential together gives
E  EK  EP
EK 
  21 EP  EP
 21 EP

1 e2
.
2 40 r
6. In the expression
E
1 e2
2 4 0 r
substitute r = k n2. This gives
E
1
e2
2 4 0 kn 2
with E proportional to 1 / n2.
7. Bohr assumed that
nh
mvr 
.
2
Dividing both sides by r gives
nh
.
mv 
2r
Squaring both sides gives
2
 nh 
(mv ) 2  
 .
 2r 
Dividing both sides by 2m gives the kinetic energy
2
EK 
1
2
mv 2 
1  nh 

 .
2m  2r 
8. Insert values e = 1.6  10–19 C, m = 9.1  10–31 kg, h = 6.6  10–34 J Hz–1 and 0 = 8.9  10–12 C2
J–1 m–1, and n = 1 in the equation
n 2h 20
r 
me2
giving:
r 

12  6.63  10 34 J Hz 1
2
 8.85  10 12 C 2 J 1 m 1

  9.11 10 31 kg  1.60  10 19 C
 0.53  10 10 m.
9.
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
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
2
En  


2
1.60  10 19 C
1
2 4 8.85  10 12 C 2 J 1 m 1 n 2 (0.53  10 10 m)

 21.7  10
19

2
J/ n .
Dividing by e = 1.6  10–19 J gives the energy in electron volts, – 13.6 eV / n2. For n = 1 the
energy is – 13.6 eV. For n = 2 the energy is – 13.6 eV / 4 = – 3.4 eV. For n = 3 the energy is –
13.6 eV / 9 = – 1.5 eV.
Products of the Big Bang
Question 80D: Data Handling
1. For the He I line:
f 
3 .00  10 8 m s 1
v


447  10  9 m
 6.71  10 14 Hz
so
E  hf
 (6.63  10 34 J s)  (6.71 1014 Hz)
 4.44  10 19 J
 4.44  10 19 J / 1.60  10 19 J eV 1
 2.78 eV.
For the H line:
f 
3 . 00  10 8 m s  1
v


486  10  9 m
 6 . 17  10 14 Hz
so
E  hf
 (6.63  10 34 J s)  (6.17  1014 Hz)
 4.09  10 19 J
 4.09  10 19 J / 1.60  10 19 J eV 1
 2.56 eV.
2. Darkening of the lines, relative to the adjoining spectral regions, is proportional to the number of
photons absorbed, so the darker hydrogen lines show that more hydrogen atoms had absorbed
photons, so there must be more hydrogen atoms present.
3. The density of darkening on the image can be measured, probably with a computer program such
as Scion Image. The total area of blackening, as well as the pixel values of the black image, will
need to be measured.
4.
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0.85 eV  ( 3.40 eV )  2.55 eV
 2.56 eV
within rounding error, and similarly
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0.84 eV  ( 3.63 eV )  2.79 eV
 2.78 eV
within rounding error.
5. The transition involved is from the first excited state, so the hydrogen atoms must be in that state
to start with.
6. There are fewer electrons in the first excited state; for cooler atoms this must mean that they are
in a lower state (the ground state).
7. There are fewer electrons in the first excited state; for hotter atoms this must mean that they are
in higher states.
8. 7 × 2 = 14
9. 14 – 2 = 12
10. 4 / (14 + 2) = 4 / 16 = 0.25 = 25%.
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