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PHYS3105 / PHYS6105, Semester 1, Physics of Matter
This paper is for all students.
Location: Physics studio, Building 38A.
Reading Time: 15 minutes, commencing at 9am
Examination Duration: 180 minutes, commencing at 9.15am
Exam Conditions:
Individual open-book examination.
Permitted Materials:
Any course-related materials (including computer/tablet with internet connection to access course materials and
internet resources).
Calculator or calculational tool (mathematica, python etc) on computer/tablet.
Choosing to rely on an internet connection is up to you – there will be no recourse if it fails.
Instructions To Students:
You may not start writing solutions to the exam questions (whether on a tablet or on paper) during the reading period,
although you may annotate the exam question sheets.
Questions to staff are preferred during the initial reading period, although later questions are also possible.
You are permitted to access materials of your choosing.
Do not submit text or images that are not your original work. All answers are to be your own work.
Do not communicate with any other person during the examination.
Do not search directly for answers to questions on the internet – this will be grounds for failure.
This examination has a total of 66 marks across 5 questions.
Marks available for each question are as indicated.
Attempt all questions.
End of Semester One, 2021 PHYS3105, Physics of Matter
Page 2 of 7
Useful constants and formulae – not inclusive of all that you might need
c = 197.3 MeV.fm e2
/(4pe0) = 1.44 MeV.fm E(I) = 2/(2á) [I(I+1) – K2] r=1.2 A1/3 fm
ΔM(A,Z) = M(A,Z) – Zmp – (A-Z)mn B.E. = – ΔM(A,Z)c2 R=IsN 1 b = 10-24 cm2
1u = 931.494102 MeV/c2
B(E1)W = 0.06446 A2/3 B(M1)W = 1.7905
B(E2)W = 0.05940 A4/3 B(M2)W = 1.6501 A2/3
B(E3)W = 0.05940 A2 B(M3)W = 1.6501 A4/3
B(E4)W = 0.06285 A8/3 B(M4)W = 1.7458 A2
B(E5)W = 0.06929 A10/3 B(M5)W = 1.9247 A8/3
B(E1) = 6.288×10-16 Eg
-3 Pg B(M1) = 5.687×10-14 Eg
-3 Pg
B(E2) = 8.161×10-10 Eg
-5 Pg B(M2) = 7.381×10-8 Eg
-5 Pg
B(E3) = 1.752×10-3 Eg
-7 Pg B(M3) = 1.584×10-1 Eg
-7 Pg
B(E4) = 5.893×103 Eg
-9 Pg B(M4) = 5.329×105 Eg
-9 Pg
B(E5) = 2.892×1010 Eg
-11 Pg B(M5) = 2.615×1012 Eg
-11 Pg
Question 1 (11 marks)
The partial level scheme of 116Sn (Z=50, N=66) to the right shows states
associated with vibrational excitations in this near spherical nucleus.
a) Why might we expect 116Sn to have a spherical shape [1 mark]
b) What sort of collective vibration gives rise to the 2+ state at
1294 keV [1 mark]
c) What sort of collective vibration gives rise to the 3- state at 2266 keV
[1 mark]
d) What is the likely spin and parity of the state at 2027 keV Explain
your answer. [2 marks]
The 3-, 2266 keV state decays via a 972 keV transition to the 2+ state at
1294 keV and a 2266 keV transition to the 0+ ground state. These
gamma-rays are in the intensity ratio of 100 to 0.154, respectively, while
the halflife of the 3-, 2266 keV state is 0.34 ps.
e) What are the multipolarities of the 972 and 2266 keV transitions
[1 mark]
f) Internal conversion is negligible for these high energy transitions in
this relatively light nucleus. Hence, ignoring internal conversion, what
are the transition strengths of the 972 and 2266 keV transitions in
Weisskopf units You should get answers of ~10-3 and ~20 W.u,
respectively. [3 marks]
g) Explain how the transition strengths from f) relate to the changes in the nuclear wavefunctions between
the initial and the final states for the 972 and 2266 keV transitions. [2 marks]
End of Semester One, 2021 PHYS3105, Physics of Matter
Page 3 of 7
Question 2 (10 marks)
Extracts from the Audi-Wapstra mass tables are shown below, with all numerical values in units of keV, and
remembering that values of c2 are often left out.
Z N Nucleus Mass Excess Binding Energy
0 1 n 8071.323 0
1 0 1H 7288.969 0
1 1 2H 13135.720 2224.573
2 2 4He 2424.9 28295.7
4 4 8Be 4941.7 56499.5
6 6 12C 0 92161.7
a) Use the values in the column of mass excesses to evaluate the binding energy of 12C and verify that the
value in the table is correct. [3 marks]
A star generates power by the fusion of light nuclei to make heavier nuclei. The first stage in the reaction is
the fusion of two protons, but this would make 2
He, which is not stable. What instead happens is the
following reaction:
1
H + 1
H -> 2
H + e+ + ne
b) For this process to happen, one of the initial protons undergoes a radioactive decay during the reaction.
What sort of radioactive decay is this [1 mark]
Through a network of further fusion reactions, the protons and deuterons eventually form 4
He nuclei,
which could, in principle, fuse to form 8
Be.
c) Evaluate the Q value for the breakup of 8
Be into two 4
He nuclei and show that this process is
energetically favourable. [2 marks]
In fact, the ground state of 8
Be is a resonance that only exists for ~10-16 seconds before breaking up into
two 4
He nuclei. For a star to progress to make heavier nuclei, we need a third 4
He nucleus to collide with a
8
He nucleus during the short period of time before it breaks up. Hence, we need to look at the Q values for
the reactions:
8
Be + 4
He -> 12C and 12C -> 4
He + 4
He + 4
He
It turns out that the exothermic fusion of 8
Be + 4
He leaves the 12C in a highly excited state with enough
energy to immediately break up into three 4
He nuclei.
Luckily, there is a special state (called the Hoyle state) in 12C at an excitation energy of 7656 keV above the
ground state that can gamma-decay to form stable 12C in its ground state, before the nucleus breaks up
into three 4
He nuclei.
d) What must the total kinetic energy of the reactants (4
He and 8
Be) be, for 12C to be formed in the
7.656 MeV state Your answer should be of the order of 300 keV. [2 marks]
e) Evaluate the Coulomb barrier between 4
He and 8
Be and show that fusion to form 12C in the Hoyle state
must involve sub-barrier tunnelling. [2 marks]
End of Semester One, 2021 PHYS3105, Physics of Matter
Page 4 of 7
Question 3 (12 marks)
Consider the level scheme shown at
right for 178Hf (Z=72, N=106).
a) Is this a deformed nucleus Give two
reasons why you might believe this,
one based on the number of protons
and neutrons and one based on the
observed level structures. [2 marks]
b) Four rotational bands are shown in
the level scheme. Explain the structure
of the nucleus that gives rise to each of
the bandhead states, that then
undergo rotation, to make each of the
bands A through D. [2 marks]
c) Evaluate the moment-of-inertia
parameter, 2
/(2á) from (i) the
energies of the 2+ and 0+ states in band
A and (ii) the energies of the 4+ and 2+
states in band C. You should get answers of the order 15 keV. [3 marks]
d) Explain why your answers in c) show that the deformation of the nucleus is similar in both bands A
and C [1 mark]
e) Using the attached Nilsson diagram, and assuming the deformation is e2=+0.25, provide a plausible two
neutron configuration that explains the structure of band B. Explain why the particle coupling gives rise to
the 8- state’s spin and parity. [2 marks]
f) The B(E2) for the 93 keV, 2+ to 0+ ground state transition is measured to be 160 W.u. Using the squared
Clebsch-Gordan coefficient <2020|00>2
=1/5, evaluate the quadrupole moment of 178Hf in the ground state
in units of e.fm2
. Your answer should be of the order 500 e.fm2
. [2 marks]
End of Semester One, 2021 PHYS3105, Physics of Matter
Page 5 of 7
Question 4 (17 marks)
The figure above shows the amount of photon absorption measured when light of different frequencies
passes through a gas made up of a diatomic molecule at room temperature. Note the scale in units of
1013 Hz.
a) Draw a level scheme showing the molecular energy levels that are involved, as well as the transitions
from specific initial to final levels, that correspond to the peaks in the spectrum. This picture does not need
to be to scale, but should give an idea of the relative energy spacings. [3 marks]
b) Using the observed spectrum, estimate the energy difference between the molecular ground state and
its first vibrational level in units of eV. [2 marks]
Denote your answer to b) to be w. Further assume that the rotational energy in the ground vibrational
state is given by E(J)=AJ(J+1), while it is E(J)=BJ(J+1) in the first excited vibrational state. The left set of
peaks correspond to absorption from the ground to excited vibrational states that change the rotational
quantum number by DJ = –1. The right set all correspond to DJ = +1.
c) Show that the energies of the DJ = –1 transitions are given by EL = w – 2A + (B-A)J2 + (B-3A)J, while the
energies of the DJ = +1 transitions are given by ER = w + 2B + (B-A)J2 + (3B-A)J. [4 marks]
d) We might expect B