“Key” questions
Tutorial 1:
SubAtomicPhysics: Nuclear Physics
In the experiment of Rutherford, Geiger and Marsden a particles of energy 7.7 MeV were incident
on a gold nucleus 197Au (Z=79). Estimate the distance of closest approach of the alpha particles to
the nucleus (ie. a “head on” collision).
1)
Later experiments used the same energy of alpha particles on 27Al(Z=13) and noted the sudden
onset of deviations from the coulomb scattering expectations for backscattered alpha particles at
~180o.
Propose why this is the case and use the given information to obtain an estimate of the size of the
aluminium nucleus. What approximations may cause inaccuracies in this estimate? Compare your
value with that obtained from R ~ R0A1/3 fm.
[ε0 = 8.85419x10-12 fm-1, e =1.60219x10-19 C]
Equate kinetic energy of alpha with the colulomb potential energy of alpha+nucleus (assuming non relativistic
kinematics). At distance of closest approach (d0) kinetic energy of incident alpha completely converted to potential
energy.
197
-12
6)
Au: T= q1q2/4πε0 : Therefore d0=(q1q2/4πε0Tα) = ( (2e)(79e) / ( 4π(8.85419x10 )(7.7x10 e) ) = 29.5 fm.
27
Al: similar calculation yields d0=4.86fm, As you are told the coulomb picture breaks down for this case you can
assume that this is the distance at which the alpha feels the strong force i.e is the potential radius of the nucleus.
Therefore can estimate size of nucleus ~0.7 fm less than this = ~4.2fm. Approximations used – don’t account for finite
size alpha or nucleus, simple approximation for average range of nuclear force,
1/3
Size of Al nucleus from R=R0A =1.3 x 27
2)
1/3
= 3.9 fm.
Use the uncertainty relation to estimate (in MeV) the minimum kinetic energy possible for an
electron confined to a sphere of diameter 2 fm
[Hint assume electrons are fully relativistic (kinetic energy ~ pc)].
What are the implications of this energy when deciding whether electrons exist in the nucleus?
∆p∆x>ħ
-34
: ∆p> ħ / ∆x > 1.054 x 10
-20
8
-15
/ 2 x 10
-11
-20
= 5.27x10
Jsm
-1
T~pc~(5.27 x 10 )(3x10 ) ~ 1.58 x 10 J = 98.7 MeV
Electrons interact via the coulomb force - this force is not sufficiently strong to confine electrons to the nucleus
(compare with typical atomic binding eneries which are of the order of eV)
3)
If a spherical drop of water (radius=0.15cm) were to have the same density as nuclear matter
(~1017kg/m3), what would its mass be?
Assume drop has constant density and is spherical : M = ρV = 1017(4/3xπx0.15x10-2))3 = 1.4x109kg !!
4)
Calculate the ratio of the gravitational potential energy to coulomb potential energy between two
protons in the nucleus.
5)
What is the parity of a nucleon arising from its orbital motion if it has a) orbital angular momentum
l=1? b) orbital angular momentum l=2?
Briefly discuss what is meant by parity and how it relates to the wavefunction properties.
l
π = (-1) if l=1 parity is negative. If l=2 parity is positive. See notes for last part
6)
Calculate the following masses in mass atomic units (u) for
a) electron (me=9.109389x10-31kg)
b) proton (mp=1.672623 x 10-27kg)
c) neutron (mn=1.6749x10-27 kg).
Hence convert these values into units of MeV/c2
-31
me=9.109 x 10
-27
/ 1.66054 x 10
(As 1u = 931.494 MeV/c
2
-4
= 5.485799 x 10 u c
therefore me = 0.511 MeV/c
2
Similarly mp=938.27 MeV/c and mn = 939.57 MeV/c
7)
2
2
2
Using the relationship between nuclear radii and atomic number (R ~ R0A1/3 fm) estimate the
density of nuclear matter and the average separation of nucleons in the nucleus.
Neutron stars can be thought of as a large assembly of neutrons packed with a density close to that
of nuclear matter – estimate the radius of a neutron star of mass 2 solar masses ( mass of sun =
1.988x1030kg)
1/3 3
3
-27
-45
17
3
ρ = M/V = AMp/[(4/3)π(1.25.A ) ] = Mp/[(4/3)π(1.25) ] = 1.67 x 10 / 8.18 x 10 = 2 x 10 kg/m
30
17
13
3
Assume neutron star is spherical: Volume = Mstar/ ρ = 2(1.988 x 10 )/2x10 = 1.99 x 10 m = 4/3πRΝ3
RΝ=17km
8)
Why are electrons a good probe of the nuclear charge distribution? Estimate the electron beam
energy required to probe the nuclear charge distribution with a spatial resolution of 0.2 fm. Using
this beam energy estimate the angle of the first diffraction minimum for scattering from a target of
27
Al(Z=13) stating any assumptions you make.
Electrons are a good probe of the nuclear charge distribution as i) their interaction is by the electromagnetic force which is well
understood and ii) The interaction strength not so strong, therefore electrons can probe the entire nuclear volume (This is not
the case for strongly interacting probes like protons / neutrons which mainly interact near the nuclear surface)
To probe with resolution of 0.2fm require the reduced wavelength (λ/2π) < 0.2fm. (Usually written as lambda bar)
E=hbar c/ lamda(bar) = 197.3/0.2 = 986.5 MeV
st
1/3
1 diffraction minimum sin(θ)=1.22 λ / 2R = 1.22 (0.2 x 2π) / 2(1.3 x28 )
o
Therefore θ =11.2
Tutorial 2:
SubAtomicPhysics: Nuclear Physics
1. Sketch the form of the nuclear potential. What does the potential radius and charge radius relate to?
Estimate the size of the Coulomb potential barrier faced by an incident proton on i) 238U(Z=92) and
ii) 12C (assume the potential radius is ~0.7 fm larger than the nuclear radius).
The fusion of a proton with 12C is one of the first steps of the CNO cycle in stellar burning –
estimate what the calculated barrier height corresponds to in terms of temperature.
R=R0A
1/3
radius of
:
238
U = 1.2(238)
12
radius of C = 1.3(12)
1/3
1/3
~7.44fm
= 2.97fm.
2
Barrier height = zZe /4πε0R
where R is the potential radius. Take the potential radius as 0.7fm larger than
the values calculated above to account for the range of the strong force. This yields a barrier height for
12
238
U = 16.3 MeV and for
C a height of 4.69 MeV.
10
Assuming this ia a monoatomic gas we can take E ~ 3/2kT. Therefore T~3.6x10 K. The centre of the sun has a temperature of
7
about 1.3x10 K. The fusion process relies on fast nuclei near to the tail of the Maxwell-Boltzmann distribution and quantum
tunnelling through the coulomb barrier in order to proceed (even with these effects you have to wait for the sun to heat up some
more (as it will in its later life) to get significant probability for this fusion reaction)
2. A contribution to the short range part of the nucleon-nucleon interaction is thought to arise from the
exchange of virtual ω mesons. Estimate the typical range over which the force would operate
(Mω =782.7 MeV/c2 )
2
2
From lecture r ~ ħc/mc ~ 197.3 (MeV fm) / 782.7 MeV/c ~ 0.25 fm
3.
In the nuclear reaction
Calculate the possible values of the orbital angular momentum quantum number for the 2 final state
nuclei.
You may assume the initial reaction occurs in s-wave. Jπ(10B)=3+, Jπ(4He)=0+, Jπ(13N)=1/2Reaction occurs in s-wave so relative orbital angular momenta of Boron and helium nuclei (l=0)
Therefore in entrance channel have total angular momentum J=3. As both Boron and Helium have even parity and l=0 (even
parity from (-1)l) therefore total parity in entrance channel is even (remember that parity is multiplicative)
As the reaction is a strong force process then expect angular momentum and parity are conserved. Therefore final state should be
J=3 and even parity. Total J of final state nuclei (neglecting orbital ang mom) is 0 or 1. Therefore the relative orbital ang mom of
final state particles must be l=3 with the first case and l=2 or 3 with the latter (to allow a total ang mom of final state of 3).
However as the parity of the nuclei in the final state is negative then only the l=3 case is allowed to ensure the total parity of the
final state is positive.
4. A nucleus with Jπ=0+ undergoes d-wave neutron emission. Specify the spins and parities of nuclear
energy levels which can be populated in the final state (A-1) nucleus.
How will this change when the initial nucleus has Jπ=1/2-?
From similar arguments as Qn above energy states of final nuclei following A→(A-1) + neutron with l=2 orbital ang mom
between neutron and (A-1) nucleus are Jπ=3/2+, 5/2+. For 1/2- nucleus the allowed states are Jπ=1-,2-.3-
5. How much energy is necessary to split up an alpha particle into its constituent nucleons
(Mn=1.008665u, MH=1.007825u, M(4He)=4.00260u).
Binding energy = [(2MH+2Mn) – M(4He)]c2 = [0.03038u]c2
as 1u = 931.494 MeV/c2 therefore c2 = 931.494 MeV/u
Therefore :
Binding Energy = ( 0.03038) (931.494) = 28.3 MeV
6. How much energy is required to remove a proton from 56Fe(Z=26)? Calculate the binding energy
per nucleon for 56Fe and compare with your value for the proton separation energy. Can you explain
any differences? ( M(56Fe) = 55.934939u, M(55Mn) = 54.938047u)
Sp = [M(A-1,Z) – M(A,Z) +MH]c2 = [54.938047u – 55.934939u + 1.007825u]c2 = [0.01093u]c2
As c2 = 931.494 MeV/u
Sp=[0.01093u] (931,393 MeV/u) = 10.184 MeV
BE = [ZMH + (A-Z)Mn - M(A,Z)]c2 = [0.528461u] (9331.494 MeV/u) = 492.26 MeV
BE/A = 492.26/56 = 8.79 MeV
SP larger than BE/A as average BE/A for 55Mn must be smaller than 56Fe (expect this as 56Fe is the most tightly bound nucleus)
7. The atomic mass of a nucleus is predicted by the semi-empirical mass formula to be given by
M(N,Z)c2 = ZMHc2 + NMn c2 - avA + asA2/3 + acZ2/A1/3 + asym(N-Z)2/A + δ
Where MH = 938.8 MeV/c2, Mn = 939.6 MeV/c2, av = 15.8 MeV, as= 18.0 MeV, ac = 0.7 MeV and
asym = 25.3 MeV.
Mirror nuclei are nuclei with the same mass number A and interchanged values of N and Z. Which
terms in the above formula give rise to differences in atomic mass between a pair of mirror nuclei?
Calculate the predicted atomic mass difference between 11C(Z=6) and 11B(Z=5) in units of MeV/c2.
From this information and any general considerations, what can you deduce about the possible decay
modes for these two nuclei?
Mirror nuclei have the same value of A but interchanged values for N,Z. The only terms in the SEMF which survive when taking
2
the mass difference between mirror nuclei is the ZM c , NM c
H
n
2
2
1/3
and a Z /A .
c
∆Mc2 = M(11B)-M(11C) = (Z1-Z2)MH c2 + (N1-N2)Mn c2 + ac/A 1/3(Z12-Z22)
Using given values of Z aand masses gives ∆Mc2) = -2.66 MeV
Difference shows 11B more stable than 11C.
11
C will decay to 11B by β+ emission
8. In the semi-empirical mass formula what experimental evidence does the pairing term δ account for?
For a given value of the mass number A, nuclear masses can be expressed as a quadratic function in
Z. Using the “minimum isobar” formula (as derived in the lecture notes) determine whether the
142
Xe(Z=54) is β+ or β- unstable.
Pairing term accounts for the observational evidence that even-even nuclei are more stable than odd-odd nuclei. It has a quantum
mechanical origin and reflects the extra binding felt by paired nucleons (with opposite spins) which can share the same spatial
states, increasing their probability to be in close proximity to each other and feel the strong force.
From formula derived in the notes Zstable = 142 / 1.972 + 0.015(142)2/3 = 60
Therefore 142Xe(Z=54) will decay by β- towards the stable nucleus at Z=60
Tutorial 3:
SubAtomicPhysics: Nuclear Physics
1) A neutron star can be crudely approximated as a large assembly (>~1055) of neutrons. Use the semi
empirical mass formula to estimate the binding energy per nucleon of a neutron star (you may
neglect the δ term). Comment on the physical implications of the value obtained.
Which additional term should also be included in the mass formula if it is to be extrapolated to
neutron stars? Propose a form for this extra term.
Yhe expression for the binding energy per nucleon (note you get this by dividing the formula for the binding energy (B) given in
the notes by A!!)
B/A = av – asA-1/3 – ac Z2/A4/3 –as(A-2Z)2/A2
As A is very large the surface and coulomb terms are negligible. The B/A is determined by the difference between the volume
term and the asymmetry term. The latter approaches if we assume the star is dominantly neutrons i.e. A>>Z
B/A = av - as = 15.6 – 23.3 MeV = -7.7 MeV. The expression is negative implying the system is unbound. Therefore to bind a star
other forces contribute. For a large object such as this the gravitational effects cannot be neglected. An attractive gravitational
potential energy term should be added to the SEMF. The form will be similar to the derivation of the coulomb potential term in
the lectures. i.e. (3/5) (G/ro) (M2/A 1/3). Using this term you can go on to calculate the minimum size needed for a stable star !!
2) Use the semi empirical mass formula to calculate the energy released when 238U fissions
symmetrically. Propose why fission products are often unstable.
Fission process total nucleons remain the same. Terms dependent on A only will not contribute.
∆Mc2 = [M(238U)- 2*M(119Pd)]c2 = [ZMH + NMn – B(238U)/c2 – 2*(Z1MH + N1Mn –B(119Pd)/c2)
In symmetric fission Z1=Z/2 . Therefore using the SEMF.
= 2* B(119Pd)/c2 - B(238U)/c2 = [2*(1856.4 – 416.13 – 301.13 – 142.74 + 0) – (3712.8 – 660.56 – 956.05 – 285.47 + 0.77) =
181 MeV
Note the differences in the SEMF terms for the 2 nuclei.
Fission products are often unstable as you have gone from a heavy nucleus where the most stable configuration favours a large
neutron excess, to a lighter nucleus where the most stable configuration has smaller (or zero) neutron excess. Therefore the fission
fragments tend to have an excess of neutrons compared to the stable configuration and are unstable.
3) Work out the spin and parity of the ground states of a) 16O(Z=8) b) 15N(Z=7) c) 23Na (Z=11) in terms
of the nuclear shell model.
4) Give possible shell model nucleon configurations which could produce the 5/2+ , 1/2+, 1/2- , 5/2- ,
3/2- and 3/2+ excited states of 17O(Z=8)
5)
207
207
208
Pb(Z=82) has a closed 3p1/2 sub-shell of neutrons and a closed 3s1/2 shell of protons. In terms of
the shell model what spin and parity is expected for ce ground state of 207Pb and 207Tl(Z=81).
Pb 1 unpaired neutron in the 3p1/2 level. Therefore Jπ=1/2Tl 1 unpaired proton in the 3s1/2 level. Therefore Jπ=1/2
6) In odd-odd nuclei, an interaction between the last odd neutron and the last odd proton must be taken
into account in order to explain the ground state spins. The coupling favours parallel spins of the odd
proton and the odd neutron. On this basis determine the ground state spin and parity for the 14N
(Z=7) nucleus, giving a sketch of the occupation of levels as prescribed by the shell model.
7) The rotational model for permanently deformed nuclei predicts excitation energies given by
E(J)=J(J+1)ħ2 / 2I with J=the spin of the nuclear state and I=moment of inertia of the nucleus about
the axis of rotation. The energies and spins of the first two excited states of 180Hf are
Energy (MeV)
J
0.093
2
0.309
4
Determine whether these values agree with the rotational model and if so
a) Predict the energy of the third excited state (J=6) and
b) Calculate the moment of inertia for the nucleus (ħ =6.582x10-22 MeVs)
Tutorial 4:
SubAtomicPhysics: Nuclear Physics
1) Calculate the Q value for 228Th(Z=90) (mass=228.0287u) to emit
a) An α particle
b) A proton
For the first case calculate the kinetic energies of the alpha particle and the daughter nucleus.
224
Ra(Z=88) = 224.0202u; 227Ac(Z=89) = 227.0278u]
4
2
α emission Qα = [ M(A,Z) – M(A-4,Z-2) - M( He) ] c = [0.059u]c
Qα=[0.059u] (931.494 MeV/u = 5.496 MeV
Q value is energy released in α decay of
228
2
As 1u = 931.494 MeV/c
2
Th. This will be split between the daughter and the alpha.
Tα = (A-4/A)Qα = (228-4/228)Qα = 5.4 MeV
1
2
Proton emission Qp = [ M(A,Z) – M(A-1,Z-1) - M( H) ] c = - 6.45 MeV
228Th
Qp value negative so energy required to remove a proton from
.
For many heavy nucei alpha particles are preferentially emitted over protons, detuterons, tritons as the 2n-2p configuration os very
tightly bound (e.g. See BE/A curve). Therefore, as can be seen in the above example, alpha emission can proceed spontaneously
with a release of energy whereas the other decay modes would require an input of energy.
2) Can 57Co(Z=27) spontaneously decay by β+ emission ?
[masses: 57Co= 56.936294u; 57Fe =224.0202u; 2me = 0.001097u]
2
2
Qβ+ = [ M(A,Z) – M(A,Z-1) – 2me ] c = [56.936294u-56.935396u-2(0.001097)u]c = -0.185 MeV
Q<0. Therefore cannot decay spontaneously.
3) The half life of Radon (222Ra) is 3.823 days. What fraction of freshly separated sample decays in a)
1day? b) 10 days?
λ=ln2/τ1/2=0.1813 day−1
1 day N(t)/N0 = e−λt = e−0.1813 = 0.834 (16.6% decayed). Similarly after 10 days 53.7% decayed
4) There are an estimated 1017 kg of uranium in the top 20km of the earth’s crust. 5.5x10-3 percent are
234
U, 0.72 percent are 235U and the rest is 238U. The decay constants for the three isotopes are
8.97x10-14s-1, 3.12x10-17s-1 and 4.92x10-18s-1 respectively. Calculate the half lives of the isotopes in
years. Calculate the activity of the total amount of uranium in the earth’s crust in Curies [Atomic
mass of Uranium = 238.029 g/mol)]
The previous existence of a natural thermal fission reactor based on Uranium deposits and moderated
by groundwater has been suggested in the Gabon. Estimate how long ago such a thermal reactor
could have operated.
17
23
No. of uranium nuclei = 10 x 6.02 x10 / 0.238029 = 2.53 x 10
-5
41
37
N(234U) = (5.5 x10 )(2.53x10 ) = 1.39 x 10 Nuclei
Activity(234) = N x lambda = 1.39 x 10
12
37 (
;
-14
41
-14
5
halflife = ln2/lamda = ln2/8.97x10 ) = 2.45 x 10 years
14
13
8.97x10 ) = 1.25 x 10 Bq = 3.4 x 10 Ci
13
Similarly A(235) = 1.5 x 10 Ci ; A(238) = 3.3 x 10 Ci
13
Total activity = Activity(234)+Activity(235)+Activity(238)= 6.9 x 10 Ci. This is a large amount of radiation – luckily we are
4
shielded from this! Note: Activity U(234) is larger than Activity(238) even though have 10 time less in the earth’s crust because
of large decay constant
For fission reactor to be moderated by water then fissile component of uranium must be 5%. (see notes). Therefore can calculate
how long ago this condition was reached.
5) The ground state of 84Be is unstable and decays into two alpha particles (42He) with a half life
τ1/2=1.0x10-16 s. Determine the relative orbital angular momentum of the two alpha particles and
estimate the particle decay width (eV) of the 84Be ground state.
6) The nuclide 12N(Z=7) decays by β+ emission with a maximum positron energy of 16.316 MeV.
Calculate the neutral atomic mass of the parent atom.
7) Which of the following heavy nuclei would you expect to have large cross sections for reactions with
thermal neutrons?
a)
251
Cf (Z=98) b)
255
Fm (Z=100) c)
250
Bk (Z=97) d) 238U(Z=92)
Fissile nuclei have large thermal neutron cross sections. These are even-odd or odd-odd. The addition of the
neutron to a fissile nucleus results in an energy benefit from the pairing energy which allows even thermal
energy neutrons to have an appreciable probability to induce fission. Nuclei a) b) and c) are fissile