## 08 February 2011

### Nuclear Physics

Problem 1.

Given the following masses, find the mass defect and the binding energy of a helium-4 nucleus.

particle mass (kg)
helium-4 nucleus 6.6447X10-27
proton 1.6726X10-27
neutron 1.6749X10-27

Solution:
a. Sum of masses of individual nucleons = 2(1.6726X10-27) + 2(1.6749X10-27)
= 6.6950X10-27 kg
mass defect = 6.6950X10-27 - 6.6447X10-27 = 0.0503X10-27 kg
b. Binding energy = (Δm)c2 = (0.0503X10-27)(3.00X108) = 4.53X10-12 J
(4.53X10-12 J)/(1.60X10-19) = 2.83X107 eV = 28.3 MeV

Problem 2.

a. Given the following masses, find the energy released when U-238 transmutes to Th-234 by alpha decay.

particle mass (amu)
U-238 238.0508
Th-234 234.0436
He-4 4.0026

b. Ignoring relativistic effects, what is the speed of the alpha particle as a result of the alpha decay?

Solution:

a. mass defect: 238.0508 - (234.0436 + 4.0026) = 0.0046 u
energy equivalent: (0.0046 u)(931.5 MeV/u) = 4.3 MeV
b. The values for part a are converted to mks units:

Conversions
quantity symbol conversion factor value
energy released
Ek
1.602x10-19 J/eV
6.88886x10-13 J
mass of Th-234
mTh
1.66053886x10-27 kg/amu
3.8864x10-25 kg
mass of He-4
mHe
1.66053886x10-27 kg/amu
6.6465x10-27 kg
speed of Th-234
vTh
-
?
speed of He-4
vHe
-
?
Conservation of momentum gives us:
mThvTh+ mHevHe = 0
and
vTh+ = -(mHe/mTh)vHe [eqn. 1]
The energy released is in the form of kinetic energy: (1/2)mThvTh2 + (1/2)mHevHe2 = Ek
Substituting eqn. 1 and simplifying:
vHe = √{[2Ek] / [(mHe)(1 + mHe/mTh)]} = 1.4x107 m/s

Problem 3.

What is the wavelength of a 0.217 MeV photon emitted during gamma decay?

Solution:

The energy is converted to Joules: (0.217x106 eV) (1.602x10-19 J/eV) = 3.48x10-14 J

The energy is converted to wavelength:
λ = hc/ΔE = (6.63x10-34 J*s)(3.00x108 m/s)/(3.48x10-14 J) = 5.72x10-12 m

Problem 4.

The radioactivity of milk in one sample was 1750 Bq/L due to iodine-131 with half-life 8.04 days. For comparison, find the activity of milk due to potassium. Assume that one liter of milk contains 1.60 g of potassium, of which 0.0117% is the isotope 40K with half-life 1.28X109 yr.

### Solution:

R = [(ln 2)/T][N/A]
where N = Avogadro number [/kmol], A[kg/kmol] is the mass number, ln 2 = 0.693, and T is the half-life in seconds.
One kilogram of a pure radioactive isotope with half-life T[sec] has activity R[Bq/kg]
The unit of activity is Becquerel (1 Bq = 1 decay/sec)
T = (1.28 x 109 yr)( 31,557,600 s/y) = 4.039 x 1016 s
R = [0.693/4.039 x 1016][ 6.0221415×1026 / 40] = 2.583 ×108 Bq/kg
R = [2.583 ×108 Bq/kg][0.00160 * 0.0117/100 kg/L] = 48.35 Bq/L

Problem 5.

(a) Use the single-particle shell model to predict the ground-state angular momenta (spin) values and parities for the following nuclei. Explain your reasoning in each case.

7Li, 33S, 44Ca, 69Ga, 80Zr, 123Sb, 175Lu.

(b) For the nuclei listed above, predict the magnetic dipole moments from the shell model.

Problem 6.

(a) The following nuclei have the spin quantum number quoted, but not the
parity. By comparing the measured dipole moments with the predictions of the
shell model, determine the parity of these states.
J <μ>
43Ca 7/2 -1.3μN
67Zn 5/2 +0.9μN
153Eu 5/2 +1.7μN
139La 7/2 +2.8μN

(b) Compare your findings in 1(b) and 2(a) with the experimental values.
Comment on any differences.