PHY303
THE UNIVERSITY OF SHEFFIELD
DEPARTMENT OF PHYSICS AND ASTRONOMY
Autumn Semester 2010 – 2011
2 hours
NUCLEAR PHYSICS
Answer question ONE (COMPULSORY) and two other questions.
A formula sheet and table of physical constants is attached to this paper.
All questions are marked out of ten. The breakdown on the right-hand side of the
paper is meant as a guide to the marks that can be obtained from each part.
[2]
(b) N197 = d/A = 1000dNA/197kg = 1.22 x 1022 atoms/m2
each neutron that travels through the foil has a probability of collision equal to the
product of the number of atoms and the cross section:
P = N197 x 7.8 x 10-24 = 0.095
This is equivalent to the attenuation of the beam.
(c)
(1 - 0.095) x intensity = 1010 so intensity = 1010/(1-0.095) = 1.105 x 1010
(d)
dN198
 R  N198
dt
(e)


Be t  R  (A  Be t ) giving A = R/ and B = -R/

so
R
N198(t)  (1 e t )

substituting for R and gives N198 = 2.64 x 1015

(3)
(c)
Coulomb energy (170 MeV) = Z1Z2e2ke/r = 1.44 (MeV-fm) Z1Z2/r(fm)
Z1 = Z2 = 46 implies [170 / (1.44 x 46 x 46)]-1 = 17.9 fm.
(e
by inserting the values above in the SEMF we get:
It ca be seen that the B/A for Br is higher (9.37 vs 8.55) so it is an energetically more
favourable nucleus.
(4)
(e) the total cross section is 1.8 + 4.5 + 2.7 = 9 eV

the debroglie wavelength is given by  = hc/pn = hc/(2mnE)1/2

x 10-15 (eV.s) x 3 x 108 (m/s) / (2 x 939.5 x 106 eV x 75 ev)1/2 = 3.30 x 10-12 m

n = [(3.3 x 10-12)2 x 4.5 x 2.7 c2 ] / [4( (75-70)2c2 + 92/4 c2)] = 1.32 x 10-22
/[425+20.25)] = 4.72 x 10-25 m2 or 4720 barns
(2.33 x 10-25)??
n = nx 1.8/2.7 = 3.14 x 10-25 m2
(e)
(5)
(c)
X does not undergo positron beta decay, as indicated, so the positrons must come form
pair production by the gammas indicated in the diagramme as along as E > 1.02 MeV
(2mec2).
A
Z

Based on this we can find the end point energies as E (gamma 1) = 4.14 -1.38 = 2.76 and
E(gamma 2) = 1.38 MeV. Hence Tmax(beta) = 2.76 -1.02 = 1.74 MeV and for the other
1.38-1.02 =0.36 MeV.