PHY303
THE UNIVERSITY OF SHEFFIELD
DEPARTMENT OF PHYSICS AND ASTRONOMY
Autumn Semester 2009 – 2010
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.
NUMERICAL ANSWERS
2
(c) ANS: 0.4 MeV
(d) ANS: around 100-200 MeV
3
(c) ANS: 1fj means we have l = 3 therefore j = l +/- 1/2 = 5/2 or 7/2
(d) ANS:
Ca I= 0+.
40
20
For 2041Ca spin I = 7/2 parity = -1.
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4
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(b) ANS: Nuclear separation = 1.2[41/3 + 2061/3] = 8.99 fm.
ke2 = (8.98 x 109) x (1.6 x 10-19)2 x (6.24 x 1018) x (1 x 10-6) x (1 x 1015) =
1.43 MeV.fm
so we have: ( 2 x 82 x 1.43)/9.0 = 26.06 MeV
(c) ANS:
5.3 MeV = (2 x 82 x 1.43 MeV.fm)/r
this gives r = 44.56 fm
The width L is then 44.56 minus the nuclear separation 9.0 fm i.e. 35.56 fm.
(d) ANS: For f use the following - 120+5.3 MeV = mv2/2 = (3727
MeV.v2)/(2c2)
So v/c = 0.259; and v = 7.78 x 107 m/s
Hence f = v/2R = 7.78 x 107 m/s / (2 x 9.0 fm x 10-15 m/fm) = 4.32 x 1021 /s
Now find  by substitution of the values into the equation above:
Firstly (h-bar)2 = ( 1.0546 x 10-34 J.s x 6.24 x 1018 ev/J = 6.58 x 10-16 ev.s )2
= 4.33 x 10-31 ev2.s2
 = [2 x 3727 x 106 x (26.0-5.3) x 106 /c2 ]/4.33 x 10-31
c = 3 x 108 m/s
so  = 1.99 x 1015
hence X = e-L = Exp -(1.99 x 1015 x 35.56 x 10-15) = 1.85 x 10-31
the half life is thus 0.693/(4.32 x 1021 x 1.85 x 10-31) = 8.67 x 108 sec
this uses T1/2 = 0.693/
5
(b) ANS: the first level will have J = 2, so there is a factor 6 in the equation
2
(i.e. 2 x (2+1)) which means
2Irot
2
x 6 = 36 keV, hence
2Irot
= 6 keV
The next levels correspond to J = 4 and 6, so we have (i) 6 keV x 4(4+1) =
120 keV, and (ii) 6 keV x 6(6 +1) = 252 keV)
 x 10-16 ev.s)2/(2 x 6 x 
(c) ANS: Irot = = (6.58
1000 ev) = 3.61 x 10-35 ev.s2
(d) ANS: use Rav = roA1/3 = 1.2 x 10-15 x 1701/3 = 6.65 x 10-15 m
3.61 x 10-35 ev.s2 = 2/5 x 170 x 931 x 106/c2 x (6.65 x 10-15)2 (1 + 0.31
Its important to get the units right i.e. to put the c2.
RHS of this gives 3.11 x 10-35 ev.s2
Hence  = ((3.61/3.11) - 1)/0.31 = 0.52
(e) ANS: 0.52 = 1.05
R
so R = 0.52 x 6.65 x 10-15/1.05 = 3.3 x 10-15 m
Rav
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