PHY303 – 2012 EXAM numerical answers
[2]
(a) (i) 5 neutrons and 4 neutrons respectively
(ii) 8Be has 4 neutrons and 4 protons. A nucleus with an even number of neutrons
and an even number of protons has zero spin and positive parity.
[book/unseen]
(b) (i) 49 Be136 C  
(ii) Q = (2603 + 12182 - 3355) x 10-6 x 931.49 MeV = 10.65 MeV
[unseen]

(c) (i) from momentum conservation
= 4.002603/13.003355 x 5.5 MeV = 1.69 MeV
[unseen]
(ii) from energy conservation we get the energy of the gamma-ray as:
(Q-value)10.65 MeV + 5.5 MeV – 1.69 MeV = 14.46 MeV
[unseen]
(d)
d
0.001295(Z1Z 2 ) 2

d E 2 (MeV )sin( /2) 4
= [0.001295 x (2 x 4)2] /5.52 sin(45/2)4
= 0.08288/0.649 = 0.128 barn

the solid angle subtended by the detector is 10 cm2/ 4 x pi x (100 cm)2 = 7.9 x 10-5
so the detection rate is: (0.05 x 6.022 x 1023)/9 = 3.345 x 1021 per cm2
106 x (0.05 x 6.022 x 1023)/9 x 0.128 barn x 1 x 10-24 x 7.9 x 10-5 =
(106 above is the rate) barn = 1 x 10-28 m2 or 10-24cm2
= 0.035 alphas per sec
[unseen]
(e) neutrons could be produced as below.
 49Be126 C  n

Energetically favoured because carbon-12 has higher binding energy (more stable)
than carbon-13 where we have an unpaired neutron.
[unseen]
[3]
(c)
(i) To obtain this we need to differentiate the equation for the mass with respect to Z,
keeping A as a constant. We can ignore terms that do not contain Z.
dm /dZ  m p  mn [2ac Z / A1/ 3  4aa (1 2Z / A)]/c 2  0
hence Z min 

[mn  m p ]c 2  4aa
2ac A 1/ 3  8aa A 1
(ii) mn – mp = 1.29 MeV
2
 divide through by c
we get [1.29 + 4 x 19]/[2 x 0.6 (101)-1/3 + (8 x 19)/101] = 77.29/1.758 = 43.96
hence A = 101 and Z = 44 is the most stable (N= 57)
[4]
For positron decay we have Qp = mN (Na) - mN(Ne) – me = mA(Na) – mA(Ne) – 2 me =
20487.686-20484.844 – 2 x 0.511 = 1.82 MeV
QEC = me + mN (Na) - mN(Ne) = mA(Na) – mA(Ne) = 2.842 MeV
[book, unseen]
(c)
This is given by
N  N0 (1  e(X / X A ) )
Where XA is the attenuation length, so: N = 106 ( 1 – e(-100/130))

=5.36 x 105 counts per second
[unseen]
(d)
the energy dumped in the body from each radiation is:
Positorn: 1.82 x 106 x 106 x 1.6 x 10-19 (J/eV) / 70 = 4.16 x 10-9 Grays/sec
EC: 2.842 x 106 x 106 x 1.6 x 10-19 / 70
= 6.50 x 10-9 Grays/sec
Gamma: 1.275 x 5.36 x 105 x 106 x 1.6 x 10-19 / 70
= 1.55 x 10-9 Grays/sec
Total in Grays is 1.22 x 10-8 Grays/sec
If we assume the quality factor Q=1 for all then this is the figure in Sv/sec
The annual dose is equivalent to a rate of 20 x 10-3 /1.22 x 10-8 = 1.64 x 106 sec = 0.052
years
[5]
(b) As below. No need to know the Q values.
[book]
(c) This comes from balancing the energy production with the energy need to get fusion
as follows
Using below. For DD we have n1 = n2 so R = n2<sigma x v>/2. Where the 2 accounts
for the fact we have the same species interacting.
For the energy in we have ndeut + nelectrons = 2n, giving 3 KT (we need to account for the
electrons in this case).
Hence
(d) (i)The speed of a 50 keV deuterium is 2.19 x 106 m s-1. Note first that on average any
two deuterons will be moving orthogonally to each other, whioch gives and average
relative speed of 21/2 x v.
from E = ½ Mv2 and remembering units we get:
v = [(2 x 0.05 MeV/c2) x md-1]1/2 = 2.19 x 106
v rel then is 21/2 x 2.19 x 106 x 0.2 x 1 x 10-28 = 6.19 x 10-23 m3s-1
[unseen]

(ii) the power output is given by:
n 2 v rel Q/2
= (1021)2 x 6.19 x 10-23 x 3.6 (MeV) x 106 x 1.6 x 10-19 / 2 = 1.78 x 107 W m-3

[unseen]
(iii) the plasma energy is 3nkT = 3 x 1021 x 50 keV = 2.40 x 107 J m-3
hence t = 2.4 / 1.78 = 1.35 sec
[unseen]