Question

A nucleus X, initially at rest, undergoes alpha-decay
according to the equation.
$ 92^{A} \rightarrow _{Z}^{228} Y + \alpha $
(a) Find the values of A and Z in the above process
(b) The alpha particle produced in the above process is found
to move in a circular track of radius 0.1 lm in a uniform
magnetic field of 3 T. Find the energy (in MeV) released
during the process and the binding energy of the parent
nucleus X.
Given that $ m(Y) = 228.03u ; m (_0^1 n) = 1.009u $
$ m (_2^4 He ) = 4.003 u ; m(_1^1 H ) = 1.008u $

Answer 1

$ (a) A - 4 = 228 $
$ \therefore $ A = 232
92 - 2 = Z or Z = 90
$ (b) $ From the relation.
$ \, \, r = \frac{\sqrt {2Km}}{Bq} $
$ K_{\alpha} = \frac{r^2B^2q^2}{2m} $
$ = \frac{(0.11)^2(3)^2(2 \times 1.6 \times 10^{-19})^2}{2 \times 4.003 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-13}} $
$ = 5.21 MeV $
From the conservation of momentum,
$ P_{\gamma;} = P_{\alpha}$ or $ \sqrt {2K_{\gamma}m_{\gamma}} = \sqrt {2K_{\alpha}m_{\alpha}} $
$ \therefore K_{\gamma} = \bigg (\frac{m_{\alpha}}{m_{\gamma}} K_{\alpha} = \frac{4.003}{228.03} \times 5.21 $
$ = 0.09MeV $
$ \therefore $ Total energy released $ = K_{\alpha} + K_{\gamma} = 5.3 MeV $
Total binding energy of daugther products
$ = [ 92 \times (mass\, \, of \, \, protons) + (232 - 92) (mass\, \, of\, \, neutron) - (m_{\gamma}) - (m_{\alpha}) ] \times 931.48 MeV $
$ = [ (92 \times 1.008) + (140) (1.009) - 228.03 - 4.003] 931.48 MeV $
$ = 1828.5 MeV $
$ \therefore $ Binding energy o f parent nucleus
= binding energy o f daughter products - energy released
$ = (1828.5 - 5.3) MeV = 1823.2 MeV $