Question

A nucleus X, initially at rest, undergoes alpha-decay
according to the equation.
$92^A{\to }_Z^{228}Y+\alpha$
(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 $m(Y)=228.03u;m{(}_0^{1}n)=1.009u$
$m{(}_2^{4}He)=4.003u;m{(}_1^{1}H)=1.008u$

Answer 1

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