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Question

(D) In the decay of a nucleus with mass number $A=236$,the parent nucleus decays into an $\alpha$-particle (mass number $4$) and a daughter nucleus (m

Vedclass · Accepted Answer

$\frac{59 E}{58}$

Vedclass · Answer

(D) In the decay of a nucleus with mass number $A=236$,the parent nucleus decays into an $\alpha$-particle (mass number $4$) and a daughter nucleus (mass number $A-4 = 232$).By the law of conservation of linear momentum,the magnitude of momentum of the $\alpha$-particle $(P_{\alpha})$ must be equal to the magnitude of momentum of the daughter nucleus $(P_{d})$,so $P_{\alpha} = P_{d} = P$.The kinetic energy $KE$ is given by $KE = \frac{P^2}{2m}$,which implies $KE \propto \frac{1}{m}$.Therefore,the ratio of kinetic energies is $\frac{(KE)_{d}}{(KE)_{\alpha}} = \frac{m_{\alpha}}{m_{d}} = \frac{4}{232} = \frac{1}{58}$.Given $(KE)_{\alpha} = E$,we find $(KE)_{d} = \frac{E}{58}$.The total energy released $(Q)$ is the sum of the kinetic energies of the products:$Q = (KE)_{\alpha} + (KE)_{d} = E + \frac{E}{58} = \frac{59E}{58}$.

An $\alpha$-particle of energy $E$ is liberated during the decay of a nucleus of mass number $236$. The total energy released in this process is

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