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(D) In the decay of a nucleus with mass number $A = 236$, the $\alpha$-particle $(m_{\alpha} = 4)$ and the daughter nucleus $(m_d = 232)$ are produced

Vedclass · Accepted Answer

$\frac{59 E}{58}$

Vedclass · Answer

(D) In the decay of a nucleus with mass number $A = 236$, the $\alpha$-particle $(m_{\alpha} = 4)$ and the daughter nucleus $(m_d = 232)$ are produced.Given the kinetic energy of the $\alpha$-particle is $(KE)_{\alpha} = E$.Since the total momentum is conserved, the magnitudes of the momenta of the $\alpha$-particle and the daughter nucleus are equal: $P_{\alpha} = P_d = P$.The kinetic energy is given by $KE = \frac{P^2}{2m}$, which implies $KE \propto \frac{1}{m}$.Therefore, the ratio of kinetic energies is $\frac{(KE)_d}{(KE)_{\alpha}} = \frac{m_{\alpha}}{m_d} = \frac{4}{232} = \frac{1}{58}$.Thus, $(KE)_d = \frac{E}{58}$.The total energy released $(Q)$ is the sum of the kinetic energies of the products: $Q = (KE)_{\alpha} + (KE)_d = E + \frac{E}{58} = \frac{59 E}{58}$.

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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