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(B) In an $\alpha$-decay,the energy released ($Q$-value) is shared between the $\alpha$-particle and the daughter nucleus to conserve momentum.The kin

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) In an $\alpha$-decay,the energy released ($Q$-value) is shared between the $\alpha$-particle and the daughter nucleus to conserve momentum.The kinetic energy of the $\alpha$-particle $(K_{\alpha})$ is given by the formula: $K_{\alpha} = \left( \frac{A - 4}{A} ight) Q$,where $A$ is the mass number of the mother nucleus.Given: $K_{\alpha} = 48 \, MeV$ and $Q = 50 \, MeV$.Substituting the values: $48 = \left( \frac{A - 4}{A} ight) 	imes 50$.Divide both sides by $50$: $\frac{48}{50} = \frac{A - 4}{A}$.Simplify the fraction: $0.96 = 1 - \frac{4}{A}$.Rearrange to solve for $A$: $\frac{4}{A} = 1 - 0.96 = 0.04$.$A = \frac{4}{0.04} = 100$.Thus,the mass number of the mother nucleus is $100$.

In an $\alpha$-decay,the kinetic energy of the $\alpha$-particle is $48 \, MeV$ and the $Q$-value of the reaction is $50 \, MeV$. The mass number of the mother nucleus is: (Assume that the daughter nucleus is in the ground state)

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