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(A) The $\alpha$-decay process is represented as: $X \rightarrow Y + \alpha$.Here, $X$ is the parent nucleus, $Y$ is the daughter nucleus, and $\alpha

Vedclass · Accepted Answer

$(m_x - m_y - m_a) c^2$

Vedclass · Answer

(A) The $\alpha$-decay process is represented as: $X \rightarrow Y + \alpha$.Here, $X$ is the parent nucleus, $Y$ is the daughter nucleus, and $\alpha$ is the $\alpha$-particle $(_{2}He^{4})$.The mass defect $\Delta m$ in the process is given by the difference between the initial mass and the final mass: $\Delta m = m_x - (m_y + m_a) = m_x - m_y - m_a$.According to Einstein's mass-energy equivalence principle, the energy released (which appears as the net kinetic energy of the products) is $Q = \Delta m c^2$.Therefore, the net kinetic energy gained is $Q = (m_x - m_y - m_a) c^2$.

$\alpha$-decay of a parent nucleus $X$ results in a daughter nucleus $Y$. If $m_x, m_y$ and $m_a$ are the masses of the parent nucleus, the daughter nucleus and the $\alpha$-particle respectively, then the net kinetic energy gained in the process is:

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