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(A) The energy released in a nuclear decay is equal to the change in binding energy of the system.Initial binding energy of $X = 6A \ MeV$.Final bindi

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$\frac{6A + 3}{A}$

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(A) The energy released in a nuclear decay is equal to the change in binding energy of the system.Initial binding energy of $X = 6A \ MeV$.Final binding energy of $Y = BE_Y$ (where $BE_Y$ is the total binding energy of nucleus $Y$).Since the electron and antineutrino have negligible binding energy, the total binding energy of the products is $BE_Y$.The energy released ($Q$-value) is given by the difference in binding energies: $Q = BE_{final} - BE_{initial}$.Given $Q = 3 \ MeV$, we have:$3 = BE_Y - 6A$$BE_Y = 6A + 3$.The binding energy per nucleon of $Y$ is $\frac{BE_Y}{A} = \frac{6A + 3}{A}$.

The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$
The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is:

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The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is: