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(B) The initial energy of the system is the sum of the mass energy of the nucleus $X^A$ and the kinetic energy of the neutron.Let $BE_X$ be the bindin

Vedclass · Accepted Answer

$\frac{8A + 3}{A + 1}$

Vedclass · Answer

(B) The initial energy of the system is the sum of the mass energy of the nucleus $X^A$ and the kinetic energy of the neutron.Let $BE_X$ be the binding energy of $X^A$,so $BE_X = 8A \ MeV$.The total energy before the reaction is $E_i = (M_X + m_n)c^2 + K_n = (M_X c^2 + m_n c^2) + 2 \ MeV$.Since $BE = Z m_p c^2 + N m_n c^2 - M c^2$,we have $M_X c^2 = Z m_p c^2 + (A-Z) m_n c^2 - 8A$.The total energy after the reaction is $E_f = M_Y c^2 + E_{\gamma 1} + E_{\gamma 2} = M_Y c^2 + 1 + 4 = M_Y c^2 + 5 \ MeV$.Equating $E_i = E_f$:$(M_X c^2 + m_n c^2) + 2 = M_Y c^2 + 5$$M_Y c^2 = M_X c^2 + m_n c^2 - 3$.Substituting the mass expressions in terms of binding energy:$(A+1) BE_Y = (A+1) (Z m_p c^2 + (A+1-Z) m_n c^2) - M_Y c^2$$(A+1) BE_Y = (A+1) (Z m_p c^2 + (A+1-Z) m_n c^2) - (M_X c^2 + m_n c^2 - 3)$Using $M_X c^2 = Z m_p c^2 + (A-Z) m_n c^2 - 8A$:$(A+1) BE_Y = 8A + 3$.Therefore,the binding energy per nucleon of $Y$ is $BE_Y = \frac{8A + 3}{A + 1} \ MeV$.

Binding energy per nucleon of a fixed nucleus $X^A$ is $8 \ MeV$. It absorbs a neutron moving with kinetic energy $2 \ MeV$ and converts into $Y$,emitting two photons of energy $1 \ MeV$ and $4 \ MeV$ respectively one after the other. The binding energy per nucleon of $Y$ (in $MeV$) is:

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