Question

Binding Energy per nucleon of a fixed nucleus $X^4$ is $6MeV.$ It absorbs a neutron moving with $KE=2MeV,$ and converts into $Y$ at ground state, emitting a photon of energy $1MeV.$ The Binding Energy per nucleon of $Y$ ($inMev$) is:

• A. $\frac{(6A+1)}{(A+1)}$
• B. $\frac{(6A-1)}{(A+1)}$
• C. $7$
• D. $\frac{7}{6}$

Answer 1

Answer: (A)

$\frac{(6A+1)}{(A+1)}$

$\begin{array}{c}\text{Given-* Binding Energy per nucleon of}{X}^A=6\text{Mev}\\ \Rightarrow \text{Binding Energy}BE\text{of}{X}^A=6A\text{MeV}\\ \text{we have -}\\ \begin{array}{c}{X}^A{+}_1^{1}n⟶Y^{A+1}+\text{Photon (l Mev)}\end{array}\end{array}$

$\begin{array}{cc}\text{Initial Energy}& =BE\text{of}{X}^A+\text{Energy of neutron}\\ & =(6A+2)MeV\\ \text{Final Energy}& =BE\text{of}{Y}^{A+1}+\text{Energy of photon}\\ & =B{E}_Y+1MeV\end{array}$

$\begin{array}{c}\text{By energy conservation-}\\ \begin{array}{c}(6A+2)meV=B{E}_Y+\text{1 MeV}\\ \Rightarrow BE\text{of}Y=(6A+1)\text{MeV}\\ \Rightarrow \text{Binding Energy per nucleon}=\\ BE=\frac{6A+1}{A+1}\text{Mev}\end{array}\\ \text{Hence, opt (A) is correct.}\end{array}$