Question

If the binding energy per nucleon in
$\begin{array}{c}7\\ 3\end{array}Li$ and $\begin{array}{c}4\\ 2\end{array}He$ nuclei are $5.60\text{MeV}$ and $7.06\text{MeV}$ respectively, then in the reaction $p+\begin{array}{c}7\\ 3\end{array}Li\to 2\begin{array}{c}4\\ 2\end{array}He$, energy of proton must be

• A. $28.24\text{MeV}$
• B. $17.28\text{MeV}$
• C. $1.46\text{MeV}$
• D. $39.2\text{MeV}$

Answer 1

The correct option is B $17.28\text{MeV}$

Energy of proton $=2\times [4\times 7.06]-7\times 5.60$
$\therefore$ Energy of proton = $17.28\text{MeV}$.