Question

If the binding energy per nucleon in ${}_3^{7}Li$ and ${}_2^{4}He$ nuclei are $5.60MeV$ and $7.06MeV$ respectively, then in the reaction
$p{+}_3^{7}Li\to 2_2^{4}He$
Energy of proton must be -

• A. $39.2MeV$
• B. $28.24MeV$
• C. $17.28MeV$
• D. $1.46MeV$

Answer 1

The correct option is C $17.28MeV$

$\begin{array}{c}\text{Given-}\ast {\text{Binding energy per neucleon}}_3^7{L}_i=5.60Mev\\ \text{* Binding energy per nucleon}\frac{4}{2}He=7.06MeV\\ \text{Total Binding Energy (BE)}\end{array}$

$\begin{array}{c}B{E}_T=B{E}_{\text{Product}}-B{E}_{\text{Reactant}}\\ BE=(\text{num of neutrons}+\text{protons})\times BE\text{per neucleons}\\ \Rightarrow B{E}_T=[8\times 7.06]-[7\times 5.60]\\ =17.28MeV\end{array}$