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

Answer 1

The correct option is B $17.28MeV$

$\begin{array}{c}\text{Given}-BE{\text{of}}_3^7{L}_i=5.6MeV,BE{\text{of}}_2^{4}He=7.06MeV\\ p+\frac{7}{3}Li\to 2\cdot \frac{4}{2}He\end{array}$

$\begin{array}{cc}\text{Energy of proton}& =2\cdot BE\text{of}He-BE\text{of Li}\\ & =2\cdot [4\times 7.06]-[7\times 5.60]\\ & =17.28MeV\end{array}$