Question

The binding energy per nucleus of ${}_1^{7}Li$ and ${}_2^{4}He$ nuclei are $5.60MeV$ and $7.06MeV$ respectively. In the nucleus reaction ${}_3^{7}Li{+}_1^{1}He{+}_2^{4}He+Q$ the value of energy $Q$ released is:

• A. $19.6MeV$
• B. $-2.4MeV$
• C. $8.4MeV$
• D. $17.3MeV$

Answer 1

The correct option is D $17.3MeV$

${}_3^{7}Li{+}_1^{1}H{\to }_4^{2}He+Q.$

$\begin{array}{c}\text{Lithium has}4\text{neutrons,}\\ \begin{array}{cc}Q& =2(4\times 7.06)-[7\times 5.6]\\ & =8\times 7.00-7\times 5.60\\ Q& =17.3MeV\end{array}\end{array}$