Question

The binding energy per nucleon of ${}_3^7\text{Li}$ and ${}_2^4\text{He}$ nuclei are $5.60\text{MeV}$ and $7.06\text{MeV}$,respectively. In the nuclear reaction ${}_3^7\text{Li}{+}_1^1\text{H}\to {}_2^4\text{He}+Q,$ the value of energy $Q$ released is

• A. $19.6\text{MeV}$
• B. $-2.4\text{MeV}$
• C. $8.4\text{MeV}$
• D. $17.3\text{MeV}$

Answer 1

The correct option is D $17.3\text{MeV}$

Binding Energy of ${}_2^4\text{He}=4\times 7.06=28.24\text{MeV}$
Bindinge energy of ${}_3^7\text{Li}=7\times 5.60=39.20\text{MeV}$

Given nuclear reaction
${}_3^7\text{Li}+{}_1^1\text{H}\to {}_2^4\text{He}+Q$
$39.2028.24\times 2$

$Q=2\times 28.24-39.20=17.28\text{MeV}$