Question

If the binding energy per nucleon of ${}_{3}L{i}^7$ and ${}_{2}H{e}^4$ nuclei are $5.60\text{MeV}$ and $7.06\text{MeV}$ respectively, then in the reaction, the energy of proton must be,

• A. $39.2\text{MeV}$
• B. $29.24\text{MeV}$
• C. $17.28\text{MeV}$
• D. $1.46\text{MeV}$

Answer 1

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

$\text{Binding Energy}(B.E_{Li})=A\times \left(\frac{\text{Binding Energy}}{\text{nucleon}}\right)$

So, binding energy of,

$B_{Li}=7\times \left(5.60\right)=39.20\text{MeV}$

$B_{He}=4\times \left(7.06\right)=28.24\text{MeV}$

$\therefore$ Energy of proton is,

$E_P=2B_{He}-B_{Li}$

$=2\left(28.24\right)-39.20$

$=56.48-39.20=17.28\text{MeV}$

Hence, $\left(C\right)$ is the correct answer.