Question

If binding energy per nucleon in $L{i}^7$ and $H{e}^4$ nuclei are $5.60$ and $7.06MeV$ , respectively then find the energy of the following reaction:
$L{i}^7+p(proton)\to 2H{e}^4$

• A. $17\cdot 3MeV$
• B. $1\cdot 73MeV$
• C. $1\cdot 46MeV$
• D. depends on binding energy of proton

Answer 1

Answer: (A)

The binding energy for proton $\left(H_1^1\right)$ is around zoro and also not given in the question. So we can ignore it.
Binding energy $=$ (Total energy of product side) $-$ (Total energy of reactant side)
Binding energy per nucleon $\left(\frac{B.E}{A}\right)$ in $L^7$ is $5.60Mev$
Total binding energy of $\left(L^7\right)$ $=7\times 5.60Mev$
Binding energy per nucleon $\left(\frac{B.E}{A}\right)$ in $H{e}^4$ is $7.60Mev$
Total binding energy of $\left(H{e}^4\right)$ $=4\times 7.06Mev$
Total energy $Q=2\left(4\times 7.06\right)-\left(7\times 5.60\right)$
$=56.48-39.2=17.28\approx 17.3Mev$