Question

If binding energy per nucleon in $Li^{7}$ and $He^{4}$ nuclei are $5.60$ and $7.06MeV$ , respectively then find the energy of the following reaction:
$Li^{7}+p\left(\right.proton\left.\right)\overset{}{ \rightarrow }2He^{4}$

• 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 $He^{4}$ is $7.60Mev$
Total binding energy of $\left(H e^{4}\right)$ $=4\times 7.06Mev$
Total energy $Q=2\left(4 \times 7 . 06\right)-\left(7 \times 5 . 60\right)$
$=56.48-39.2=17.28\approx17.3Mev$