Q 12-13-123JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
Consider the nuclear fission $\text{Ne}^{20} \rightarrow 2\,\text{He}^4 + \text{C}^{12}$. Given that the binding energy per nucleon of $\text{Ne}^{20}$, $\text{He}^4$ and $\text{C}^{12}$ are $8.03\ \text{MeV}$, $7.07\ \text{MeV}$ and $7.86\ \text{MeV}$ respectively, identify the correct statement:
Answer: (B) Energy of $9.72\ \text{MeV}$ has to be supplied.
Total binding energies:
$$\text{Ne}^{20}: 20\times8.03 = 160.60\ \text{MeV}$$
$$2\,\text{He}^4 + \text{C}^{12}: 8\times7.07 + 12\times7.86 = 56.56 + 94.32 = 150.88\ \text{MeV}$$
The products are less tightly bound, so $160.60 - 150.88 = 9.72\ \text{MeV}$ has to be supplied.
Solution by Sreeraj P, M.Sc Physics