Find the energy required for separation of a nucleus into two alpha-particles and a nucleus if it is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Total BE of $\text{Ne}^{20} = 20 \times 8.03 = 160.6 \text{ MeV}$. Total BE of $2\text{He}^4 = 2 \times 4 \times 7.07 = 56.56 \text{ MeV}$. Total BE of $\text{C}^{12} = 12 \times 7.68 = 92.16 \text{ MeV}$. Energy required to separate is the difference in binding energies: $\Delta E = BE(\text{Ne}^{20}) - (BE(2\text{He}^4) + BE(\text{C}^{12})) = 160.6 - (56.56 + 92.16) = 160.6 - 148.72 = 11.88 \text{ MeV}$. Therefore, correct answer is C.