Find the energy required for separation of a Ne^20 nucleus into two α-particles and a C^12 nucleus i — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Find the energy required for separation of a Ne^20 nucleus into two α-particles and a C^12 nucleus if it is known that the binding energies per nucleon in Ne^20, He^4 and C^12 nuclei are equal to 8.03, 7.07 and 7.68 MeV, respectively.
💡 Solution & Explanation
Step 1 - Write the Separation Reaction $$\ce{^{20}_{10}Ne -> 2^{4}_{2}He + ^{12}_{6}C}$$ Step 2 - Use Binding Energy Approach Energy required = (Total BE of products) $-$ (Total BE of reactant) Given binding energies per nucleon: - $\text{BE/nucleon for Ne}^{20} = 8.03\ \text{MeV}$ → Total BE of Ne$^{20}$ = $20 \times 8.03 = 160.6\ \text{MeV}$ - $\text{BE/nucleon for He}^{4} = 7.07\ \text{MeV}$ → Total BE of 2 He$^4$ = $2 \times 4 \times 7.07 = 56.56\ \text{MeV}$ - $\text{BE/nucleon for C}^{12} = 7.68\ \text{MeV}$ → Total BE of C$^{12}$ = $12 \times 7.68 = 92.16\ \text{MeV}$ Step 3 - Calculate Energy Required Energy released when Ne forms from nucleons = 160.6 MeV. Energy released when products form from same nucleons = $56.56 + 92.16 = 148.72\ \text{MeV}$. Energy required to break Ne into the products: $$E = BE(\text{Ne}^{20}) - BE(2\text{He}^4) - BE(\text{C}^{12}) = 160.6 - 56.56 - 92.16 = \boxed{11.88\ \text{MeV}}$$ Step 4 - Evaluate Options - **(A) 6.72 MeV**: Incorrect. - **(B) 40.16 MeV**: Incorrect. - **(C) 11.88 MeV**: Matches our calculation. **Correct.** - **(D) 5.8 MeV**: Incorrect. $$\boxed{\text{Answer: C — 11.88 MeV}}$$