Q131 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
Two deuterons are fused to form one alpha particle. If binding energy per nucleon of deuterium is 1.05 MeV and that of alpha particle is 7 MeV, what is the energy released in the formation of one alpha particle from the fusing of two deuterons?
- A$26.8\ \text{MeV}$
- B$28.3\ \text{MeV}$
- Correct answerC$23.8\ \text{MeV}$
- D$24.8\ \text{MeV}$
Solution
The fusion reaction is given by:
$2\ ^{2}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He}$
Total binding energy of two deuterons = $2 \times (2 \times 1.05) = 4.2\ \text{MeV}$
Total binding energy of one alpha particle = $4 \times 7 = 28\ \text{MeV}$
Energy released = Total binding energy of products - Total binding energy of reactants
Energy released = $28 - 4.2 = 23.8\ \text{MeV}$
Answer: $23.8\ \text{MeV}$
$2\ ^{2}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He}$
Total binding energy of two deuterons = $2 \times (2 \times 1.05) = 4.2\ \text{MeV}$
Total binding energy of one alpha particle = $4 \times 7 = 28\ \text{MeV}$
Energy released = Total binding energy of products - Total binding energy of reactants
Energy released = $28 - 4.2 = 23.8\ \text{MeV}$
Answer: $23.8\ \text{MeV}$
Hence, the correct answer is option: 'C'.