JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2026
- Medium
- MCQ
The energy released if hydrogen atoms are combined to form \(^{4}_{2}\text{He}\) is __________ MeV. (Take binding energies per nucleon of \(^{2}_{1}\text{H}\) and \(^{4}_{2}\text{He}\) as \(1.1\) MeV and \(7.2\) MeV, respectively)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(24.4\)
Step-by-step Solution
Detailed explanation
The fusion reaction is given by:
\(2 ^{2}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}\)
Binding energy of one \(^{2}_{1}\text{H}\) nucleus = \(2 \times 1.1 = 2.2\) MeV
Total initial binding energy = \(2 \times 2.2 = 4.4\) MeV
Binding energy of \(^{4}_{2}\text{He}\) nucleus = \(4 \times 7.2 = 28.8\) MeV
Energy released \(\Delta E\) = Final binding energy - Initial binding energy
\(\Delta E = 28.8 - 4.4 = 24.4\) MeV
Answer: \(24.4\)
\(2 ^{2}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}\)
Binding energy of one \(^{2}_{1}\text{H}\) nucleus = \(2 \times 1.1 = 2.2\) MeV
Total initial binding energy = \(2 \times 2.2 = 4.4\) MeV
Binding energy of \(^{4}_{2}\text{He}\) nucleus = \(4 \times 7.2 = 28.8\) MeV
Energy released \(\Delta E\) = Final binding energy - Initial binding energy
\(\Delta E = 28.8 - 4.4 = 24.4\) MeV
Answer: \(24.4\)
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