JEE Mains · Physics · STD 12 - 13. Nuclei
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)
- A \(6.1\)
- B \(24.4\)
- C \(26.6\)
- D \(5\)
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 =…
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