Atoms & Nuclei

JEE Main level, single correct, medium

179 of 239

Question 179 of 239

The energy released if hydrogen atoms are combined to form 24He^{4}_{2}\text{He} is __________ MeV. (Take binding energies per nucleon of 12H^{2}_{1}\text{H} and 24He^{4}_{2}\text{He} as 1.11.1 MeV and 7.27.2 MeV, respectively)

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Answer: (B) 24.424.4

The fusion reaction is given by:

212H→24He2 ^{2}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}

Binding energy of one 12H^{2}_{1}\text{H} nucleus = 2×1.1=2.22 \times 1.1 = 2.2 MeV

Total initial binding energy = 2×2.2=4.42 \times 2.2 = 4.4 MeV

Binding energy of 24He^{4}_{2}\text{He} nucleus = 4×7.2=28.84 \times 7.2 = 28.8 MeV

Energy released ΔE\Delta E = Final binding energy - Initial binding energy

ΔE=28.8−4.4=24.4\Delta E = 28.8 - 4.4 = 24.4 MeV

Answer: 24.424.4

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