Atoms & Nuclei

JEE Main level, single correct, medium

231 of 239

Question 231 of 239

Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron=1.1MeV=\text{1.1}MeV and for helium=7.0MeV=\text{7.0}MeV)

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Answer: (A) 23.6MeV\text{23.6}MeV

Q=∑B.E.ofproducts−∑B.E.ofreactantsQ=\sum B.E.ofproducts-\sum B.E.ofreactants

The equation for two deuterons combining to form Helium nucleus is given as H2+1H2→2H4{H}^{2}{+}_{1}{H}^{2}{\rightarrow }_{2}{H}^{4}

Energyreleased=Q=4[B.E.(2H4)]−4[B.E.(1H2)]Energyreleased=Q=4\left[B.E.\left({}_{2}{H}^{4}\right)\right]-4\left[B.E.\left({}_{1}{H}^{2}\right)\right]
=4×7−4×1.1=4\times 7-4\times \text{1.1}
=28−4.4=28-\text{4.4}
The energy released in this process is =23.6MeV=\text{23.6}MeV

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