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Question: The binding energy of deuteron <sub>1</sub>H<sup>2</sup> is 1.112 MeV per nucleon and a-particle <su...

The binding energy of deuteron 1H2 is 1.112 MeV per nucleon and a-particle 2He4 has a binding energy of 7.047 MeV per nucleon. Then in the relation :

1H2 + 1H2®2He4 + Q the energy Q released is :

A

1 MeV

B

11.9 MeV

C

23.8 MeV

D

931 MeV

Answer

23.8 MeV

Explanation

Solution

Released energy = BE of RHS – LHS

= (7.047 × 4) – (4 × 1.112)