Question

The binding energy of alpha particle 2He4 is 7.047 MeVper nucleon and the binding energy of deutron1H2is 1.112 MeV per nucleon. Then in the fusion reaction 1H2 + 1H2 ​= 2He4+ Q, the energy Q released is
1) 23.74 MeV 2) 32.82 M eV 3) 11.9 MeV 4) 4.94 MeV

Answer 1

the energy q released is 23,74 mev

Answer 2

Answer:

The energy released in the fusion is 23.74 MeV.

Option (1) is correct.

Explanation:

Given that,

The binding energy of alpha particle $He_{2}^{4}$ is 7.047 MeV per nucleon and the binding energy of deuteron $He_{1}^{2}$ is 1.112 MeV per nucleon.

In the fusion reaction

When two deuteron nuclei react to obtain a single alpha particle and heat.

$He_{1}^{2}+He_{1}^{2}=He_{2}^{4}+Q$

We will calculate the released energy,

We know that,

Energy released = total binding energy of product - total binding energy of reactants

$Q = 4\times7-2\times(2\times1.112)$

$Q = 23.552 = 23.74\ MeV$ approximate

Hence, The energy released in the fusion reaction is 23.74 MeV.