The binding energy of deuteron 21H is 1-15 MeV per nucleon and an alpha particle 42He has a binding energy of 7-1 MeV per nucleon- Then in the reaction21H - 21H - 42 He - Qthe energy Q released is

Binding energy of ^2_1 H = 1.15 × number of nucelons = 1.15 × 2 = 2.3 MeV. Total binding energy of reactants = 2.3 + 2.3 = 4.6 MeV. Binding energy of ^4_2He = 7 ...

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Standard XII

Physics

Mechanism of Nuclear Fission

The binding e...

Question

The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$
the energy Q released is

1 MeV

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11.9 MeV

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23. MeV

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931 MeV

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Solution

The correct option is C
23. MeV

Binding energy of $_{1}^{2} H = 1.15 \times$ number of nucelons = 1.15 $\times$ 2 = 2.3 MeV. Total binding energy of reactants = 2.3 + 2.3 = 4.6 MeV. Binding energy of $_{2}^{4} H e = 7.1 \times$ number of nucleons = 7.1 $\times$ 4 = 28.4 MeV. Therefore, Q = 28.4 - 4.6 = 23.8 MeV.
Hence the correct choice is (c)

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The binding energy of deuteron (21H) is 1.15 MeV per nucleon and an alpha particle (42He) has a binding energy of 7.1 MeV per nucleon. Then in the reaction 21H+21H→42He+Q the energy Q released is

The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$
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