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Binding Energy

The binding e...

Question

The binding energy per nucleon for $_{1}^{2} H$ and $_{2}^{4} H e$ respectively are $1.1$ MeV and $7.1$ MeV. The energy released in MeV when two $_{1}^{2} H$ nuclei fuse to form $_{2}^{4} H e$ is

A

4.4

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B

8.2

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C

24

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D

28.4

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Solution

The correct option is C $24$
The chemical reaction of process is:
$2_{1}^{2} H ⟶_{2}^{4} H e$
Energy released $= 4 \times (7.1) - 4 (1.1) = 24 e V$

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