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Binding Energy

Binding energ...

Question

Binding energy per nucleon of $_{1} H^{2}$ and $_{2} H e^{4}$ are $1.1$ MeV and $7.0$ MeV respectively. Energy released in the process $_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}$ is

A

20.8

MeV

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B

16.6

MeV

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C

25.2

MeV

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D

23.6

MeV

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Solution

The correct option is D $23.6$ MeV
Energy released
$δ E$ = Total binding energy of $_{2} H e^{4} - 2$ (total binding energy of $_{1} H^{2})$
$= 4 \times 7.0 - 2 (1.1) (2) = 23.6 M e V$

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