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Sub-Atomic Particles

The binding e...

Question

The binding energy per nucleon for $_{1}^{2} H$ and $_{2}^{4} H e$ respectively are $1.1 M e V$ and $7.1 M e V$ . The energy released in $M e V$ 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 A 24
The process is $2_{1}^{2} H \to_{2}^{4} H e$ .
Energy released $= 4 \times (7.1) - 4 (1.1) = 24 e V$

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