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>> $The binding energy per nucleon for 1^2H$

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

$4.4$

B

$8.2$

C

$24$

D

$28.4$

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Updated on : 2022-09-05

Solution

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Correct option is C)

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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