The binding energy of deutron is 2-2 MeV and that of 42 He is 28 MeV- If two deutrons are fused to form one 42 He then the energy released is-

The correct option is B 23.6 MeV _1D^2→ _2He^4 Energy released =28 2 × 2.2 = 23.6 MeV Hence,option (B) is correct answer. ...

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Standard XII

Physics

Binding Energy

The binding e...

Question

The binding energy of deutron is $2.2 M e V$ and that of $_{2}^{4} H e$ is $28 M e V$ . If two deutrons are fused to form one $_{2}^{4} H e$ then the energy released is:

25.8 M e V

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23.6 M e V

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19.2 M e V

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30.2 M e V

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Solution

The correct option is B $23.6 M e V$
$_{1} D^{2} \to_{2} H e^{4}$
Energy released $= 28 - 2 \times 2.2 = 23.6 M e V$
Hence,
option $(B)$ is correct answer.

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The binding energy of deutron is 2.2 MeV and that of 42He is 28 MeV. If two deutrons are fused to form one 42He then the energy released is:

The correct option is B 23.6 MeV1D2→2He4Energy released =28−2×2.2=23.6MeVHence,option (B) is correct answer.

The binding energy of deutron is $2.2 M e V$ and that of $_{2}^{4} H e$ is $28 M e V$ . If two deutrons are fused to form one $_{2}^{4} H e$ then the energy released is:

The correct option is B $23.6 M e V$
$_{1} D^{2} \to_{2} H e^{4}$
Energy released $= 28 - 2 \times 2.2 = 23.6 M e V$
Hence,
option $(B)$ is correct answer.