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Mechanism of Nuclear Fission

The binding e...

Question

The binding energy per nucleon of deuteron $(_{1} H^{2})$ and helium nucleus $(_{2} {H e}^{4})$ is $1.1$ $M e V$ and $7$ $M e V$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is

13.9 M e V

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26.9 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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Solution

The correct option is D $23.6 M e V$
Energy required to break $4$ nucleons in the two deuterons $= 4 \times 1.1 M e V$
Energy released in formation of $4$ nucleons in the single helium nucleus $= 4 \times 7 M e V$
Hence, total energy released $= (28 - 4.4) M e V = 23.6 M e V$

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