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Nuclear Fusion

If the bindin...

Question

If the binding energy per nuclear in ${L i}^{7}$ and ${H e}^{4}$ nuclei are respectively $5.60 M e V$ and $7.06 M e V$ , then energy of reactor ${L i}^{7} + P \to 2_{2} {H e}^{4}$ is:

19.6 M e V

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2.4 M e V

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8.4 M e V

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17.6 M e V

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Solution

The correct option is D $17.6 M e V$
BE of ${L i}^{7} = 39.20 M e V$
and ${H e}^{4} = 28.24 M e V$
Hence binding energy of $2 \times_{2} {H e}^{4} = 56.84 M e V$
Energy of reaction $= 56.84 - 39.20 = 17.64 M e V$

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