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The binding energy per nucleon number ...

The binding energy per nucleon number for deuteron `(_(1) H^(2) ) `and belium `(_(2) Hx^(4))` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `(_(2) He^(4)) ` is ……..

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The correct Answer is:
B, C
`_(1)^(2) H + _(2)^(4)H e `
Binding energy of two deuterons
` = 2 [ 1.1 xx 2 ] = 4.4 MeV`
Binding of the belium nucleue ` = 4 xx 7.0 = 28 MeV`
The energy released ` = 28 - 4.4 = 23.6 MeV`
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