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The binding energy per nucleon number ...

The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..

A

20.8 MeV

B

16.6MeV

C

25.2MeV

D

23.6MeV

Text Solution

Verified by Experts

The correct Answer is:
d
Energy released would be
`Delta E` =total energy of `._(2)He^(4)`
`- 2xx` (total binding energy of `._(1)_(He)^(4)`
` 4 xx 7.0 -2(1.1)(2)`
`=23.6 MeV` .
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