The binding energy of deuteron is `2.2` MeV and that of `._(2)^(4)He` is `28` MeV. If two deuterons are fused to form one `._(2)^(4)He`, th `n` the energy released is

Options

A. `30.2 meV`
B. `25.8` MeV
C. `23.6` MeV
D. `19.2` MeV

Answer

C

Explanation

`"" _(1) ^(2) H + "" _(1) ^(2) H to "" _(2) ^(4) He + `energy
energy released = B.E of
` "" _(2) ^(4) He -2 (B.E of "" _(1) ^(2)H)`
`= 28- (2x 2.2 ) = 28-4.4 =23 6 MeV`