The binding energy of deuteron is 2.2 Me...

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

`23.6 MeV`

`19.2 MeV`

`30.2 MeV`

`25.8 MeV`

Text Solution

Verified by Experts

The correct Answer is:

`2 "Deuteron rarr ._(2)^(4)He`
`Q = "BE of Product" - "BE of reactant"`
`= [28 MeV - 2(2.2 MeV)]`
`= [28-4.4 ] = 23.6 MeV`

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