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Calculate the energy released in fusion ...

Calculate the energy released in fusion reaction :
`4._(1)^(1)Hto._(2)^(4)He+2._(+1)^(0)e`
Given : mass of `._(1)^(1)H=1.007825u`, mass of `._(2)^(4)He=4.00260` u and 1u=931.5 MeV
Neglect the mass of positron `(._(+1)^(0)e)`.

Text Solution

Verified by Experts

The correct Answer is:
`26.7 MeV`
Mass defect `trianglem=` mass of `4._(1)^(1)H-` mass of `._(2)^(4)He=4(1.007825u)-4.00260u=0.0287u`
`=0.0287xx931.5MeV=26.7MeV`.
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