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Calculate the energy released in joules ...

Calculate the energy released in joules and MeV in the following nuclear reaction :
`._(1)^(2) H+_(1)^(2) rarr _(2)^(3)He + _(0)^(1)n`
Assume that the masses of `._(1)^(2)H, _(2)^(3)He` and neutron (n) respectively are `2.020, 3.0160 ` and `1.0087` in amu.

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The correct Answer is:
`[Delta E =2.28 xx 10^(-12)J =14.25 MeV ]`
`DeltaE=Delta m xx931.5 Me rArrDeltam=SigmaM_(R)-SigmaM_(P) rArr Deltam=2(2.020)-(3.0160+1.0087)`
`rArrDeltaE=Deltamxx931.5=14.25 MeV`
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