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Calculate the energy released in MeV dur...

Calculate the energy released in MeV during the reaction `._(3)^(7)Li + ._(1)^(1)H to 2[._(2)^(4)He]` if the masses of `._(3)^(7)Li, ._(1)^(1)H` and `._(2)H_(4)He` are 7.018, 1.008 and 4.004 amu respectively.

Text Solution

Verified by Experts

The correct Answer is:
`7.45`
Here , we are given atomic masses . In the given nuclear reaction , masses of 4 electrons before and after reaction cancel out . Therefore , we can use there masses as corresponding nuclear masses .
`Q =Deltam xx 931 `MeV
`=[m(Li)+m(H)-2m(He)]xx931`MeV
`=[7.016 xx 1.008 -2 xx 4.004 ]xx931`
`= 0.061 xx 931 = 14.896 `MeV
This energy is shared equally between two alpha particles .
` :. ` Energy of each `alpha `- particle
`= (14.896)/2 approx 7.45 `MeV .
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