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The binding energy per nucleon of .(3)^(...

The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released is

A

19.6 MeV

B

`-2.4` MeV

C

8.4 MeV

D

17.3 MeV

Text Solution

Verified by Experts

The correct Answer is:
D
Q= Binding energy of product - Binding energy of reactant.
`=2(4 xx B.E. of He) -(7 xx B.E. of Li)`
`=8 xx 7.06 - 7 xx 5.60 = 56.48 - 39.2=17.28` MeV
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