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

Options

A. `19.6` Me V
B. `-2.4` Me V
C. `8.4` Me V
D. `17.3` Me V

Answer

D

Explanation

The binding energy for `""_(1)H^(1)` is around zero and also not given in the question so we can igonor it
`Q=2(4xx7.06)-7xx(5.60)=(8xx7.06)-(7xx5.60)`
`=56.48-39.2=17.28MeV~=17.3 MeV`