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. `17.3` MeV
B. `19.6` MeV
C. `-2.4` MeV
D. `8.4` MeV

Answer

A

Explanation

Binding energy of `"" _(3) ^(7) Li` nucleus = `7 xx 5.60 MeV = 39 . 2 MeV`
Binding energy of `"" _(2) ^(4) He` nucleus =
`4 xx 7.06 MeV = 28.24 MeV`
The reaction is
`"" _(3) ^(7) Li + "" (1) ^(1) H to 2 )("" _(2) ^(4) He) + Q`
`therefore Q = 2 (Be of "" _(2) ^(4) He) - (Be of "" _(3) ^(7) Li)`
`= 2 x 28.24 MeV - 39.2 MeV`
`= 56 48 MeV -39.2 MeV`
`= 17.28 MeV ~~ 17.3 MeV`