In the nuclear reaction `""_(3)^(7)Li + ""_(1)^(1)H to 2""_(2)^(4)He`, the mass loss is nearly 0.02 amu. Hence, the energy released (in units of million kcal/mol) in the process is approximately.

Options

A. 430
B. 220
C. 120
D. 50

Answer

A

Explanation

Mass defect ` = 0.02` amu ` = (0.02)/(N_(0))` g `"atom"^(1) = (0.02)/(N_(0)) xx N_(0)g "mol"^(-1) = 0.02 g "mol"^(-1)`
`E = mc^(2) = 0.02 xx 9 xx 10^(20)`erg `"mol"^(-1) = (0.02 xx 9 xx 10^(20) xx 0.002)/(8.314 xx 10^(7) xx 10^(6)) = 430` million kcal `"mol"^(-1)`
`(1 "million" = 10^(6)`
`8.314 xx 10^(7) "erg" = 0.02 "kcal")`