Question

In nuclear reaction
${}_3^{7}Li+{}_1^{1}H⟶2_2^{4}He$
the mass loss is neraly 0.02 amu. Hence, the energy released (in units of million kcal/mol) in the process is approximate:

• A. 430
• B. 220
• C. 120
• D. 50

Answer 1

The correct option is A 430

Mass defect $=0.02amu$
$=\frac{0.02}{N_0}gato{m}^{-1}$
$=\frac{0.02}{N_0}gmol-1$
$=0.02gmo{l}^{-1}$
$E=m{c}^2=0.02\times 9\times {10}^{20}ergmo{l}^{-1}$
$=\frac{0.02\times 9\times {10}^{20}\times 0.002}{8.314\times {10}^7\times {10}^6}millionkcalmo{l}^{-1}$
$=430\left(\begin{array}{c}1million={10}^6\\ 8.314\times {10}^{7}erg=0.002kcal\end{array}\right)$