Question

The binding energy per nucleon for deuterium and helium is $1.1MeV$ and $7MeV$ respectively. Find how much energy is released by the fusion of $1g$ of deuterium.

• A. $5.68\times {10}^{11}J$
• B. $1.72\times {10}^{7}J$
• C. $9.68\times {10}^{5}J$
• D. $5.68\times {10}^{3}J$

Answer 1

Answer: (A)

$5.68\times {10}^{11}J$

$\begin{array}{c}B.E{\text{of}}_1^{2}H=1.1\times 2MeV=2.2MeV\\ \text{B.E. of}{}_2^{4}He=7\times 4MeV=28MeV\end{array}$

$\begin{array}{c}\text{Nuclear reaction:}\\ 2_1^{2}H{⟶}_2^{4}He\end{array}$

$\begin{array}{cc}{\text{No. of moles of}}_2^4\text{He formed}& =\frac{1}{2}n\left({}_1^{2}H\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\text{mol}\\ \text{Energy released per reaction}& =28-(2.2\times 2)MeV\\ & =23.6MeV\end{array}$

$\text{Total energy released}=n\left({\begin{array}{}\end{array}}_2^{4}He\text{formed}\right)\times E\left(\text{released per reaction}\right)$

$\begin{array}{c}=(2.36\times {10}^6\times 1.6\times {10}^{-19})\times (\frac{1}{4}\times 6.023\times {10}^{23})J\\ =5.68\times {10}^{11}J\end{array}$