It is proposed to use the nuclear fusion reaction:
 ${ }_1^{2}H{+}_1^{2}H\to {}_2^{4}He$
in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with 25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of  ${ }_1^{2}Hand{}_2^{4}He$ are 2.0141 atomic mass units and 4.0023 atomic mass units respectively).

Mass defect in the given nuclear reaction :
$$\begin{gathered} \Delta \text{m = 2}\left(\text{mass\hspace{0.17em}of\hspace{0.17em}deuterium}\right)-\left(\text{mass\hspace{0.17em}of\hspace{0.17em}helium}\right) \\ =2\left(2.0141\right)-\left(4.0026\right)=0.0256 \end{gathered}$$
Therefore, energy released
$$\begin{gathered} \Delta E=\left(\Delta m\right)\left(931.48\right)MeV=23.85MeV \\ =23.85\times 1.6\times {10}^{-13}J=3.82\times {10}^{-12}J \end{gathered}$$
Efficiency is only 25% therefore,
$25\%of\Delta E=\left(\frac{25}{100}\right)(3.82\times {10}^{-12})J=9.55\times {10}^{-13}J$
i.e, by the fusion of two deuterium nuclei,  $9.55\times {10}^{-13}J$ energy is available to the nuclear reactor.
Total energy required in one day to run the reactor with a given power of 200 MW:
 $E_{total}=200\times {10}^6\times 24\times 3600=1.728\times {10}^{-13}J$
  Total number of deuterium nuclei required for this purpose
$n=\frac{E_{total}}{\Delta E/2}=\frac{2\times 1.728\times {10}^{13}}{9.55\times {10}^{-13}} =0.362\times {10}^{26}$

Mass of deuterium required = (Number of g-moles of deuterium required) $\times 2g$

$=\left(\frac{0.362\times {10}^{26}}{6.02\times {10}^{23}}\right)\times 2=120.26g.$