Question

In the process of nuclear fission of $1\, g$ of uranium, the mass lost is $0.90\, mg$. The efficiency of fission reactor of power house is $20 \%$. To obtain $400\, MW$ power from the power house, how much uranium (in gram) is required per hour?

Answer 1

Power $=400\, MW$
$\therefore $ Energy obtained per hour $=400 \times 10^{6} \times 3600$
$=144 \times 10^{10} J$
Input energy of power house
$=\frac{144 \times 10^{10} \times 10}{20}=72 \times 10^{11} J$
Now, $\Delta m c^{2}=72 \times 10^{11}$
$\Delta m=\frac{72 \times 10^{11}}{9 \times 10^{16}} kg =0.08\, g$
As mass lost in $1\, g$ uranium is $0.9 \times 10^{-3} g$,
$\therefore $ Required mass $=0.08 /\left(0.9 \times 10^{-3}\right)$
$=8.89\, g$