Question

A nuclear reactor generates $500MW$ of electrical energy using the fission of ${}_{92}{U}^{235}$. Find the mass of ${}_{92}{U}^{235}$ required to produce electricity for 30 days at the same rate.

(0.1% of the mass of ${}_{92}{U}^{235}$ gets converted into energy).

• A. 15.6 kg
• B. 14.4 kg
• C. 18.2 kg
• D. 7.2 kg

Answer 1

The correct option is B 14.4 kg

The electrical energy produced in one second is $500MJ$. The electrical energy generated in 30 days is $E=500\times 60\times 60\times 24\times 30\times {10}^6=1.3\times {10}^{15}J$. (approximately)
To generate above energy the mass required can be found using the formula $E=m{c}^2$
or $m=\frac{E}{c^2}=\frac{1.3\times {10}^{15}}{{(3\times {10}^8)}^2}=0.144\times {10}^{-1}kg$
$=1.44\times {10}^{-2}kg$
Let $M$ be the mass of uranium required to generate energy $E$.
Then ${}^{\prime }{m}^{\prime }$ is 0.1% of $M$.
i.e., $m=\frac{0.1}{100}\times M\Rightarrow M=1000m$
$=1000\times 1.44\times {10}^{-2}=14.4kg$.