Question

When ${}_{92}{U}^{235}$ undergoes fission, $0.1\%$ of its mass is converted into energy. How much mass of ${}_{92}{U}^{235}$ will be required to produce $10\text{MJ}$ of energy?

• A. $1.11\times {10}^{-5}\text{kg}$
• B. $4.44\times {10}^{-7}\text{kg}$
• C. $1.11\times {10}^{-7}\text{kg}$
• D. $4.44\times {10}^{-5}\text{kg}$

Answer 1

The correct option is C $1.11\times {10}^{-7}\text{kg}$

We know that, $E=\Delta m{c}^2$

Here, $\Delta m=0.1\%\text{of}M={10}^{-3}M$

Where, $M$ is the total mass of ${}_{92}{U}^{235}$ undergoing fission reaction.

$\therefore 10\times {10}^6={10}^{-3}M\times (3\times {10}^8{)}^2$

$\therefore M=1.11\times {10}^{-7}\text{kg}$

Hence, $\left(C\right)$ is the correct answer.