Question

It is estimated that the energy released in the explosion of the atomic bomb at Hiroshima was $9\times {10}^{13}\text{J}$. If an average $200\text{MeV}$ of energy is released in the fission of one ${}_{92}{\text{U}}^{235}$. Then

• A. The mass of uranium used for the bomb is nearly $1.1\text{k}g$.
  
• B. The mass of uranium used for the bomb is nearly $9.1\text{k}g$.
  
• C. The number of uranium atoms undergoing fission are $2.81\times {10}^{24}$.
• D. The number of uranium atoms undergoing fission are $2.81\times {10}^{29}$.

Answer 1

Answer: (A), (C)

The number of uranium atoms undergoing fission are $2.81\times {10}^{24}$.

Energy released in one fission of one atom,

$E_1=200\text{MeV}$

$\Rightarrow E_1=200\times 1.6\times {10}^{-13}\text{J}=3.2\times {10}^{-11}\text{J}[\because e=1.6\times {10}^{-19}\text{C}]$

The number of atom or fission required to release the energy of $E=9\times {10}^{13}\text{J}$ is,

$n=\frac{E}{E_1}$

$\Rightarrow n=\frac{9\times {10}^{13}}{3.2\times {10}^{-11}}=2.81\times {10}^{24}$

We know that, $235\text{g}$ of ${\text{U}}^{235}$ contains $6.023\times {10}^{23}{\text{U}}^{235}$ atoms.

Let, $m\text{g}$ of ${\text{U}}^{235}$ contains $2.81\times {10}^{24}{\text{U}}^{235}$ atoms,

$\Rightarrow m=\frac{235\times 2.81\times {10}^{24}}{6.023\times {10}^{23}}$

$\Rightarrow m=1096.3\text{g}$

$\Rightarrow m\approx 1.1\text{kg}$

Hence, $\left(A\right)$ and $\left(C\right)$ are the correct answers.

Why this question? This question gives an idea how to solve real life numerical.