Question

An explosion of atomic bomb release an energy of $7.6\times {10}^{13}\text{J}$. If $200\text{MeV}$ energy is released in fission of one ${\text{U}}^{235}$ atom, then the number of uranium atoms undergoing fission and the mass of uranium used in the tom bomb

• A. $2.375\times {10}^{24};926.66\text{g}$
• B. $3.375\times {10}^{24};926.66\text{g}$
• C. $4.375\times {10}^{24};826.66\text{g}$
• D. $5.375\times {10}^{24};826.66\text{g}$

Answer 1

The correct option is A $2.375\times {10}^{24};926.66\text{g}$

Total energy released,

$T.E=7.6\times {10}^{13}\text{J}$

Energy released in one fission,

$E_1=200\text{MeV}=200\times 1.6\times {10}^{-13}\text{J}$

Total number of uranium,

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

$\Rightarrow n=\frac{7.6\times {10}^{13}}{200\times 1.6\times {10}^{-13}}$

$\Rightarrow n=\frac{7.6}{3.2}\times {10}^{24}=2.375\times {10}^{24}$

Mass in terms of Avogadro number and number of atoms is given by,

$m=\frac{n}{N_A}M$

Here, $M=235\text{g}$

$\Rightarrow m=\frac{2.375\times {10}^{24}\times 235\text{g}}{6.023\times {10}^{23}}=926.66\text{g}$

Hence, option $\left(A\right)$ is correct.