Chapter 31, Problem 54GP

(a)

To determine

Mass of $\text{U}$ actually fissioned in the first atomic bomb.

Answer to Problem 54GP

Mass of $\text{U}$ actually fissioned in the first atomic bomb is $\mathrm{m}=1.22\text{ kg}$

Explanation of Solution

Given:

Equivalent energy is $\mathrm{E}=20\text{ kilotons of TNT}$

Energy released by 1 kiloton of TNT is $5\times {10}^{12}\text{J}$

  $1\text{ eV}=1.602\times {10}^{-19}\text{J}$

Formula used:

Number of particles can be given by: $\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

Calculation:

According to the statement, the energy released in the fission process can be given as:

  $\begin{array}{l}\mathrm{E}=20\text{ kilotons of TNT}\\ \mathrm{E}=20\left(5\times {10}^{12}\text{J}\right)\\ \mathrm{E}=\frac{{10}^{14}\text{J}}{1.602\times {10}^{-19}}\times {10}^{-6}\text{MeV}\\ \mathrm{E}=6.24\times {10}^{26}\text{MeV}\end{array}$

Energy released in each fission reaction is $\text{200 MeV}$ . So, the number of particles can be given by:

  $\begin{array}{l}\mathrm{n}=\frac{6.24\times {10}^{26}\text{MeV}}{200\text{ MeV}}\\ \mathrm{n}=3.12\times {10}^{24}\end{array}$

Hence, the actual mass fissioned in the first atomic bomb can be given as:

  $\begin{array}{l}\mathrm{m}=\mathrm{n}\times \mathrm{M}\\ \mathrm{m}=\left(3.12\times {10}^{24}\right)\left(\frac{235\times {10}^{-3}\text{kg}}{6.023\times {10}^{23}}\right)\\ \mathrm{m}=1.22\text{ kg}\end{array}$

Conclusion:

Hence, the actual mass fissioned in the process is $\mathrm{m}=1.22\text{ kg}$

(b)

To determine

Actual mass transformed to energy.

Answer to Problem 54GP

Actual mass transformed is $\mathrm{m}=1.11\times {10}^{-3}\text{kg}$

Explanation of Solution

Given:

Equivalent energy is $\mathrm{E}=20\text{ kilotons of TNT}$

Energy released by 1 kiloton of TNT is $5\times {10}^{12}\text{J}$

  $1\text{ eV}=1.602\times {10}^{-19}\text{J}$

Calculation:

We know that that mass-energy equivalence equation can be given as:

  $\mathrm{E}=\mathrm{m}{\mathrm{c}}^2$

So, the actual mass transformed is given by:

  $\begin{array}{l}\mathrm{m}=\frac{\mathrm{E}}{{\mathrm{c}}^2}\\ \mathrm{m}=\frac{20\left(5\times {10}^{12}\text{J}\right)}{{\left(3\times {10}^8\text{m/s}\right)}^2}\\ \mathrm{m}=1.11\times {10}^{-3}\text{kg}\end{array}$

Conclusion:

The actual mass transformed is $\mathrm{m}=1.11\times {10}^{-3}\text{kg}$