Chapter 6, Problem 1P

To determine

Find the mass of ${}^{235}\text{U}$ atom.

Answer to Problem 1P

The mass of ${}^{235}\text{U}$ atom is $\underset{\_}{4.2\times {10}^{-4}\text{kg}}$.

Explanation of Solution

Refer Appendix V “Energy Content of Fuels” in the textbook,

The thermal energy content of one tonne of coal is $3.10\times {10}^{10}\text{J}$.

Consider the energy per fission for ${}^{235}\text{U}$ is 180 MeV.

Convert the energy per fission MeV into Joules.

$\begin{array}{c}180\text{MeV}=180\text{MeV}\times \frac{1.6\times {10}^{-13}\text{J}}{\text{MeV}}\\ =2.88\times {10}^{-11}\text{J}\end{array}$

Find the number of fissions required to produce the same energy as one tonne of coal.

$\begin{array}{c}\text{Number of fissions}=\frac{\text{Thermal energy content of one tonne of coal}}{\text{The energy per}\text{fission}}\\ =\frac{3.10\times {10}^{10}\text{J}}{2.88\times {10}^{-11}\text{J}}\\ =1.08\times {10}^{21}\text{fissions}\end{array}$

Find the mass of ${}^{235}\text{U}$ atom using the relation:

$\begin{array}{c}\text{Mass of}{}^{235}\text{U} \text{atom}=\left(\frac{\text{Number of fissions}}{6.02\times {10}^{23}{\text{mol}}^{-1}}\right)\times \left(\frac{235\text{g/mol}}{1,000\text{g/kg}}\right)\\ =\left(\frac{1.08\times {10}^{21}\text{fissions}}{6.02\times {10}^{23}{\text{mol}}^{-1}}\right)\times \left(\frac{235\text{g/mol}}{1,000\text{g/kg}}\right)\\ =4.2\times {10}^{-4}\text{kg}\end{array}$

Thus, the mass of ${}^{235}\text{U}$ atom is $\underset{\_}{4.2\times {10}^{-4}\text{kg}}$.