Chapter 14, Problem 14.55PAE

Interpretation Introduction

To determine: The amount of energy released in the fusion of $1kg$ of ${}^{235}U$ according to the equation

${}_{92}^{235}U{+}_0^{1}n{\to }_{52}^{137}Te{+}_{40}^{97}Zr+2_0^{1}n$

Explanation of Solution

${}_{92}^{235}U{+}_0^{1}n{\to }_{52}^{137}Te{+}_{40}^{97}Zr+2_0^{1}n$

Apply Einstein’s equation

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

$\Delta m=$ Mass of product $-$ mass of reactant

C: speed of light.

Mass of $1$ atom of $U=235.0439231u\left(\text{given}\right)$

Mass of $1$ neutron $=1.0086649u\left(\text{given}\right)$

Mass of $1$ atom of $Te=136.92532u\left(\text{given}\right)$

Mass of $1$ atom of $Zr=96.91095u\left(\text{given}\right)$

Mass of reactant

$\begin{array}{l}=m_U+m_n\\ =235.0439231u+1.0086649u\\ =236.0525880u\end{array}$

Mass of product

$\begin{array}{l}=m_{Te}+m_{Zr}+2\left(m_n\right)\\ =136.92532u+96.91095u+\left(2\times 1.0086649\right)u\\ =235.8535989u\end{array}$

$\begin{array}{c}\Delta m=\text{mass of product}-\text{mass of reactants}\\ \text{=235}\text{.85359984u}-236.0525880u\\ =-0.1989882u\end{array}$

Energy released in the fusion of the U-atom $=\left(-0.1989882u\right)\times C^2$

$\begin{array}{c}1u=1.66053886\times {10}^{-27}kg\\ C=3\times {10}^{8}m/s\end{array}$

$\begin{array}{c}=\left(\Delta m\right)C^2\\ =\left(0.1989882\times 1.66053886\times {10}^{-27}kg\right)\times {\left(3\times {10}^{8}m/s\right)}^2\\ =2.96973535\times {10}^{-11}J\end{array}$

No of moles in $1$ kg U $=\frac{\text{mass of U in gram}}{\text{molar mass}}$

$=\frac{1\times 1000gram}{235.0439231g/mol}$

So no of atoms in $1$ kg of $U=\text{no}\text{. of moles }\times {\text{ N}}_A\left({\text{ N}}_A=\text{Avogadro's no}\right)$

${\text{ N}}_A=6.0221415\times {10}^{23}\text{ atom/mol}$

(Putting values)

$\begin{array}{l}=\frac{1000\text{gram}}{235.0439231\text{g/mol}}\times 6.0221415\times {10}^{23}\text{atom/mol}\\ \text{=}\frac{100\times 6.0221415\times {10}^{23}}{235.0439231}\text{atoms}\end{array}$

So total amount of energy released

$\begin{array}{l}=\left(\text{amount of energy released in fusion one atom}\right)\times \text{total number of atoms}\\ \text{=2}\text{.96973535}\times {\text{10}}^{-11}J/atom\times \frac{1000\times 6.0221415\times {10}^{23}}{235.0439231}atoms\\ =7.6089\times {10}^{13}J\end{array}$

Hence energy released per kg of U $=7.6089\times {10}^{13}J$

Conclusion

Energy released by U $=7.6089\times {10}^{13}J$