Question

The average energy released in the fission of a ${}^{235}\mathbf{U}$ nucleus is about 200 MeV. Suppose the conversion of this energy to electrical energy is 40% efficient. What mass of ${}^{235}\mathbf{U}$is converted to its fission products in a year’s operation of a 1000-megawatt nuclear power station? Recall that 1 Wis$\mathbf{1}{\mathbf{Js}}^{-1}$

Short answer

Mass of uranium consumed in a year $=9.6\times {10}^5\mathrm{g}$

Step-by-step solution

Energy required to run the power station in MeV

Power required to run the power station = $1000\mathrm{MW}=1\times {10}^9\mathrm{J}/\mathrm{s}$

Energy consumed by the power station in a year$=1\times {10}^9{\mathrm{Js}}^{-1}\times 365\times 24\times 60\times 60\mathrm{s}=3.15\times {10}^{16}\mathrm{J}$

Conversion of energy from Joules to MeV

$\begin{array}{l}1.6\times {10}^{-13}\mathrm{J}=1\text{\hspace{0.17em}}\mathrm{MeV}\\ 3.15\times {10}^{16}\mathrm{J}=\frac{3.15\times {10}^{15}\mathrm{J}}{1.6\times {10}^{-13}}=1.98\times {10}^{29}\text{}\mathrm{MeV}\end{array}$

Mass of uranium consumed

Energy released in a fission reaction of one atom$=200\mathrm{MeV}$

Let the total number of atom decaying in a year be x.

$\begin{array}{l}\frac{40}{100}\times 200\times \mathrm{x}=1.98\times {10}^{29}\\ \mathrm{x}=\frac{1.98\times {10}^{28}}{80}=0.02475\times {10}^{29}\end{array}$

Mass of $0.02475\times {10}^{28}$atoms $=\frac{0.02475\times {10}^{29}}{6.023\times {10}^{23}}\times 235g=9.6\times {10}^{5}g$