Question

Suppose India had a target of producing by $2020AD,200000$$MW$ of electric power, $10$ $\%$ of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e.conversion to electric energy) of thermal energy produced in a reactor was $25$ $\%$ How much amount of fissionable uranium would our country need per year by $2020$ $?$ Take $235$ $U$ to be about $200$ $MeV$ .

Answer 1

Total electric power which is aimed to be produced by 2020 = $2\times {10}^{5}MW$

Power obtained from the nuclear power plant = $10\text{\% of}2\times {10}^{5}MW=2\times {10}^{4}MW=2\times {10}^{10}W$

Energy required from the nuclear plant in one year= $\text{Power}\times \text{time}=2\times {10}^{10}\times 365\times 24\times 60\times 60=6.3\times {10}^{17}J$

Available electric energy per fission= $25\text{\% of}200MeV=(0.25\times 200\times 1.602\times {10}^{-13})J=8\times {10}^{-12}J$

Required no. of fission per year = $\frac{6.3\times {10}^{17}}{8\times {10}^{-12}}=0.7875\times {10}^{29}$

Now, $6.023\times {10}^{23}$ nuclei of ${}_{92}^{235}U$ have mass = $235g$

$\therefore$ Mass required to produce $7.9\times {10}^{28}$ nuclei = $\frac{235}{6.023\times {10}^{23}}\times 3.95\times {10}^{28}g\approx 3.084\times {10}^{4}kg$