Suppose India Had a Target of Producing, by 2020 AD, 200,000 MW of Electric Power, Ten Percent of Which Was to be Obtained from Nuclear Power Plants. Suppose We are Given That, on an Average, the Efficiency of Utilization (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 the Heat Energy Per Fission of ${}^{235}U$ to be about \[\mathbf{200MeV}\].
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Ans: We are given the following:
Amount of electric power to be generated, \[P=2\times {{10}^{5}}MW\], 10% of this amount has to be obtained from nuclear power plants.
Amount of nuclear power, ${{P}_{1}}=\frac{10}{100}\times 2\times {{10}^{5}}=2\times {{10}^{4}}MW$
$\Rightarrow {{P}_{1}}=2\times {{10}^{4}}\times {{10}^{6}}J/s=2\times {{10}^{10}}\times 3600\times 24\times 365J/y$
Heat energy released per fission of a ${}^{235}U$ nucleus, \[E=200\text{ }MeV\]
Efficiency of a reactor \[=25%\]
Hence, the amount of energy converted into the electrical energy per fission is calculated as:
$\frac{25}{100}\times 200=50MeV=50\times 1.6\times {{10}^{-19}}\times {{10}^{6}}=8\times {{10}^{-12}}J$
The number of atoms required for fission per year would be:
$\frac{2\times {{10}^{10}}\times 60\times 60\times 24\times 365}{8\times {{10}^{-12}}}=78840\times {{10}^{24}}atoms$
1 mole, i.e., 235g of ${{U}^{235}}$contains \[6.023\times {{10}^{23}}atoms\]
That is, the mass of \[6.023\times {{10}^{23}}atoms\]of ${{U}^{235}}=235g=235\times {{10}^{-3}}kg$
Also, the mass of:$78840\times {{10}^{24}}atoms\text{ of }{{\text{U}}^{235}}=\frac{235\times {{10}^{-3}}}{6.023\times {{10}^{23}}}\times 78840\times {{10}^{24}}=3.076\times {{10}^{4}}kg$
Hence, the mass of uranium needed per year is found to be,\[3.076\times {{10}^{4}}kg\].