Question

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 200 MeV.

Answer 1

Firstly, we will compute the amount of electric power to be obtained from the nuclear power plants, then, we will find the heat energy released from Uranium-235. Using both the values, we will compute the number of atoms of Uranium required for fission. Finally, using the number of atoms and the mass of Uranium, we will compute the amount of fissionable uranium needed.

Complete step-by-step solution:
From the given information, we have the data as follows.
India had a target of producing 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. 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%. The heat energy per fission of ${}^{235}U$to be about 200 MeV.
Amount of electric power to be obtained from the nuclear power plants,

& {{P}_{1}}=\dfrac{10}{100}\times 2\times {{10}^{5}}MW \\\ & \Rightarrow {{P}_{1}}=2\times {{10}^{4}}MW \\\ & \Rightarrow {{P}_{1}}=2\times {{10}^{4}}\times {{10}^{6}}{J}/{s}\; \\\ & \therefore {{P}_{1}}=2\times {{10}^{10}}\times 60\times 60\times 24\times 365{J}/{y}\; \\\ \end{aligned}$$ As the heat energy released per fission of $${}^{235}U$$nucleus is, 200 MeV and the efficiency of a rector is 25%, so, we get, $$\begin{aligned} & \dfrac{25}{100}\times 200=50MeV \\\ & =50\times 1.6\times {{10}^{-19}}\times {{10}^{6}} \\\ & =8\times {{10}^{-12}}J \\\ \end{aligned}$$ The number of atoms required for fission per year: $$\dfrac{2\times {{10}^{10}}\times 60\times 60\times 24\times 365}{8\times {{10}^{-12}}}=78840\times {{10}^{24}}\text{atoms}$$ 1 mole, that is, 235 g of Uranium-235 contains$$6.023\times {{10}^{23}}\text{atoms}$$. Thus, the mass of $$6.023\times {{10}^{23}}\text{atoms}$$of Uranium-235 equals, $$235\times {{10}^{-3}}\text{kg}$$. Therefore, the mass of $$78840\times {{10}^{24}}$$atoms of Uranium-235 equals, $$\begin{aligned} & =\dfrac{235\times {{10}^{-3}}}{6.023\times {{10}^{23}}}\times 78840\times {{10}^{24}}\text{kg} \\\ & \text{=3}\text{.076}\times {{10}^{4}}\text{kg} \\\ \end{aligned}$$ $$\therefore $$ $$\text{3}\text{.076}\times {{10}^{4}}\text{kg}$$ of fissionable uranium would our country need per year by 2020. **Note:** 1 mole, that is, 235 g of Uranium-235 contains$$6.023\times {{10}^{23}}\text{atoms}$$. Thus, the mass of $$6.023\times {{10}^{23}}\text{atoms}$$of Uranium-235 equals, $$235\times {{10}^{-3}}\text{kg}$$. The units of the parameters should be taken care of, as, in this case, the unit of the mass is converted from gram to kg.