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Suppose India has a target of producing by 2020AD, 200 GW of electric power, ten precent of which is to be obtained from nuclear power plants. Suppose the efficiency remained at 25%, what amount of fuel may be required per year by 2020? Take the heat energy per fission of .^(235)U to be about 200 MeV. |
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Answer» Solution :Nuclear power to be generated `=10% "of 200 hw=20 GW"` At `25%` efficiency actual required `=(100)/(25)xx20GW=80GW` Total energy required for 1 YEAR production of power `=80xx10^(9)xx365xx8.64xx10^(4)J` `"i.e.," E=2.523xx10^(18)J` Energy per fission `=200 MeV` `=200xx1.6xx10^(-13)J` `=320.0xx10^(-13)J` `E=3.2xx10^(-11)J` Total number of fission atoms required `=(2.523xx10^(18))/(3.2xx10^(-11))=7.884xx10^(28)` But `6.023xx10^(23)` atoms of Uranium have a mass of `235g`. `therefore 7.884xx10^(30)` atoms of Uranium have a mass of xg. `"i.e.," x=(235xx7.884xx10^(28))/(6.023xx10^(23))` `x=307.6xx10^(5)` Hence,`m=307.6xx10^(2) KG` `=3.076xx10^(4)kg` or`m=30.76`tons. |
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