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Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten per cent 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 the heat energy per fission of""^(235)U to be about 200 MeV. |
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Answer» Solution :Target of power production by 2020 AD=`2 xx 10^(5) MW =2xx10^(11)`W Power from nuclear plants = 10% of `2 xx 10^(11) W =2xx10^(10)`W which when converted into electric energy in one year, will be equivalent to `E =2xx10^(10)Wxx1 year =2 xx 10^(10) W xx 3.154 xx 10^(7) s = 6.308 xx 10^(17) `J Energy RELEASED in one act of fission = 200 MeV Energy available to ACTUALLY produce electrical power per act of fission = 25% of 200 MeV= 50 MeV =`50xx 1.6xx 10^(-13)J =8.0xx10^(-12)`J Number of `" "^(235)U` fissions required in one year=`("Total energy needed")/("Energy available per act of fission")= (6.308 xx 10^(17))/( 8.0 xx 10^(-12))= 7.885 xx 10^(28)` `therefore` Number of MOLES of `" "^(235)U` required per year `= (7.889 xx10^(28))/(6.023 xx 10^(23))=1.31xx10^(5)` mole `therefore` Mass of `" "^(235)U` required per year = number of moles `xx` mass number` = 1.31 xx 10^(5) xx 235 g = 3.078 xx 10^(7) g = 3.078 xx 10^(4) KG`. |
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