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A reactor is developing nuclear energy at a rate of 32,000 kilowatt. How many kg of U^(235) undergo fission per second? How many kg of U^(235)would be used up in 1000 hour of operation ? Assume an average energy of 200 MeV released per fission ? Take Avogadro.s number as6xx10^(23) and MeV = 1.6 xx 10^(-13)joule |
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Answer» Solution :Power developed by reactor = 32, 000 kilowatt `=3.2XX10^7` watt `:.` Energy RELEASED by reactor per sec `=3.2xx10^7` joule `=(3.2xx10^7)/(1.6xx10^(-13))MeV = 2 xx10^(20)MeV`. Number of fissions occurring in the reactor per second `= (2xx10^(20))/200 =10^(18)`( `:.` Energy released per fission = 200 MeV) The number of atoms of `U^(235)` consumed in 1000 hour Now mass of `U^(235)` consumed in 1000 hour `=10^(18)xx(1000xx3600)=36xx10^(23)` `(36xx10^(23))/(6xx10^(23))xx235 =1410 " GM " = 1.41` kg |
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