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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 as 6 xx 10^23and 1 MeV =1.6xx10^(-13) joule |
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Answer» Solution :Power developed by reactor = 32,000 kilowatt `=3.2xx10^7` watt `THEREFORE` Energy released by reactor per sec = `3.2xx10^7` joule `=(3.2xx10^7)/(1.6xx10^(-13))` MeV= `2xx10^20` MeV Number of FISSIONS occurring in the reactor per second `=(2xx10^20)/200=10^18` (`because` Energy released per fission = 200 MeV) The number of ATOMS of `U^235` CONSUMED in 1000 hour `=10^18xx(1000xx3600)=36xx10^23` Now mass of `U^235` consumed in 1000 hour `=(36xx10^23)/(6xx10^23)xx235`=1410 gm = 1.41 kg. |
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