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No. of uranium 235 nuclei required to undergo fission to give 9xx10^(13) joule of energy is

In a nuclear reactor, the number of U^(235) nuclei undergoing fissions per second is 4xx10^(20). If the energy releases per fission is 250 MeV, then the total energy released in 10 h is (1 eV= 1.6xx10^(-19)J)

In a nuclear reactor, the number of U^(235) nuclei undergoing fissions per second is 4xx10^(20). If the energy releases per fission is 250 MeV, then the total energy released in 10 h is (1 eV= 1.6xx10^(-19)J)

Which product of the nuclear fission of uranium-235 is utilised to cause further fission of its nuclei ?

Energy released during the fission of one Uranium-235 nucleus is 200MeV. Energy released by the fission of 500gm of U-235 nuclei will be about

In a nuclear reactor, ""^(235)U undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 years, find the total mass of uranium required.

If 200 MeV of energy is released in the fission of 1 nucleus of ._(92)U^(235) , the number of nuclei that undergo fission to produce energy of 10 kWh in 1 s is

Barium-141 has a half-life of 18 minutes.The half-life of Krypton-92 is 3.0 s .In the fission reaction of a mass of Uranium - 235 , equual numbers of barium and krypton nuclei are produced .Estimate the time tken after the fission of the sample of uranium for the ratio ("number of Barium" - 141 "nuclei")/("number of" "Krypton"-92 "nuclei") . to be approximately equal to 8.