To generate power of 3.2 MW, the number of fissions of `U^(235)` per minute is (Energy released per fission 200 MeV, `1 eV = 1.6 xx 10^(-13)` J.

To generate a power of 3.2 mega watt, the number of fissions of U^235 per minute is. (Energy released per fission = 200 MeV, 1 eV = 1.6 xx 10^-19 J ).

To generate a power of 3.2 mega watt, the number of fissions of U^235 per minute is. (Energy released per fission = 200 MeV, 1 eV = 1.6 xx 10^-19 J ).

Calculate the energy released by the fission 1 g of .^235U in joule, given that the energy released per fission is 200 MeV. (Avogadro.s number = 6.023 xx 10^23 )

Calculate the energy released by the fission 1 g of .^(235)U in joule, given that the energy released per fission is 200 MeV . (Avogadro's number =6.023xx10^(23))

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)

Calculate the energy released by fission from 2 gm of ._92U^235 in KWH. Given that the energy released per fission is 200 Mev.