The power output of a`_92U^(235)` reactor if it takes 30 days to use up 2 kg of fuel and if each fission gives 185 MeV of energy (Avogadro number`=6xx10^(23)//mole) will be -

What is the power output of a ._(92) U^(235) reactor if it is takes 30 days to use up 2 kg of fuel, and if each fission gives 185 MeV of usable energy ?.

What is the power output of a ._(92) U^(235) reactor if it is takes 30 days to use up 2 kg of fuel, and if each fission gives 185 MeV of usable energy ?.

What is the power output of a ._(92) U^(235) reactor if it is takes 30 days to use up 2 kg of fuel, and if each fission gives 185 MeV of useable energy ?.

""_(92)^(235)U reactor takes 30 days to use 2kg of fuel il each ission gives 185 MeV of usable energy the power output of the reactor is K(10^(7)) watt. Where K=

If reactor takes 30 day to consume 4 kg of fuel and each fission gives 185 MeV of usable energy , then calculate the power output.

In reactor 2, kg of ""_(92) U^(235) fuel is fully used up in 30 days . The energy released per fission is 200 MeV. Given that the Avogadro number ,N=6.023 xx 10^(26) per kilo mole and 1e V=1.6 xx 10^(-19)J. The power output of the reactor is close to

In reactor 2, kg of ""_(92) U^(235) fuel is fully used up in 30 days . The energy released per fission is 200 MeV. Given that the Avogadro number ,N=6.023 xx 10^(26) per kilo mole and 1e V=1.6 xx 10^(-19)J. The power output of the reactor is close to

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))