In a reactor, $$2kg$$ of $$92U235$$ fuel is fully used up in $$30$$ days. The energy released per fission is $$200$$ MeV. Given that the Avogadro number, $$N=6.023$$×$$1026$$ per kilo mole and $$1$$ $$eV=1.6$$×$$10$$−$$19J$$. The power output of the reactor is close to

Question

In a reactor, $$2kg$$ of   $$92U235$$ fuel is fully used up in $$30$$ days. The energy released per fission is $$200$$ MeV. Given that the Avogadro number, $$N=6.023$$×$$1026$$ per kilo mole and $$1$$ $$eV=1.6$$×$$10$$−$$19J$$. The power output of the reactor is close to

Infinity Learn · Accepted Answer

Total  $$energy=massmolar$$  mass×avogadro  no×$$200$$×$$106$$×e J​$$Power=Total$$  $$energyTime=2$$×$$103$$×$$6.023$$×$$1023$$×$$200$$×$$106$$×$$1.6$$×$$10$$−$$19235$$×$$30$$×$$24$$×$$3600W$$​$$=63$$×$$106W$$​≃$$60$$ MW

In a reactor, 2kg 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, $N = 6.023 \times 10^{26}$ per kilo mole and 1 $e V = 1.6 \times 10^{- 19} J$ . The power output of the reactor is close to

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In a reactor, 2kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N=6.023×1026 per kilo mole and 1 eV=1.6×10−19J. The power output of the reactor is close to

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In a reactor, 2kg 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, $N = 6.023 \times 10^{26}$ per kilo mole and 1 $e V = 1.6 \times 10^{- 19} J$ . The power output of the reactor is close to