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Assuming 100% efficiency in converting this energy to electricity , how much 235U (in grams) would...

Assuming 100% efficiency in converting this energy to electricity , how much 235U (in grams) would be necessary to light a 50 watt light bulb for one year ? 235U + 1n ----> 142Ba + 91kr+3 1n

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Answer #1

50 watt or 50J/sec of poweris being utilized from the 235U
In 1 year how much such power is being utilized by that bulb ?
365 * 24 * 60 * 60 * 50 J/year power is being used by that bulb
On an average it is said that 5*109 J of power can be generated from 1 kg of U235
5*109J ------------------------------------ 1 Kg of 235U
365 * 24 * 60 * 60 * 50 J ----------------------------------- X
cross multiply
X = 0.31536 Kg = 315.36 gms
Therefore 315.36 gms of 235U is required to light a bulb for 1 year with 100% efficiency.

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