Question

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 ? ^{235}U + ^{1}n
----> ^{142}Ba + ^{91}kr+3 ^{1}n

Answer #1

^{235}U

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*10^{9} J of power can be
generated from 1 kg of U^{235}

5*10^{9}J ------------------------------------ 1 Kg of
^{235}U

365 * 24 * 60 * 60 * 50 J -----------------------------------
X

cross multiply

X = 0.31536 Kg = 315.36 gms

Therefore 315.36 gms of ^{235}U is required to light a bulb
for 1 year with 100% efficiency.

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