Question

An all-electric home uses approximately 1910 kWh of electric energy per month. How much uranium-235 would be required to provide this house with its energy needs for one year? Assume 100% conversion efficiency and 208 MeV released per fission.

Answer 1

U235 - 968 mg required

Step-by-step explanation:

given data

electric energy Ee = 1910 kWh per month 1910 ×12×3.6×
`10^(6)` = 8.25 ×
`10^(10)` J

conversion efficiency = 100%

fission energy Ef = 208 MeV = 3.33 ×
`10^(-11)` J

to find out

How much uranium-235 required

solution

we find no of fission required for U235 that is

no of fission = Ee/ Ef

no of fission =
`(8.25*10^(10))/(3.33*10^(-11))`

no of fission = 2.48 ×
`10^(21)`

and

mass =
`(2.48*10^(21))/(6.023*10^(23))` × 235 gm

we know avogadro no is 6.023 ×
`10^(23)`

mass = 968 mg

so that 968 mg required