Answer:
U235 - 968 mg required
Explanation:
given data
electric energy Ee = 1910 kWh per month 1910 ×12×3.6×[tex]10^{6}[/tex] = 8.25 ×[tex]10^{10}[/tex] J
conversion efficiency = 100%
fission energy Ef = 208 MeV = 3.33 ×[tex]10^{-11}[/tex] 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 = [tex]\frac{8.25*10^{10}}{3.33*10^{-11}}[/tex]
no of fission = 2.48 ×[tex]10^{21}[/tex]
and
mass = [tex]\frac{2.48*10^{21}}{6.023*10^{23}}[/tex] × 235 gm
we know avogadro no is 6.023 ×[tex]10^{23}[/tex]
mass = 968 mg
so that 968 mg required
