# Intro nuclear phyics

In a nuclear power plant, the energy produced by fission of (235,92)U has Q= 2 MeV/fission and an efficiency or 30%. If a 1500 kg supply of (235,92)U is consumed every 1.20 years, find the plant’s power output. Assume the atomic mass of 235U is 235 u/nuclei and there is one (235,92)U nucleus consumed per fission.

Answer

The molar mass of 235U is M =235 g

in $1500*10^3 g$ 235U there are

$n= 1500*10^3/235 =6383 moles$

In one mole of 235U there are $Na =6.023*10^23 atoms$ (Avogadro number).

total number of atoms in 1500 kg is

$N =n*Na =6383*6.023*10^23 =3.84*10^27 atoms$

total energy released by fission for 100% efficiency is

$E0 =N*Q =200 MeV *3.84*10^{27} =7.689*10^{35} eV =1.23*10^{17} J$

for 30% efficiency

$E =0.3*E0 =3.69*10^{16} J$

Total power is

$P =E/time = 3.69*10^{16}/1.2/365/24/3600 =975*10^6 =9.75*10^8 W$