When an atom of ${ }^{235} \mathrm{U}$ undergoes fission in a reactor, about 200 MeV of energy is liberated. Suppose that a reactor using uranium235 has an output of $700 \mathrm{MW}$ and is 20 percent efficient. (a) How many uranium atoms does it consume in one day? (b) What mass of uranium does it consume each day?
(a) Each fission yields
$$
200 \mathrm{MeV}=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right) \mathrm{J}
$$
of energy. Only 20 percent of this is utilized efficiently, and so Usable energy per fission $=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right)(0.20)=6.4 \times 10^{-12} \mathrm{~J}$
Because the reactor's usable output is $700 \times 10^{6} \mathrm{~J} / \mathrm{s}$, the number of fissions required per second is
$$
\begin{array}{c}
\text { Fissions } / \mathrm{s}=\frac{7 \times 10^{8} \mathrm{~J} / \mathrm{s}}{6.4 \times 10^{-12} \mathrm{~J}}=1.1 \times 10^{20} \mathrm{~s}^{-1} \\
\text { and Fissions/day }=(86400 \mathrm{~s} / \mathrm{d})\left(1.1 \times 10^{20} \mathrm{~s}^{-1}\right)=9.5 \times 10^{24} \mathrm{~d}^{-1}
\end{array}
$$
(b) There are $6.02 \times 10^{26}$ atoms in $235 \mathrm{~kg}$ of uranium-235.