A city requires $10^7 watts$ of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency $20\; \%$. Using $_{92}{U}^{235}$ as the fuel source, calculate the amount of fuel required per day (Energy released per fission $_{92}{U}^{235} = 200\;MeV$).

Given,
Output power, $P_{out} = 10^7\;W$
Time, $t = 1\;day = 24 * 60 * 60\;Sec = 86400\;Sec$
Efficiency, $\eta = 20 \% = 0.2$
Energy per fission of $_{92}{U}^{235}, E = 200\;MeV = 200 * 1.6 * 10^{-19} * 10^6 = 3.2 * 10^{-11}\;J$
Mass of the Uranium, $M = ?$

Now,

$\eta = $ $\frac{P_{out}}{P_{in}} $ $\; * \; 100 \%$

or, $0.2 = $ $\frac{10^7}{P_{in}}$        ⇒   $P_{in} = $ $\frac{10^7}{0.2}$ $ = 5 * 10^7\;J$

Again,

$P_{in} = $ $\frac{Input \; Energy \,(Q)}{Time \, (t)}$      ⇒   $Q = P_{in} * t = 5 * 10^7 * 86400 = 4.32 * 10^{12}\;J$

But again we have,

$Q = N * E$

⇒ $N = $ $\frac{Q}{E} = \frac{4.32 \; * \; 10^{12}}{3.2 \; * \; 10^{-11}}$ $= 1.35 * 10^{23}$ atoms

Now,

$6.023 * 10^{23} \; atoms = 235\;g$

⇒ $1\; atoms = \frac{235}{6.023 \; * \; 10^{23}}$

⇒ $1.35 * 10^{23}\;atoms = \frac{235 \; * \; 1.35 \; * \; 10^{23}}{6.023 \; * \; 10^{23}} = 52.7\;gm = 0.0527\;kg$

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