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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