The energy liberated in the fission of a single uranium-235 atom is $3.2 * 10^{-11}\;J$. Calculate the power production corresponding to the fission of $1\;gm$ of uranium per day. Assume Avogadro constant as $6 * 10^{23}\;mol^{-1}$.

Given,

Energy liberated in the fission of a single uranium-235 atom = $3.2 * 10^{-11}\;J$
Avogadro constant $(N_A) = 6 * 10^{23}\;mol^{-1}$
Time $(t) = 1\; day = 24 * 60 * 60 = 86400 \; Secs$
Power Produced = ?

Now, We have

1 mole = 235 gm of Uranium $= 6 * 10^{23}\; atoms$

1 gm of Uranium $= \frac{6 \; * \; 10^{23}}{235}\;atoms$ $= 2.56 * 10^{21}\;atoms$

$\therefore$ Total energy released by the fission of $1\;gm$ Uranium $= 3.2 * 10^{11} * 2.56 * 10^{21} = 8.20 * 10^{10}\;J$

Again,

Power produced  = $\frac{Energy\; Released}{Time\;Taken} = \frac{8.20 \; * \; 10^{10}}{86400}$  $= 9.49 * 10^5 \; Watt$

Hence, the required power is $= 9.49 * 10^5 \; Watt$.

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