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An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of a uranium atom U²³⁸ is 170 MeV. The number of uranium atoms fissioned per hour will be

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$30 \times 10^{25}$

$10 \times 10^{20}$

$4 \times 10^{22}$

$5 \times 10^{15}$

answer is B.

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Detailed Solution

$P = \frac{NE}{t} \Rightarrow 300 \times 10^{6}$

$= \frac{N \times 170 \times 10^{6} \times 1 .6 \times 10^{- 19}}{t}$

$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$

Number of atoms per hour

$= 1.02 \times 10^{19} \times 3600$

$= 3.97 \times 10^{22}$

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