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Question

If the energy released in the fission of one nucleus is 200 MeV, then the number of nuclei required per second in a power plant of 16 kW will be:


A
0.5×1014
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B
0.5×1022
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C
5×1012
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D
5×1014
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Solution

The correct option is D $$5 \times 10^{14}$$
According to the question,
Released energy in fission of one nucleus $$=\,200\,MeV$$
$$ = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J$$
$$ = 3.2 \times {10^{ - 11}}J$$
$$P = 16KW$$
$$ = 16 \times {10^3}\,watt$$
So, the number of  required $$nuclei/second$$ is 

$$n = \dfrac{P}{E} = \dfrac{{16 \times {{10}^3}}}{{3.2 \times {{10}^{ - 11}}}}$$

$${ = 5 \times {{10}^{14}}}$$

Physics
NCERT
Standard XII

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