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
B
0.5×1022
C
5×1012
D
5×1014

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 $$n = \dfrac{P}{E} = \dfrac{{16 \times {{10}^3}}}{{3.2 \times {{10}^{ - 11}}}}$$$${ = 5 \times {{10}^{14}}}$$PhysicsNCERTStandard XII

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