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Question: If the energy released in the fission of one nucleus is 200 MeV. Then the number of nuclei required ...

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

0.5×10140.5 \times 10^{14}

B

0.5×10120.5 \times 10^{12}

C

5×10125 \times 10^{12}

D

5×10145 \times 10^{14}

Answer

5×10145 \times 10^{14}

Explanation

Solution

Energy released in the fission of one nucleus = 200 MeV

=200×106×1.6×10−19J=3.2×10−11J= 200 \times 10^{6} \times 1.6 \times 10^{- 19}J = 3.2 \times 10^{- 11}J

P=16KW=16×103WattP = 16KW = 16 \times 10^{3}Watt

Now, number of nuclei required per second

n=PE=16×1033.2×10−11=5×1014n = \frac{P}{E} = \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}} = 5 \times 10^{14}.