200 MeV of energy can be obtained per fission in a reactor generating 1000 kW. Find the number of nuclei under going the fission per second.

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Solution

Energy generated per fission is 200MeV
Power produced is $P = \frac{E n}{t}$
$\frac{n}{t}$ is number of fission
P=1000kw
$\frac{n}{t} = \frac{1000 \times 10^{3}}{200 \times 1.6 \times 10^{- 13}} = 3.125 \times 10^{16}$

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200 MeV of energy can be obtained per fission in a reactor generating 1000 kW. Find the number of nuclei under going the fission per second.

Energy generated per fission is 200MeVPower produced is P=Entnt is number of fission P=1000kwnt=1000×103200×1.6×10−13=3.125×1016

Energy generated per fission is 200MeV
Power produced is $P = \frac{E n}{t}$
$\frac{n}{t}$ is number of fission
P=1000kw
$\frac{n}{t} = \frac{1000 \times 10^{3}}{200 \times 1.6 \times 10^{- 13}} = 3.125 \times 10^{16}$