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Wave Optics

Question

If the energy released in the fission of one nucleus is 3.2x10^-11J, then number of nuclei required per second in a power plant of 16 kW is

$0.5 \times 10^{14}$

$0.5 \times 10^{12}$

$5 \times 10^{12}$

$5 \times 10^{14}$

Solution

$5 \times 10^{14}$

The number of nuclei required per second is

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