If 200 MeV of energy is released in the fission of one nucleus of 92U235, The number of nuclei that must undergo fission to produce energy of 1000J in 1 sec is

p=nEt⇒103=n×32×10−121n=10332×10−12=3.125×1013sec

If 200 MeV of energy is released in the fission of one nucleus of 92U235, The number of nuclei that must undergo fission to produce energy of 1000J in 1 sec is

p=nEt⇒103=n×32×10−121n=10332×10−12=3.125×1013sec

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If 200 MeV of energy is released in the fission of one nucleus of ₉₂U²³⁵, The number of nuclei that must undergo fission to produce energy of 1000J in 1 sec is

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3.125 × 10¹³

3.125 × 10¹⁴

6.25 × 10¹³

12.5 × 10¹³

answer is A.

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$\begin{array}{l} p = \frac{nE}{t} \Rightarrow 10^{3} = \frac{n \times 32 \times 10^{- 12}}{1} \\ n = \frac{10^{3}}{32 \times 10^{- 12}} = 3 .125 \times 10^{13} \sec \end{array}$

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If 200 MeV of energy is released in the fission of one nucleus of 92U235, The number of nuclei that must undergo fission to produce energy of 1000J in 1 sec is

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If 200 MeV of energy is released in the fission of one nucleus of ₉₂U²³⁵, The number of nuclei that must undergo fission to produce energy of 1000J in 1 sec is