(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.Converting this energy into Joules:$E_1 = 200 \times 1.6 \times 10^{-13}

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(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.Converting this energy into Joules:$E_1 = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.We need to find the number of nuclei $(n)$ required to release a total energy $E_{total} = 1000 \text{ J}$.The relationship is $E_{total} = n \times E_1$.Therefore, $n = \frac{E_{total}}{E_1} = \frac{1000}{3.2 \times 10^{-11}}$.$n = \frac{10^3}{3.2 \times 10^{-11}} = \frac{1}{3.2} \times 10^{14} = 0.3125 \times 10^{14} = 3.125 \times 10^{13}$ nuclei.

Class 12 Physics Nuclei Nuclear Fission, Fusion and Nuclear Reactor

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If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$, the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

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A
$3.125 \times 10^{13}$
B
$6.25 \times 10^{13}$
C
$12.5 \times 10^{13}$
D
$3.125 \times 10^{14}$

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Column-$A$	Column-$B$
$A$. Rocket propulsion	$P$. Bernoulli's principle in fluid dynamics
$B$. Aeroplane	$Q$. Total internal reflection of light
$C$. Optical fibres	$R$. Newton's laws of motion
$D$. Fusion test reactor	$S$. Magnetic confinement of plasma
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If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$, the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

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