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PhysicsModern PhysicsNuclear physicsHard2 minPYQ_2023

PhysicsHardnumerical

The energy released per fission of nucleus of ${}^{240}X$ is $200 MeV$ . The energy released if all the atoms in $120 g$ of pure ${}^{240}X$ undergo fission is _____ $\times 10^{25} MeV$ .

(Given $N_{A} = 6 \times 10^{23}$ )

Answer:

6.00

Solution:

Energy released per fission is $200 MeV$ .

The number of atoms $n$ in $120 g$ :

$n = \frac{120}{240} \times 6 \times 10^{23} atoms .$

Total energy released, $E = n \times 200 \times 10^{6}$

$= 6 \times 10^{25} MeV$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2023

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