In a hypothetical fission reaction 92 X 236 → 56 Y 141 + 36 Z 92 + 3 R The identity of emitted particles ( R ) is :

Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsNuclear physicsMedium2 minPYQ_2024

PhysicsMediumsingle choice

In a hypothetical fission reaction $_{92} X^{236} \to_{56} Y^{141} +_{36} Z^{92} + 3 R$ The identity of emitted particles ( $R)$ is :

Answer:

Solution:

$Z in LHS = 92 Z in RHS = 56 + 36 = 92 A in LHS = 236 A in RHS = 141 + 92 = 233$
So 3 neutrons are released.

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2024

Loading discussions...