JEE Main202120 Jul 2021Morning ShiftPhysicsNuclear PhysicsActual
A nucleus of mass M emits γ -ray photon of frequency v . The loss of internal energy by the nucleus is: [Take c as the speed of electromagnetic wave]
Options
- Ah v
- B0
- Ch v 1 - h v 2 M c 2
- Dh v 1 + h v 2 M c 2
Correct answer
D. h v 1 + h v 2 M c 2
Step-by-step solution
Energy of γ ray E γ = h v Momentum of γ ray P γ = h λ = h v c Total momentum is conserved. P → γ + P → Nu = 0 where P → Nu = momentum of decayed nuclei ⇒ P γ = P Nu ⇒ h v c = P Nu ⇒ KE of nuclei = 1 2 M v 2 = P Nu 2 2 M = 1 2 M h v c 2 Loss in internal energy = E γ + KE Nu = h v + 1 2 M h v c 2 = h v 1 + h v 2 M c 2