JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
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]
- (a) hv
- (b) hv[ 1 + hv/2Mc² ]
- (c) hv[ 1 - hv/2Mc² ]
- (d) 0
Answer and solution
Answer (b) hv[ 1 + hv/2Mc² ]
Solution
Physics sir · 4 steps
The photon carries momentum, so the nucleus recoils, and that recoil energy also comes from inside it.
- Momentum conservation. The nucleus starts at rest, so total momentum is zero. A photon of energy hν carries momentum hν/c, so the nucleus must recoil with momentum p = hν/c in the opposite direction.
- Recoil kinetic energy. The nucleus is heavy and slow, so its kinetic energy is p²/2M = (hν)²/2Mc².
- Energy conservation. The internal energy lost by the nucleus pays for both the photon and the recoil.
Δ E = hν + (hν)²/2Mc² = hν(1 + hν/2Mc²)
- So. The loss of internal energy is hν(1 + hν/2Mc²), which is option (b).
The idea to keep. Whenever something at rest emits a photon or a particle, conserve momentum first. The emitter recoils, and the energy released must cover the emitted energy plus the recoil kinetic energy.
Answer from the NTA final answer key. Solution by PrepWiser.