JEE Main 2022 · 26 June · morning · Single correct · +4 / −1
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Eₑ and pₑ and that of photon are Eₚₕ and pₚₕ respectively. Which of the following is correct?
- (a) Eₑ/Eₚₕ = 2c/v
- (b) Eₑ/Eₚₕ = v/2c
- (c) pₑ/pₚₕ = 2c/v
- (d) pₑ/pₚₕ = v/2c
Answer and solution
Answer (b) Eₑ/Eₚₕ = v/2c
Solution
Physics sir · 4 steps
Same wavelength means same momentum, so compare energies through p alone: pv/2 against pc.
- Same momentum. The de Broglie wavelength is λ = h/p for both the electron and the photon. Equal wavelengths give equal momenta, so pₑ = pₚₕ = p and pₑ/pₚₕ = 1. That rules out (c) and (d).
- Electron energy. For the electron, Eₑ = 1/2mv² and p = mv. So Eₑ = 1/2(mv)v = pv/2.
- Photon energy. A photon has no rest mass, and its energy and momentum are linked by Eₚₕ = pc.
Eₑ/Eₚₕ = (pv/2)/pc = v/2c
- So. The energy ratio is v/2c, which is option (b).
The idea to keep. Equal de Broglie wavelengths always mean equal momenta. Then use E = pv/2 for a slow particle and E = pc for a photon. The factor of 2 comes only from the particle.
Answer from the NTA final answer key. Solution by PrepWiser.