JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
- (a) 1/c( 2m/E )^(1/2)
- (b) 1/c( E/2m )^(1/2)
- (c) ( E/2m )^(1/2)
- (d) c(2mE)^(1/2)
Answer and solution
Answer (b) 1/c( E/2m )^(1/2)
Solution
Physics sir · 4 steps
Same energy, different momentum rules: use p = √2mE for the electron and p = E/c for the photon.
- Electron wavelength. For a non-relativistic electron, E = p²/2m, so p = √2mE. Then λₑ = h/p = h/√2mE.
- Photon wavelength. A photon has no rest mass and E = pc, so p = E/c. Then λₚ = h/p = hc/E.
λₑ/λₚ = h/√2mE × E/hc = 1/cE/√2mE = 1/c(E/2m)^(1/2)
- Check units. E/m has units of speed squared, so √(E/2m) is a speed. Dividing by c leaves a pure number, as a ratio of two wavelengths must be.
- So. The ratio is 1/c(E/2m)^(1/2), which is option (b).
The idea to keep. Both wavelengths come from λ = h/p. Only the link between E and p changes: p = √2mE for a particle with mass, p = E/c for a photon.
Answer from the NTA final answer key. Solution by PrepWiser.