JEE Main 2020 · 7 January · evening · Single correct · +4 / −1
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)
- (a) 1/c( 2E/m )^(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
Write both wavelengths as h over momentum; the only difference is how each momentum depends on E.
- Electron wavelength. A few eV is far below the electron's rest energy, so it is non-relativistic: E = p²/2m, giving pₑ = √2mE. Then λₑ = h/√2mE.
- Photon wavelength. A photon has E = pc, so pₚₕ = E/c and λₚₕ = hc/E.
λₑ/λₚₕ = h/√2mE × E/hc = 1/c E/√2mE = 1/c(E/2m)^(1/2)
- Check units. √(E/m) has units of speed, and dividing by c makes the ratio dimensionless, as a ratio of two lengths must be.
- So. The ratio is 1/c(E/2m)^(1/2), which is option (b).
The idea to keep. Same energy does not mean same momentum. For a slow particle p = √2mE, for a photon p = E/c. A ratio of wavelengths must be dimensionless, so check units.
Answer from the NTA final answer key. Solution by PrepWiser.