JEE Main 2021 · 20 July · evening · Single correct · +4 / −1
An electron having de-Broglie wavelength λ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :
- (a) 0
- (b) (2m²c²λ ²)/h²
- (c) (2mcλ ²)/h
- (d) hc/mc
Answer and solution
Answer (c) (2mcλ ²)/h
Solution
Physics sir · 3 steps
The shortest X-ray carries the electron's entire kinetic energy, so get K from λ, then use hc/K.
- Energy from λ. The de Broglie relation gives the electron's momentum, p = h/λ. Its kinetic energy is then K = p²/2m = h²/2mλ².
- Cut-off condition. The cut-off (minimum) wavelength comes from an electron that gives all its kinetic energy to a single photon. So hc/λ₀ = K.
λ₀ = hc/K = (hc × 2mλ²)/h² = 2mcλ²/h
- So. The cut-off wavelength is 2mcλ²/h, which is option (c).
The idea to keep. Cut-off wavelength always means one photon takes the full kinetic energy: λₘᵢₙ = hc/K. For an electron, write K = p²/2m with p = h/λ.
Answer from the NTA final answer key. Solution by PrepWiser.