Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ₀) of the emitted X-ray is

Options

(a) λ₀= 2mcλ²/h
(b) λ₀= 2h/mc
(c) λ₀= 2m²c²λ³/h²
(d) λ₀= λ

Correct Answer:

λ₀= 2mcλ²/h

Explanation:

λ = h / p ⇒ p = h / λ

KE of electrons = E = p² / 2m = h² / 2mλ²

Also in X-ray, λ₀ = hc / E = 2mcλ² / h