Q: Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

(A) $\large \lambda_0 = \frac{2 m c \lambda^2}{h} $

(B) $\large \lambda_0 = \frac{2 h}{m c} $

(C) $\large \lambda_0 = \frac{2 m^2 c^2 \lambda^3}{h^2} $

(D) $\large \lambda_0 = \lambda $

Ans: (A)

Sol: Kinetic Energy of Striking Electrons is

$\large K = \frac{p^2}{2 m} $

As , $\large p = \frac{h}{\lambda}$

$\large K = \frac{h^2}{2 m \lambda^2} $

This is the maximum energy of X-ray photons .

Therefore , $\large \frac{h c}{\lambda_0} = \frac{h^2}{2 m \lambda^2} $

$\large \lambda_0 = \frac{2 m c \lambda^2}{h} $