Question

Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. What is the cut-off wavelength λo of the emitted X-rays?

In this question, I'm not able to understand what is meant by cut off wavelength. Please explain that and also how to solve this question. Thanks!

P.S. The answer is (a) part.

Answer 1

Below

Explanation:

The kinetic energy of the incoming electron is:

`T = p^2/(2m) = (h/lambda)^2/(2m) = h^2/(2 lambda^2 m)`

Assuming that all of this energy is converted into a photon in the Bremsstrahlung process, that photon will have energy:

`(h c)/lambda_o = h^2/(2 lambda^2 m)`

Or: `lambda_0 = (2 mc lambda^2)/h`

Most photons will have lower energies, which is why it is a cut off.