Electrons with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

Options

A. `lamda_0 = (2mclamda^2)/(h)`
B. `lamda_0 = (2h)/(mc)`
C. `lamda_0 = (2m^2c^2 lamda^2)/( h^2)`
D. `lamda_0 = lamda`

Answer

D

Explanation

If E is the kinetic energy of the electrons, then their de broglie wavelength is
`lamda = (h)/(sqrt(2mE)) therefore lamda^2 = (h^2)/(2mE) therefore E = (h^2)/(2mlamda^2) " " ..(1)`
and for the emitted X rays, the cut off wavelength
`lamda_0 = (hc)/(E)`
`therefore lamda_0 = (hcxx2mlamda^2) /(h^2)= (2mclamda^2)/(h)`