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

Options

A. `lamda_0=(2mclamda^2)/h`
B. `lamda_0=(2h)/(mc)`
C. `lamda_0 = (2m^2 c^2 lamda^3)/h^2`
D. `lamda_0=lamda`

Answer

A

Explanation

`lamda=h/(sqrt(2mk))impliesk=h^2/(2malamda^2)`
Since, `lamda_0=(hc)/k`
`:.lamda_0=(2mclamda^2)/h`