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)=(2m^(2)c^(2)lamda^(2))/(h^(2))`
B. `lamda_(0)=lamda`
C. `lamda_(0)=(2mclamda^(2))/h`
D. `lamda_(0)=(2h)/(mc)`

Explanation

`lamda=h/pimpliesp=h/(lamda)`
K.E of electron `=E=(p^(2))/(2m)=(h^(2))/(2mlamda^(2))`
Also in X-ray `lamda_(0)=(hc)/Eimplieslamda_(0)=(2mclamda^(2))/h`