Electrons of mass m with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength `(lambda_(0))` of the emitted X-ray is

Explanation

If `lamda`=de Broglie wavelength, then momentum `p=(h)/(lamda)`
K.E of the striking electrons `K=(p^(2))/(2m)=(h^(2))/(2mlamda^(2))`
This is equal to maximum energy of x - ray
photons. `(hc)/(lamda_(0))=(h^(2))/(2mlamda^(2))implieslamda_(0)=(2mlamda^(2)c)/(h)`