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

The cut off wavelength is given byλ0=hceV                                              ……(i)According to de Broglie equationλ=hp=h2meV⇒λ2=h22meV⇒V=h22meλ2         ……(ii)From (i) and (ii),λ0=hc×2meλ2eh2=2mcλ2h