Electrons of mass m with de-Broglie wavelength λ fall on the target in an X -ray tube. The cutoff wavelength λ0 of the emitted X -ray is
λ0=2mcλ2h
λ0=2hmc
λ0=2m2c2λ3h2
λ0=λ
Kinetic energy of electrons
K=p22m=(h/λ)22m=h22mλ2
So, maximum energy of photon =K
hcλ0=h22mλ2 ∴λ0=2mcλ2h