A material whose K absorption edge is 0.15 Å is irradiated with 0.1 Å X-rays. The maximum kinetic energy of photoelectrons that are emitted from K-shell is -

41 KeV 51 KeV 61 KeV 71 KeV

41 KeV

$|E_K|=\frac{hc}{λ_K}=\frac{12.4KeVÅ}{0.15Å}= 82.7 KeV$ The energy of incident photon $E_v=\frac{hc}{λ}=\frac{12.4}{0.1}=124KeV$ The maximum kinetic energy is $K_{max} = E_v – |E_K| = 41.3 KeV ≈41 KeV$