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

$E_K=\frac{hc}{λ_K}=\frac{12.4KeVÅ}{0.15Å}=82.7KeV$

The energy of incident photon

$E_v=\frac{hc}{λ}=\frac{12.4}{0.1}=124 KeV$

The maximum kinetic energy is

$K_{max}=E_v-|E_K|=41.3KeV≈41KeV$