A material whose $K$ absorption edge is $0.2 \stackrel{\circ }{A}$ is irradiated by X-rays of wavelength $0.15 \stackrel{\circ }{A}$. Find the maximum energy of the photoelectrons that are emitted from the $K$ shell. (Take $hc=12400eV\stackrel{\circ }{A}$ )

The binding energy for $K$ shell in $eV$ is

       $E_k=\frac{hc}{{\lambda }_k}=\frac{12400}{0.2}eV=62keV$

The energy of the incident photon in $eV$ is

       $E=\frac{hc}{\lambda }=\frac{12400}{0.15}=83keV$

Therefore, the maximum energy of the photoelectrons emitted from the $K$ shell is

        $E_{\max }=E-E_k=(83-62)keV=21keV$