A material whose `K` -absorption edge is `0.2 Å` is irradiated by X-ray of wavelength `3644 Å` , find the maximum energy of the photoelectrons that are emitted from the `K` shell.

Explanation

The binding energy for `K` shell in e V is
`E_(k) = (hc)/(lambda_(k)) = (12431)/(0.2) e V = 62.155 keV`
The energy of the inclined photon in `eV` is `E = (hc)/(lambda) = (12431)/(0.15) = 82.873 keV`
Therefore , the maximum energy of the photoelectrons emitted from the `K` shell is
`E_(max) = E - E_(K) = (82.873 - 62.155) keV`
`= 20.718`