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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. |
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Answer» 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` |
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