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Find the kinetic energy and the velocity of the photoelectrons liberated by K_(alpha) radiation of zinc from the K shell of iron whose K band absorption edge wavelength is lambda_(k)=174p m. |
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Answer» Solution :The energy of the `K_(alpha)` radiation of Ƶn is ` ħ omega=(3)/(4) ħR(Ƶ-1)^(2)` where `Ƶ=` atomic number of Zinc`= 30`. The binding energy of the `K` electrons in iron is obtained from the wavelength of `K` absorption edge as `E_(k)= 2PI ħc//lambda_(k)` Hence by Einstein equation `T=(3)/(4) ħR(Ƶ-1)^(2)-(2pi ħc)/(lambda_(k))` SUBSTITUTION GIVES `T=1.463keV` This corresponds to a velocity of the photo electrons of `v= 2.27xx10^(6)m//s` |
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