1.

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`.

Answer» 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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