In an experiment on photoelectric emission of `gamma`-rays on platinum, the energy distribution of photoelectron exhibits peaks at a number of descrete energies : 270keV, 339keV and 354keV. The binding energy of K,L and M shell in platinum are known to be 77keV, 13keV and 3.5keV approx., what is the wavelength of `gamma`-rays with which the data are consistent.

Explanation

Energy of incident photon =Binding energy (B) +energy of emitted photoelectron (E)
`:. Hv=B+E`
Now for K shell, `B+E=270+77=347keV`
L shell `B+E=339+13=352keV`
M shell, `B+E=354+3.5=357.5keV`
The values of (B+E) are equivalent within about 3%
`:.` Average value of B+E
`=1/3[347+352+357.5]=352keV`
`=352xx10^(3)xx1.6xx10^(-19)J`
Now, `B+E=hv=(hc)/lambda`
or `lambda=(hc)/(B+E)`
`=((6.63xx10^(-34))xx(3xx10^(8)))/(352xx10^(3)xx1.6xx10^(-19))=3.5xx10^(-12)m`