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The wavelength of a resonat mercury line is lambda==253.65nm. The mean life time of mercury atoms in the state of resonance excition is tau=.15 mu s. Evaluate the ration of the Doppler line broadening to the the natural linewidth at a gas temperature T=300K |
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Answer» Solution :A short lived state of mean life `T` has an uncertainity in ENERGY of `DeltaE~( ħ)/(T)` which is transmitted to the PHOTON it emits as natural broadening. Then `Delta omega_(nat)=(1)/(T)` so `Delta_(lambda)_(nat)=(lambda^(2))/(2pic tau)` The Dopper broadening on the other hand ARISES from the thermal motion of radiating atoms. The effect is non-relativistic and the maximum brodening can be written as `Delta(lambda_(Dop))/(lambda)=2 beta=(2v_(pr))/(c )` Thus `(Deltalambda_(Dopp))/(Delta lambda_(nat))=(4piv_(pr)tau)/(lambda)` substitution gives USING `v_(pr)=sqrt((2RT)/(M))= 157 m//s` `(Delta lambda_(Dopp))/(Delta lambda_(nat))~~1.2xx10^(3)` Note:- Our formula is an order of magnitude estimate. |
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