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Atomic lithium of concentration n=3.6.10^(16)cm^(-3) is at a temperature T=1500K. In this case the power emmitted at the resonat line's wavelength lambda=671nm(2Prarr 2S) per unit volume of gas is equal to P=0.30W//cm^(3). Find the mean lifetime of Li atoms in the resonance excitation state. |
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Answer» Solution :The number of excited atoms PER unit volume of the gase in `2P` state is `N=n(g_(p))/(g_(s))e^(-2pi hc//lambda kT)` Here `g_(p)=` degeneracy of the 2p state`=6,g_(s)=` degeneacy of the `2s` state`=2` and `lambda=` WAVELENGTH of the resonant line `2p rarr 2s`. The rate of decay of these atoms is `(N)/(TAU)` per sec. per unit volume, since each such atom EMITS light of wavelength `lambda`, we MUST have `(1)/(tau)(2pi ħc)/(lambda)n(g_(p))/(g_(s))e^(-2pi h c//lambda kT)=P` Thus `tau=(1)/(P)` Thus `tau=(1)/(P)(2pi ħc)/(lambda)n(g_(p))/(g_(s))e^(-2pi h c//lambdakT)=65.4xx10^(-9)s=65.4ns` |
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