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Determine the wavelength of spectral lines appearing on transition of exicted Li atoms from the state 3S down to the ground state 2S. The Rudberg corrections for the S and P terms are -0.41 and -0.04. |
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Answer» Solution :The enrgy of the `3S` STATE is `E(3S)=-(ħR)/((3-0.41)^(2))=-2.03eV` The ENERGY of a `2S` state is `E(2S)= -(ħR)/((2-0.41)^(2))= -5.39eV` The energy of a `2P` state is `E(2P)=(ħR)/((2-.04)^(2))=-3.55eV` We see that `E(2S) lt E(2P) lt E(3S)` The transitions are `3S rarr2P` and `2Prarr2S`. Direct `3Srarr2S` transition is forbidden by selection rules. The wavelength are determined by `E_(2)-E_(1)= DeltaE=(2piħc)/(lambda)` Substitution gives `lambda= 0.816mu m(2Srarr2P)` and `lambda=0.67mu m(2Prarr2S)` |
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