1.

The first line of the sharp series of atomic cesium is a doublet with wavelengths `1358.8` and `1469.5 nm`. Find the frquency intervals (in rads/s units) between the components of the sequent lines of that series.

Answer» The sharp series arise from the transitions `ns rarr mp`. The `s` lines are unsplit so the splitting is due entirely to the `p` level. The frequency difference between sequen t lines is `(DeltaE)/( ħ)` and is the same for all lines of the sharp series. It is
`(1)/( ħ)((2pi ħc)/(lambda_(1))-(2pi ħc)/(lambda_(2)))=(2pi cDeltalambda)/(lambda_(1)lambda_(2))`
Evaluation gives
`1.645xx10^(14)rad//s`


Discussion

No Comment Found

Related InterviewSolutions