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Find the ratio of the number of the atoms of gaseous sodium in the state 3P to that in the ground state 3S at a temperature T==2400K. The spectral line corresponding to the transition 3P rarr 3S is known to have the wavelength lambda=589nm. |
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Answer» Solution :We have `(N)/(N_(0))=(g)/(g_(0))E^(-h omega//kT)=(g)/(g_(0))e^(-2pi h c//lambdakT)` Here `g=` degeneracy of the `3P` state`=6, g_(0)=` degeneracy of the `3S` state `=2` and `lambda=` wavelength of the `3P rarr 3S` line `((2pi ħc)/(lambda))=` energy DIFFERENCE between `3P & 3S` levels. Substitution gives `(N)/(N_(0))= 1.13xx10^(-4)` |
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