InterviewSolution
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InterviewSolution
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Rarefield Hg gas whose atoms are practically all in the ground state was lighted by a mercury lamp emitting a resonanace line of wavelength lambda= 253.65nm. As a result, the radiation power of Hg gas at that wavelength turned out to be P=35mW. Find the number of the atom in the state of resonance exitation whose mean lifetime is tau=0.15 mu s. |
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Answer» Solution :As a result of the lighting by mercury lamp a number of ATOMS are pumped to the EXCITED state. In equilibrium the number of such atoms is `N`. Since the mean life TIME of the atom is `T`, the number decaying per UNIT is `(N)/(tau)`. Since a photon of energy `(2piħc)/(lambda)` results from each decay, the total RADIATED power will be `(2piħc)/(lambda).(N)/(tau)`. This must equal `P`. Thus `N=P tau//(2piħc)/(lambda)=(P tau lambda)/(2 piħ)=6.7xx10^(9)` |
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