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InterviewSolution
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In a hydrogen atom, an electron jumps from third orbit to the orbit. Find out the spectral line. |
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Answer» Step I. Calculation of the wave length of the spectral line According to Rydbergy formula : `(1)/(lambda)=bar(v)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] , R_(H)=109678 cm^(_1) , n_(1)=1, n_(2)=3`. `:. " " (1)/(lambda)=109678 [(1)/(1^(2))-(1)/(3^(2))]=109678 [(9-1)/(9)]cm^(-1)`. `"or" " " lambda=(9)/(109678 xx8) cm =(9xx10^(-2)m)/(109678 xx 8)=1.0257 xx 10^(-7) m`. Step II. Calculation of the frequency of the spectral line `c=v lambda "or" v=c//lambda , c=3.0 xx 10^(8)ms^(-1), lambda =1.0257 xx 10^(-7) m`. `:. " " v=((3.0 xx 10^(8)ms^(-1)))/((1.0257 xx 10^(-7)m))=2.9248 xx 10^(15) s^(-1)`(Hz). |
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