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Calculate the shortest and longest wavelength in hydrogen spectrum of Lyman series. |
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Answer» For Lyman series `n_(1) = 1` For shortest wavelength in Lyman series (i.e, series limit), the energy difference in two states showing transition should be maximum, i.e., `n_(2) = oo` `(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((oo)^(2))] = R_(H)` `lambda = (1)/(3xx 109678) = 9.11 xx 10^(-6) cm = 911. 7 A^(@)` For longest wavlength in Lyman series (i,e. first line the energy difference in two states showing transition should be minimum, i.e., `n_(2) = 2` `(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((2)^(2))] = (3)/(4) R_(H)` `lambda =(4)/(3) (1)/(R_(H)) = (4)/(3 xx 109677)` `= 1215.7 xx 10^(-8) cm = 1215.7 A^(@)` |
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