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InterviewSolution
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What are the two longest wavelength lines (in nanometers) in the Lyman series of the hydrogen spectrum? Strategy: The Lyman series is given by the Balmer-Rydberg equation with `n = 1` and `mgt1`. Since the left side of Eq. is a fraction that has `lambda` in the denominator, the value of `lambda` (the wavelength) increases as the value of the term on the right side of the equation decreases. Since the value of `1//n^(2)` is now fixed and we need to subtract `1//m^(2)` from this, the wavelength `lambda` is the greatest when `1//m^(2)` is the largest or when `m` is the smallest, i..e., when `m = 2` and `m = 3`. |
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Answer» Solving Eq. first for `n =2` gives `(1)/(lambda) = R((1)/(1^(2))-(1)/(2^(2)))` `= 1.097xx10^(7)m^(-1)(1-(1)/(4))` `= 0.8228 xx 10^(7)m^(-1)` `lambda = (1)/(0.8228 xx 10^(7)m^(-1)) = 1.215xx10^(-7)m = 121.5nm` Solving next for `m = 3` gives `(1)/(lambda) = R ((1)/(1^(2))-(1)/(3^(2)))` `= 1.097xx10^(7)m^(-1) (1-(1)/(9))` `= 0.9751xx10^(7)m^(-1)` `lambda = (1)/(0.9751 xx 10^(7)m^(-1)) = 1.026xx10^(-7)m = 102.6 nm` |
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