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Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum. |
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Answer» SOLUTION :The Rydberg formula is `(hc)/(lambda_(l))=(me^(4))/(8epsi_(0)^(2)H^(2))(1)/(n_(f)^(2)-(1)/(n_(i)^(2)))` The wavelengths of the first four lines in the Lyman SERIES correspond to transitions from `n_(1) = 2, 3, 4, 5, "to" n_(f) =1` . We know that `(me^(4))/(8epsi_(0)^(2)h^(2))=13.6eV=21.76xx10^(-19)J` `:. lambda_(s)=(hc)/(21.76xx10^(-19))((1)/(1)-(1)/(n_(i)^(2)))m` `=(6.625xx10^(-34)xx3xx10^(8)xx n_(i)^(2))/(21.76xx10^(-19)xx(n_(i)^(2)-1))` `=(0.9134 n_(i)^(2))/((n_(i)^(2)-1))xx10^(-7)m` `=(913.4n_(i)^(2))/((n_(i)^(2)-1)Å)` Substituting `n_(i)=2,3,4,5` we get `lambda_(21)=1218Å`, `lambda_(31)=1028Å, lambda_(41)=974.3Å, and lambda_(51)=951.4019Å` |
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