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According to the Rydberg formula, the wavelengths of the first two spectral line in the Lyman series of the hydrogen spectrum, areA. `912 Å` & `3684 Å`B. `684 Å` & `811 Å`C. `1218 Å` & `1028 Å`D. `915.4 Å` & `974.3 Å` |
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Answer» The Rydberg formula is `hc//lambda_(g)=(me^(4))/(8epsilon_(0)^(2)h^(2))((1)/(n_(f)^(2)-n_(i)^(2)))` The wavelength of the first four lines in the Lyman series correspond to transitions from `n_(i)=2,3,4,5` to `n_(f)=1`. We know that `(me^(4))/(8epsilon_(0)^(2)h^(2))=13.6eV=21.76xx10^(-19)J` Therefore, `lambda_(i1)=(hc)/(21.76xx10^(-19)((1)/(1)-(1)/(n_(i)^(2))))m=(0.9134n_(i)^(2))/((n_(i)^(2)-1))xx10^(-7)m` `=913.4n_(i)^(2)//(n_(i)^(2)-1)A` Substituting `n_(i)=2,3` we get `lambda_(21) 1218 Å`, `lambda_(31)1028Å` |
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