Question

Using the Rydberg formula, calculate the wavelength of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Answer 1

The Rydberg formula is
`hc//lambda_(if)=(me^(4))/(8epsilon_(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_(i)` = 2,3,4,5 to nf = 1. We know that
`(me^(4))/(8epsilon_(0)^(2)h^(2))=13.6 eV = 21.76xx10^(-19)J`
Therefore,
`lambda_(i1)=(hc)/(21.76xx10^(-19)((1)/(1)-(1)/(n_(i)^(2))))m`
`=(6.625 xx 10^(-34)xx3xx10^(8)xxn_(i)^(2))/(21.76xx10^(-19)xx(n_(i)^(2)-1))m=(0.9134n_(i)^(2))/((n_(i)^(2)-1))xx10^(-7)m`
`=913.4 n_(i)^(2)//(n_(i)^(2)-1)Å`
Substituting `n_(i)` = 2,3,4,5, we get `lambda_(21)` = 1218 `Å`, `lambda_(31)=1028 Å,lambda_(41) = 974.3 Å`, and `lambda_(51) = 951.4 Å`.