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

Explanation

The Rydberg formula is
`hc//lambda_(if) = (me^4)/(8 epsilon_0^2 h^2) (1)/(n_f^2) - 1/(n_t^2)`
The wavelengths 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)/(8 epsilon_0^2 h^2) = 13.6 eV = 21.76 xx 10^(-19) J`
Therefore,
`lambda_(t1) = (hc)/(21.76 xx 10^(-19)(1/1 - 1/(n_t^2)))m`
`= (6.625 xx 10^(-34) xx 3 xx 10^8 xx n_t^2)/(21.76 xx 10^(-19) xx (n_t^2 - 1)) m = (0.9134 n_t^2)/((n_t^2 - 1)) xx 10^(-7) m`
`= 913.4 m_t^2//(n_i^2 - 1) Å`
Substituting `n_i = 2,3,4,5,` we get `lambda_(21) = 1281 Å, lambda_(31) = 1028 Å, lambda_41 = 974.3Å` and `lambda_(51) = 951.4 Å`.