Answer: \(\dfrac{1}{\lambda} = R\left(\dfrac{1}{m^2} - \dfrac{1}{n^2}\right)\), where \(R = \dfrac{me^4}{8\varepsilon_0^2h^3c}\) is Rydberg’s constant.
Ionization energy: The minimum energy required to remove an electron from the ground state of an atom to infinity (so that the atom becomes an ion) is called its ionization energy. For hydrogen it is 13.6 eV.
Wavelength of spectral lines: According to Bohr’s theory, the energy of the electron in the \(n^{th}\) orbit of the hydrogen atom is
\[E_n = -\frac{me^4}{8\varepsilon_0^2h^2n^2}\]
When the electron jumps from a higher orbit \(n\) to a lower orbit \(m\) (\(n \gt m\)), the energy difference is emitted as a photon of frequency \(\nu\):
\[h\nu = E_n - E_m = \frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{m^2} - \frac{1}{n^2}\right)\]
Using \(\nu = c/\lambda\):
\[\frac{hc}{\lambda} = \frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{m^2} - \frac{1}{n^2}\right) \Rightarrow \frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2h^3c}\left(\frac{1}{m^2} - \frac{1}{n^2}\right)\]\[\boxed{\frac{1}{\lambda} = R\left(\frac{1}{m^2} - \frac{1}{n^2}\right)}\]
where \(R = \dfrac{me^4}{8\varepsilon_0^2h^3c} = 1.097\times10^7\ \mathrm{m^{-1}}\). For \(m = 1, 2, 3, 4, 5\) this gives the Lyman, Balmer, Paschen, Brackett and Pfund series respectively.
