Calculate the shortest wavelength of Paschen series and the longest wavelength of Balmer series for hydrogen atom.

Answer: Paschen series shortest \(\lambda\approx8204\) Å; Balmer series longest \(\lambda\approx6563\) Å

\(\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)\), \(R=1.097\times10^7\) m^–1.

Paschen series (\(n=3\)), shortest wavelength (series limit, \(m=\infty\)):

\(\dfrac{1}{\lambda_{min}}=R\left(\dfrac{1}{9}-0\right)\Rightarrow\lambda_{min}=\dfrac{9}{R}=\dfrac{9}{1.097\times10^7}=8.204\times10^{-7}\) m \(=8204\) Å.

Balmer series (\(n=2\)), longest wavelength (\(m=3\)):

\(\dfrac{1}{\lambda_{max}}=R\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=\dfrac{5R}{36}\Rightarrow\lambda_{max}=\dfrac{36}{5\times1.097\times10^7}=6.563\times10^{-7}\) m \(=6563\) Å.