Question

Find the ratio of the longest wavelength in the Paschen series to the shortest wavelength in the Balmer series.

Answer 1

\(\cfrac1{\lambda}\) = R \(\left(\cfrac{1}{n^2}-\cfrac1{m^2}\right)\)

For the paschen series, n = 3. For the longest wavelength (\(\lambda_{p\alpha}\)), m = 4.

For the Balmer series, n = 2. For the short wavelength limit (\(\lambda_{B\infty}\)), m = ∞.