Question

Using Rydberg formula, calculate the longest wavelength belonging to Lyman and Balmer series. In which region of hydrogen spectrum do these transitions lie?
[Given R = 1.1 ✕ 10⁷ m⁻¹]

Answer 1

The wavelength λ of the spectral line of Lyman series can be calculated using the following formula:

\[\frac{1}{\lambda} = R\left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]\]

The longest wavelength is the first line of the series for which

\[n_2 = 2\]

Now,

\[\frac{1}{\lambda} = 1 . 1 \times {10}^7 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right]\]

\[ \Rightarrow \frac{1}{\lambda} = 1 . 1 \times {10}^7 \times \frac{3}{4}\]

\[ \Rightarrow \lambda = \frac{4}{1 . 1 \times {10}^7 \times 3}\]

\[ \Rightarrow \lambda = 1 . 2121 \times {10}^{- 7} m\]

\[ \Rightarrow \lambda = 121 . 21 nm\]

This transition lies in the ultraviolet region.
For the Balmer series, the wavelength is given by

\[\frac{1}{\lambda} = R\left[ \frac{1}{2^2} - \frac{1}{n_2^2} \right]\]

The longest wavelength is the first line of the series for which

\[n_2 = 3\]

Now,

\[\frac{1}{\lambda} = 1 . 1 \times {10}^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]\]

\[ \Rightarrow \frac{1}{\lambda} = 1 . 1 \times {10}^7 \times \frac{5}{36}\]

\[ \Rightarrow \lambda = \frac{36}{1 . 1 \times {10}^7 \times 5}\]

\[ \Rightarrow \lambda = 6 . 545 \times {10}^7 m\]

\[ \Rightarrow \lambda = 654 . 5 \text { nm} \]

This transition lies in the visible region.