Question

Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and of the Balmer series.

Answer 1

The Rydberg formula for the spectrum of the hydrogen atom is given below:

\[\frac{1}{\lambda} = R\left[ \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right]\]

Here, 

\[\lambda\]  is the wavelength and R is the Rydberg constant.
R = \[1 . 097 \times {10}^7\] m^-1

For the first member of the Lyman series:

\[n_1 = 1 \]

\[ n_2 = 2\]

Now,

\[\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right]\]

\[\Rightarrow \lambda = 1215 A^o\]

For the first member of the Balmer series:

\[n_1 = 2 \]

\[ n_2 = 3\]

Now,

\[\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]\]

\[\Rightarrow \lambda = 6563 A^o\]