The Rydberg Formula and Balmer’s Formula

Calculating a Balmer Series Wavelength

The first step in the calculation is to find the principle quantum number for the transition you’re considering. This simply means putting a numerical value on the “energy level” you’re considering. So the third energy level has ​n​ = 3, the fourth has ​n​ = 4 and so on. These go in the spot for ​n​₂ in the equations above.

Start by calculating the part of the equation in brackets:

\frac{1}{2^2}-\frac{1}{n_2^2}

All you need is the value for ​n​₂ you found in the previous section. For ​n​₂ = 4, you get:

\frac{1}{2^2}-\frac{1}{n_2^2}=\frac{1}{2^2}-\frac{1}{4^2}=\frac{1}{42}-\frac{1}{16}=\frac{3}{16}

Multiply the result from the previous section by the Rydberg constant, ​R_H​ = 1.0968 × 10⁷ m⁻¹, to find a value for 1/​λ​. The formula and the example calculation gives:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{n_2^2})=1.0968\times 10^7 \times \frac{3}{16}=2056500\text{ m}^{-1}

Find the wavelength for the transition by dividing 1 by the result from the previous section. Because the Rydberg formula gives the reciprocal wavelength, you need to take the reciprocal of the result to find the wavelength.

So, continuing the example:

\lambda = \frac{1}{2056500}=4.86\times 10^{-7}\text{ m} = 486\text{ nanometers}

This matches the established wavelength emitted in this transition based on experiments.