Calculating Wave Interference of a Hydrogen Line

For an Integrated Concept problem, we must first identify the physical principles involved. In this example, we need to know (a) the wavelength of light as well as (b) conditions for an interference maximum for the pattern from a double slit. Part (a) deals with a topic of the present chapter, while part (b) considers the wave interference material of Wave Optics.

Hydrogen spectrum wavelength. The Balmer series requires that $n_{\text{f}}=2$. The first line in the series is taken to be for $n_{\text{i}}=3$, and so the second would have $n_{\text{i}}=4$.

The calculation is a straightforward application of the wavelength equation. Entering the determined values for $n_{\text{f}}$ and $n_{\text{i}}$ yields

Inverting to find $\lambda$ gives

This is indeed the experimentally observed wavelength, corresponding to the second (blue-green) line in the Balmer series. More impressive is the fact that the same simple recipe predicts all of the hydrogen spectrum lines, including new ones observed in subsequent experiments. What is nature telling us?

Double-slit interference (Wave Optics). To obtain constructive interference for a double slit, the path length difference from two slits must be an integral multiple of the wavelength. This condition was expressed by the equation

where $d$ is the distance between slits and $\theta$ is the angle from the original direction of the beam. The number $m$ is the order of the interference; $m=1$ in this example. Solving for $d$ and entering known values yields

This number is similar to those used in the interference examples of Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).