# Spectroscopy analysis

1. May 16, 2004

### UrbanXrisis

In a lab, our class looked at a hydrogen lamp and calculated the wavelengths of the color red, blue and green. I needed to verify that only Balmer lines are visible using the equation:

1/lambda=R(1/m^2)-(1/n^2)

It says that R is Rydbergs constant and I need to find n. It says that m is the proper value for the Balmer series. What is m supposed to be?

2. May 16, 2004

### Staff: Mentor

Balmer series: m = 2

The emission spectrum for the hydrogen atom is given by:
$$\frac{1}{\lambda} = R(\frac{1}{m^2} - \frac{1}{n^2})$$
This represents wavelengths emitted when the atom transitions from a higher state (n) to a lower state (m). (n > m) When the lower state is m = 2, the spectrum is called the Balmer series. It has several wavelengths in the visible range.

Several other spectral series have names (after the experimentalists who discovered them). For example: The Lyman series (ultraviolet) come from transitions to the ground state (m = 1) and the Paschen series (infrared) has m = 3.