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Chemistry 352

The Hydrogen Atom Spectrum

Discrete emission lines

Why hydrogen does not emit light like a hot object.

When an electric current passes through low-pressure hydrogen gas, the emitted light does not form a continuous spectrum. Instead, hydrogen emits light at a small number of very specific wavelengths, producing a line spectrum.

This behavior is fundamentally different from that of a blackbody emitter. A blackbody produces a continuous distribution of wavelengths that depends only on temperature. Hydrogen, by contrast, emits light only at discrete, sharply defined wavelengths.

No classical model of matter or radiation could explain why hydrogen emits light at only certain wavelengths.


Balmer’s empirical discovery

In 1885, Johann Jakob Balmer discovered a simple mathematical formula that accurately described the visible emission lines of hydrogen. Remarkably, this formula was found without any underlying physical model of the atom.

Balmer showed that the wavelengths of the visible hydrogen lines follow the pattern

\[ \frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), \qquad n = 3,4,5,\dots \]

This set of lines is now called the Balmer series. Each value of \(n\) corresponds to a different emission line.


Energy form of the Balmer formula

Using \(E = hc/\lambda\), Balmer’s result can be rewritten in terms of photon energies:

\[ E = R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right), \]

where \( R_H = 2.18\times10^{-18}\,\text{J} \) is the Rydberg energy for hydrogen.

This equation correctly predicts the energies of the photons emitted by hydrogen atoms in the visible region of the spectrum.


Why this result was so striking

Balmer’s formula works extraordinarily well, yet it was purely empirical. It was not derived from any model of atomic structure, forces, or electron motion.

Balmer himself emphasized that the simplicity of the formula suggested a deeper physical reason for its success, even though such a reason was unknown at the time.

The existence of discrete emission lines — and the ability to describe them with such a simple mathematical pattern — was a major clue that energy in atoms must be quantized.

Big idea: the hydrogen emission spectrum cannot be explained classically. Balmer’s empirical formula revealed an underlying order that would later demand a quantum-mechanical explanation.

Your turn

Problem 1
Using the Balmer energy expression \( E = R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right) \) with \( R_H = 2.18\times10^{-18}\,\text{J} \), calculate the photon energy for the transition \( n=4 \rightarrow 2 \).
\( 4.09\times10^{-19}\,\text{J} \) \( 1.36\times10^{-18}\,\text{J} \) \( 5.45\times10^{-19}\,\text{J} \) \( 2.73\times10^{-19}\,\text{J} \) \( 8.18\times10^{-19}\,\text{J} \)
Problem 2
A Balmer-series photon has energy \( E = 3.03\times10^{-19}\,\text{J} \). Using \( E = R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right) \) with \( R_H = 2.18\times10^{-18}\,\text{J} \), what is the upper-state quantum number \(n\)?
\( n=3 \) \( n=4 \) \( n=5 \) \( n=6 \) \( n=8 \)
Problem 3
The transition \( n=3 \rightarrow 2 \) (Balmer series) produces a photon with wavelength about \( \lambda \approx 656~\text{nm} \). Which region of the electromagnetic spectrum is this?
Ultraviolet Visible Infrared Microwave

Key points (one glance)