5.3 · Light & matter

Atomic spectra

Line spectra are fingerprints of energy levels.

By the end you should be able to:

Calculate atomic transitions and interpret emission and absorption spectra

Key idea

Emission and absorption spectra

Emission: an excited atom drops to a lower level and emits a photon. A hot or electrically excited gas gives bright lines on a dark background.
Absorption: an atom takes in a photon and jumps to a higher level. White light passed through a cool gas gives dark lines in an otherwise continuous spectrum.

Only photons with $h\nu = |\Delta E|$ for a pair of allowed levels take part, so every element has its own pattern of lines, a fingerprint. Emission and absorption between the same two levels occur at the same wavelength.

Formula

Line positions for hydrogen

$|\Delta E| = Z^2R_H\left|\frac{1}{n_f^2} - \frac{1}{n_i^2}\right|, \qquad \lambda = \frac{hc}{|\Delta E|}$

with $R_H = 2.179\times10^{-18}$ J and $Z = 1$ for hydrogen. The equivalent Rydberg form works directly in wavelength:

$\frac{1}{\lambda} = Z^2R_\infty\left|\frac{1}{n_f^2} - \frac{1}{n_i^2}\right|, \qquad R_\infty = \frac{R_H}{hc} = 1.097\times10^{7}\ \mathrm{m^{-1}}$

Key idea

The hydrogen series

A series is every line that ends on (emission) or starts from (absorption) the same lower level.

Series	Lower level	Region	Longest-wavelength line
Lyman	$n = 1$	ultraviolet	$2 \to 1$ , 122 nm
Balmer	$n = 2$	visible and near UV	$3 \to 2$ , 656 nm
Paschen	$n = 3$	infrared	$4 \to 3$ , 1875 nm
Brackett	$n = 4$	infrared	$5 \to 4$ , 4050 nm

The four visible Balmer lines are 656 nm (red, $3 \to 2$ ), 486 nm (blue-green, $4 \to 2$ ), 434 nm (violet, $5 \to 2$ ) and 410 nm (violet, $6 \to 2$ ). Lines in a series crowd together toward the series limit ( $n_i \to \infty$ ): 91.2 nm for Lyman and 365 nm for Balmer.

Method

Calculating a spectral line

Identify $n_i$ and $n_f$ (and $Z$ for a one-electron ion).
$\Delta E = -Z^2R_H\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ : negative for emission, positive for absorption.
$\lambda = hc/|\Delta E|$ ; convert to nm.
Place it: below 400 nm UV, 400 to 750 nm visible, above 750 nm IR. The lower level names the series.
The number of different emission lines possible from level $n$ down to $n = 1$ is $\dfrac{n(n-1)}{2}$ .

Common mistake

Spectrum traps

Wrong: expecting Balmer lines in the absorption spectrum of room-temperature hydrogen. Right: almost every atom is in $n = 1$ , so cool hydrogen absorbs only Lyman (UV) lines.

Wrong: "the biggest jump gives a visible line." Right: every line ending on $n = 1$ is UV; the lower level decides the region.

Wrong: reporting a negative photon energy or wavelength for emission. Right: $\Delta E$ is negative, but the photon energy is $|\Delta E|$ .

Wrong: thinking absorption $2 \to 4$ and emission $4 \to 2$ occur at different wavelengths. Right: the same energy gap gives the same wavelength (486 nm).

Worked example

Worked example: hydrogen, n = 4 → 2

$\Delta E = -(2.179\times10^{-18})\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = -(2.179\times10^{-18})(0.1875) = -4.086\times10^{-19}\ \mathrm{J}$

The sign is negative, so a photon of $4.086\times10^{-19}$ J is emitted:

$\lambda = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{4.086\times10^{-19}} = 4.862\times10^{-7}\ \mathrm{m}$

Answer: 486 nm, the blue-green line of the Balmer series (visible).

Worked example

Worked example: identify the transition

Hydrogen emits a line at 1875 nm. Which transition produces it?

$E = \frac{hc}{\lambda} = \frac{1.986\times10^{-25}}{1.875\times10^{-6}} = 1.059\times10^{-19}\ \mathrm{J}$

The line is in the infrared, so try the Paschen series ( $n_f = 3$ ):

$\frac{1}{3^2} - \frac{1}{n_i^2} = \frac{1.059\times10^{-19}}{2.179\times10^{-18}} = 0.0486 \;\Rightarrow\; \frac{1}{n_i^2} = 0.1111 - 0.0486 = 0.0625 \;\Rightarrow\; n_i = 4$

Answer: $n = 4 \to n = 3$ , the first line of the Paschen series.

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