Lesson 2 · 25 min

Light and spectra: the fingerprints of atoms

Light’s energy comes in packets: weigh Planck’s constant with the photoelectric effect, then read the lines of hydrogen, helium, neon, sodium and mercury — and the colours of flames.

In this lesson

Convert between wavelength, frequency and photon energy with $c = \lambda\nu$ and $E = h\nu$ (in eV and kJ/mol)
Use the photoelectric effect to argue that light comes in photons, and compute threshold wavelengths and maximum kinetic energies
Explain why atoms emit only particular wavelengths, and turn a spectral line into an energy-level gap
Identify elements from emission spectra and flame colours — and know which flame colours come from molecules, not atoms

Sprinkle table salt into a blue gas flame and it flares bright yellow. Seen through a spectroscope, that yellow is not a smear of colour but two very close, thin lines: 589.0 nm and 589.6 nm. Whether the salt came from the sea or a mine, the two lines always sit in exactly the same place. Why do atoms emit only these particular kinds of light?

Light is a wave — whose energy comes in packets

Light is an electromagnetic wave. Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and γ-rays are the same thing with different wavelengths; the eye sees only the narrow band from 380 to 750 nm. All of them travel through vacuum at the speed of light $c$ , so a shorter wavelength means a higher frequency.

In 1900 Planck, explaining the colours of hot objects, boldly assumed that light exchanges energy only in packets; in 1905 Einstein went further: light itself is made of photons, and each photon’s energy is set by its frequency alone.

c = \lambda\nu \qquad E = h\nu = \frac{hc}{\lambda} \qquad E\,[\mathrm{eV}] = \frac{1239.84}{\lambda\,[\mathrm{nm}]}

For a mole of photons multiply by Avogadro’s number: 1 eV per photon = 96.485 kJ/mol.

$c$: speed of light, 299 792 458 m/s
$\lambda$: wavelength
$\nu$: frequency (Hz)
$h$: Planck’s constant, 6.626 × 10⁻³⁴ J·s
$E$: energy of one photon

Predict first

A very bright red beam (700 nm) shines on potassium metal, whose work function is 2.30 eV. Will any electrons be knocked out?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Yes — and the brighter the light, the faster they fly Yes, after a while, once enough energy builds up Not a single one Only after the metal warms up

The photoelectric effect

Light falling on a metal can knock electrons out — the photoelectric effect. Classical wave theory says brighter light delivers more energy, so electrons should come out faster, and any colour should work if you wait long enough. Experiments gave three “strange” results instead:

Below a threshold frequency no electrons come out, however bright the light;
The maximum kinetic energy of the electrons grows with frequency only, not with intensity;
Intensity changes only the number of electrons per second — and they appear at once, with no build-up time.

Einstein’s photons explain all three at once: one photon gives its whole energy $h\nu$ to one electron; part of it frees the electron from the metal (the work function $\varphi$ ) and the rest becomes kinetic energy. In 1916 Millikan set out to disprove the idea — and instead used it to measure Planck’s constant directly, matching the value Planck had found from thermal radiation. Einstein’s 1921 Nobel Prize singled out his discovery of the law of the photoelectric effect.

\mathrm{KE}_\mathrm{max} = h\nu - \varphi \qquad \nu_0 = \frac{\varphi}{h} \qquad \lambda_0 = \frac{hc}{\varphi}

Plot

\mathrm{KE}_\mathrm{max}

against

\nu

: every metal gives a straight line of slope

h

, shifted by its threshold.

$\mathrm{KE}_\mathrm{max}$: kinetic energy of the fastest photoelectrons
$\varphi$: work function: the least energy that frees an electron from the surface
$\nu_0$: threshold frequency
$\lambda_0$: threshold (longest usable) wavelength

One photon · one electron · Einstein 1905

Photoelectric effect bench

Wavelength 400 nm · violet Brightness (photons per second) 60 %

Photon energy E = hν 3.10eV 299 kJ/mol

Frequency ν 7.49×10¹⁴ Hz λ = 400 nm

Work function φ · Na 2.75eV Threshold λ₀ = 451 nm

Fastest electrons’ energy 0.35eV stopping voltage 0.35 V · speed 351 km/s

Photocurrent (relative) 60 %

hν > φ: each photon frees one electron and the extra 0.35 eV becomes kinetic energy. Brighter light gives more electrons, not faster ones.

Einstein’s line: KEmax against frequency

Show every metal’s line

Measure points at several wavelengths (two or more): the slope of the line is Planck’s constant — exactly how Millikan measured h in 1916.

\mathrm{KE_{max}} = h\nu - \varphi = 3.10\,\mathrm{eV} - 2.75\,\mathrm{eV} = 0.35\,\mathrm{eV}

“Measure a point” at several wavelengths and check whether the fitted slope is Planck’s constant. Switch metals: the line shifts but keeps its slope. Then turn the brightness to maximum — does the kinetic energy change?

Worked example

How much energy is in one photon of sodium light?

Sodium’s yellow light has a wavelength of 589.0 nm. Find its frequency, the energy of one photon (in J and eV) and of a mole of photons, and compare with the average C–C bond energy (about 348 kJ/mol).

Line spectra: each atom sings only a few notes

A hot solid (a lamp filament, molten iron) glows with a continuous spectrum: every colour is there. But put a thin gas in a discharge tube, or a salt in a flame, and you see separate bright lines — a line spectrum. The reason is that an atom’s electrons can occupy only certain energy levels. When an electron drops from a higher level to a lower one it emits one photon whose energy equals the gap between them. Each element has its own set of levels, so the positions of its lines are its fingerprint.

In 1868 astronomers found a yellow line at 587.6 nm in the spectrum of a solar prominence that matched no known element. Only in 1895 was the gas that emits it found on Earth — named helium, after the Sun (helios).

h\nu = E_2 - E_1 \qquad \frac{1}{\lambda} = R_\mathrm{H}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Hydrogen’s levels are

E_n = -13.6\ \mathrm{eV}/n^2

; jumps ending on

n = 2

(the Balmer series) fall in the visible.

$E_2,\ E_1$: energies of the higher level before the jump and the lower level after it
$R_\mathrm{H}$: Rydberg constant for hydrogen, 1.09678 × 10⁷ m⁻¹
$n_2 > n_1$: principal quantum numbers of hydrogen’s upper and lower levels (the right-hand formula is for hydrogen only)

Emission spectra · the fingerprints of elements

Spectroscope

White light (incandescent): a continuous spectrum

What the eye sees pink-magenta computed from the visible lines through the CIE eye model

Wavelength (in air): 656.29 nm vacuum 656.47 nm, from the Rydberg formula
Frequency: 4.567 × 10¹⁴ Hz
Photon energy: 1.889 eV 182.2 kJ/mol
Emitter: H I atomic line
Transition: n = 3 → n = 2
Colour: red
Relative intensity: 100.0 %

Hydrogen energy levels (eV)

n = 1 sits far below, at −13.6 eV

Intensity basis: Wavelengths computed from the Rydberg formula (then converted to air); intensities borrow the Case B recombination ratios at 10 000 K — a real tube’s ratios change with pressure and current.

Start with hydrogen: select each Balmer line and compare its photon energy with the level gap in the diagram. Then compare helium, neon, sodium and mercury — why does neon glow orange-red while mercury looks blue-violet? Use the white-light continuum at the top to see which colours the lines fall on.

Why does hydrogen have exactly these lines? The story Why hydrogen sings only a few notes builds the level diagram step by step. To run flame tests yourself, filter sodium’s glare with cobalt glass and identify unknown samples, open the flame and discharge-tube lab. Note that calcium, strontium, barium and copper flames get their colours mainly from broad bands of molecules formed in the flame (CaOH, SrCl, BaCl, CuCl), not from atomic lines.

Concept check

Which change makes the ejected photoelectrons faster?

Make the light brighter Use light of shorter wavelength Use a larger metal plate Shine the light for longer

Try the calculation

Copper’s work function is 4.65 eV. What is the longest wavelength that can free electrons from copper?

Try the calculation

In hydrogen an electron drops from $n = 4$ to $n = 2$ . What is the photon energy in eV? ( $E_n = -13.6\ \mathrm{eV}/n^2$ )

Key takeaways

Electromagnetic waves obey $c = \lambda\nu$ ; each photon carries $E = h\nu = hc/\lambda$ , $E\,[\mathrm{eV}] = 1239.84/\lambda\,[\mathrm{nm}]$ , and $N_\mathrm{A}$ times that per mole.
Photoelectric effect: $\mathrm{KE}_\mathrm{max} = h\nu - \varphi$ . No electrons below the threshold frequency; intensity changes their number, not their energy.
Atoms occupy only certain energy levels; an emitted photon’s energy equals the level gap, so each element has a unique line spectrum.
Flame tests identify elements: Li crimson, Na yellow, K lilac (atomic lines); Ca, Sr, Ba and Cu colours come mainly from molecular bands formed in the flame.

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Data and references

A. Einstein, “Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt”, Ann. Phys. 17, 132 (1905).
R. A. Millikan, “A Direct Photoelectric Determination of Planck’s ‘h’”, Phys. Rev. 7, 355 (1916).
E. Tiesinga et al., CODATA Recommended Values of the Fundamental Physical Constants: 2018, Rev. Mod. Phys. 93, 025010 (2021).
H. B. Michaelson, “The work function of the elements and its periodicity”, J. Appl. Phys. 48, 4729 (1977); CRC Handbook of Chemistry and Physics, “Electron Work Function of the Elements”.
A. Kramida, Yu. Ralchenko, J. Reader and NIST ASD Team, NIST Atomic Spectra Database (ver. 5.11), https://physics.nist.gov/asd.
J. E. Sansonetti & W. C. Martin, “Handbook of Basic Atomic Spectroscopic Data”, J. Phys. Chem. Ref. Data 34, 1559 (2005).
R. W. B. Pearse & A. G. Gaydon, The Identification of Molecular Spectra, 4th ed., Chapman & Hall (1976).
P. Atkins, J. de Paula & J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford (2018), Topics 7A and 8A.