Light and Atomic Spectra: Photons, Energy Levels and the Hydrogen Lines

Wavelength, frequency and the speed of light

Light is an electromagnetic wave. Its wavelength λ is the distance between crests, and its frequency ν is the number of crests passing a point each second, in hertz (s⁻¹). In a vacuum every wavelength travels at the same speed, c = 2.99792458 × 10⁸ m/s, an exact defined value listed by NIST. The three quantities are linked by c = λν, so wavelength and frequency are inversely proportional: halve one and the other doubles.

Visible light runs from roughly 400 nm (violet) to 700 nm (red). Ultraviolet light has shorter wavelengths and higher frequencies; infrared has longer wavelengths and lower frequencies. Convert nanometers to meters before using c = λν: 1 nm = 10⁻⁹ m.

Photon energy and the photoelectric effect

Light is absorbed and emitted in packets called photons. Each photon carries energy E = hν = hc/λ, where h = 6.62607015 × 10⁻³⁴ J·s is the Planck constant, also exact. Shorter wavelength means more energy per photon. Brighter light of the same color means more photons, not more energetic ones.

Worked example 1. The yellow light of a sodium lamp is near 589.0 nm. Its frequency is ν = c/λ = (2.998 × 10⁸ m/s)/(589.0 × 10⁻⁹ m) = 5.090 × 10¹⁴ Hz. One photon carries E = hν = (6.626 × 10⁻³⁴ J·s)(5.090 × 10¹⁴ s⁻¹) = 3.373 × 10⁻¹⁹ J. Multiply by Avogadro's number, 6.022 × 10²³ mol⁻¹, for a mole of photons: 2.031 × 10⁵ J, or 203.1 kJ/mol. That is comparable to the energy of a weak chemical bond, which is why visible and ultraviolet light can drive photochemical reactions.

The photoelectric effect shows the photon picture directly. Light ejects electrons from a metal only if each photon carries at least the metal's work function, Φ. Any extra energy becomes the electron's kinetic energy: KEmax = hν − Φ.

Worked example 2. Suppose a metal surface has a threshold wavelength of 540 nm (a given value for this exercise). Its work function is Φ = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s)/(540 × 10⁻⁹ m) = 3.679 × 10⁻¹⁹ J. Violet light at 405 nm has photons of 4.905 × 10⁻¹⁹ J, so KEmax = 4.905 × 10⁻¹⁹ J − 3.679 × 10⁻¹⁹ J = 1.226 × 10⁻¹⁹ J. Rounded to three significant figures, that is 1.23 × 10⁻¹⁹ J, or 0.765 eV. Red light at 650 nm, however intense, ejects no electrons, because each photon has only 3.06 × 10⁻¹⁹ J.

Why atoms give line spectra instead of rainbows

A hot solid, such as a lamp filament, emits a continuous spectrum. A hot, low-pressure gas of atoms emits only certain wavelengths. Electrons in an atom can occupy only certain energy levels. When an electron drops from a higher level to a lower one, the atom emits a single photon whose energy equals the gap: ΔE = hν. Because only certain gaps exist, only certain wavelengths appear.

Absorption is the reverse. Shine white light through a cool gas and the atoms remove exactly the wavelengths that match their upward jumps, leaving dark lines in an otherwise continuous spectrum. A jump between the same two levels has the same wavelength whether the atom absorbs or emits. Each element's set of gaps is unique, so spectra identify elements in flames, lamps and distant stars.

Hydrogen energy levels and the Rydberg equation

For a one-electron atom, the energy levels follow a simple formula. OpenStax Chemistry 2e (Section 6.2) gives Eₙ = −kZ²/n², with k = 2.179 × 10⁻¹⁸ J, Z the nuclear charge and n = 1, 2, 3, …. The energy is negative because the bound electron is lower in energy than a free electron at rest, which is set to zero. As n grows, the levels crowd together toward zero.

The photon emitted when the electron falls from nᵢ to a lower n_f has energy ΔE = kZ²(1/n_f² − 1/nᵢ²). Dividing by hc gives the Rydberg equation for hydrogen, 1/λ = R(1/n_f² − 1/nᵢ²), with R = 1.097 × 10⁷ m⁻¹.

Worked example 3. Find the wavelength of hydrogen's n = 3 → n = 2 line. ΔE = (2.179 × 10⁻¹⁸ J)(1/4 − 1/9) = (2.179 × 10⁻¹⁸ J)(5/36) = 3.0264 × 10⁻¹⁹ J, keeping one extra digit until the end. With hc = 1.9864 × 10⁻²⁵ J·m, λ = hc/ΔE = 6.564 × 10⁻⁷ m, or 656.4 nm. The NIST Handbook of Basic Atomic Spectroscopic Data lists this red line at 656.3 nm in air. The last-digit difference comes from rounding k and from the slightly slower speed of light in air.

Worked example 4. The ionization energy of hydrogen is the energy to move the electron from n = 1 to n = ∞, which is k itself: 2.179 × 10⁻¹⁸ J per atom. Per mole, (2.179 × 10⁻¹⁸ J)(6.022 × 10²³ mol⁻¹) = 1.312 × 10⁶ J, or 1312 kJ/mol. The NIST Atomic Spectra Database value, 13.598 eV, converts to 1312.0 kJ/mol.

Lyman, Balmer and Paschen series

Lines that end on the same lower level form a series. All of them get closer together as nᵢ increases and approach a series limit, where the electron starts free.

• Lyman series, ending at n = 1: ultraviolet. The 2 → 1 line is at 121.6 nm, and the series limit is near 91.2 nm.
• Balmer series, ending at n = 2: the four visible hydrogen lines, at about 656, 486, 434 and 410 nm for nᵢ = 3, 4, 5 and 6. The series limit is near 365 nm, in the ultraviolet.
• Paschen series, ending at n = 3: infrared. The 4 → 3 line is near 1875 nm.

Spectra mistakes and quick checks

Most errors in these problems are unit errors or sign errors. Two checks catch most of them: a visible photon should carry roughly 2.8 to 5.0 × 10⁻¹⁹ J, and a larger energy gap must give a shorter wavelength.

• Leaving wavelength in nanometers when using c = λν or E = hc/λ. Convert to meters first.
• Using the Rydberg equation with nᵢ and n_f swapped, which gives a negative wavelength.
• Expecting every hydrogen line to be visible. The whole Lyman series is ultraviolet and the Paschen series is infrared.
• Thinking brighter light ejects faster photoelectrons. Intensity changes the number of electrons, not their maximum energy.
• Applying Eₙ = −kZ²/n² to helium or any atom with more than one electron. It holds only for one-electron species such as H, He⁺ and Li²⁺.
• Forgetting Avogadro's number when converting a per-photon energy to kJ/mol, or applying it twice.

What the Bohr picture explains and where it stops

Bohr's model of the hydrogen atom, with the electron in circular orbits of fixed energy, reproduces the hydrogen lines and the Z² scaling for ions such as He⁺. For He⁺, every hydrogen transition energy is four times larger, so its 2 → 1 line is near 30.4 nm.

The model fails for atoms with two or more electrons, because electron–electron repulsion changes the levels. It also pictures electrons on definite paths, which quantum mechanics rules out. Modern theory keeps Bohr's energy levels for hydrogen but replaces orbits with orbitals, which describe where an electron is likely to be found.