Quantum theory & atomic models
The idea
Quantum theory replaces the planetary picture of the atom you met in high school with a model built on two experimental facts: energy exchange between light and matter is quantized (E = hν), and electrons show wave behavior. Bohr captured the first fact for hydrogen — electrons sit in discrete energy levels, and a photon is emitted or absorbed only when the electron jumps between them — which is why atomic spectra are sharp lines rather than continuous bands.
The full quantum model goes further: solving the Schrödinger equation gives orbitals, three-dimensional standing waves whose square tells you the probability of finding the electron in each region of space. The working habit to build is energy bookkeeping: every spectral line corresponds to ΔE between two allowed levels, and ΔE = hc/λ converts between energy and wavelength. For hydrogen, Eₙ = −2.18 × 10⁻¹⁸ J/n², with the negative sign meaning the electron is bound.
The classic misconception is to keep imagining the electron circling the nucleus on a track. An orbital is not an orbit — it has no defined path, only a probability cloud with a characteristic shape (spherical s, dumbbell p), and the electron does not spiral or travel between levels; it transitions with no intermediate energies allowed.
Worked example
A hydrogen atom relaxes from the n = 3 level to the n = 2 level. Using Eₙ = −2.18 × 10⁻¹⁸ J/n², find the energy and wavelength of the emitted photon (h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s), and state which region of the spectrum it falls in.
- Evaluate the two levels first: E₃ = −2.18 × 10⁻¹⁸/9 = −2.42 × 10⁻¹⁹ J and E₂ = −2.18 × 10⁻¹⁸/4 = −5.45 × 10⁻¹⁹ J. Both are negative because a bound electron has less energy than a free one.
- The atom loses energy in the drop, and that loss becomes the photon: ΔE = E₃ − E₂ = (−2.42 + 5.45) × 10⁻¹⁹ = 3.03 × 10⁻¹⁹ J carried away by the light.
- Convert energy to wavelength with E = hc/λ, rearranged to λ = hc/E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(3.03 × 10⁻¹⁹) = 1.99 × 10⁻²⁵/3.03 × 10⁻¹⁹ = 6.56 × 10⁻⁷ m.
- Interpret: 656 nm sits in the visible range (400–700 nm), at the red end. That is exactly the behavior expected for a Balmer-series transition (any drop ending at n = 2 lands in or near the visible).
Answer. The photon carries 3.03 × 10⁻¹⁹ J with wavelength about 656 nm — visible red light, the famous red line of hydrogen.
Check your understanding
- Why must the emitted photon energies of hydrogen form a discrete set rather than a continuum, and what would the spectrum look like if energy levels were not quantized?
- How would you explain to a friend the difference between an orbit and an orbital without using equations?
- What happens to the spacing between adjacent energy levels as n grows, and how does that explain where each spectral series converges?
- If a photon arrives with slightly less energy than the n = 1 to n = 2 gap, what does the atom do with it — and why?
Build the foundations first
Quantum theory & atomic models builds on these concepts. If any feel shaky, start there.