Selection Rules for Hydrogen Transitions

Introduction

The energy-level formula for hydrogen tells us where spectral lines could appear: any pair of levels gives a photon of energy ΔE = hν. Yet a high-resolution spectrum shows that many of the conceivable pairings never produce a line at all. An electron in the 2s orbital, for example, cannot simply drop to 1s by emitting one photon, even though the energy gap is the same as for 2p → 1s. The rules that decide which jumps actually happen are called selection rules , and they follow directly from the shapes and symmetries of the wavefunctions.

Core explanation

The transition dipole moment. Light interacts with an atom mainly through its oscillating electric field, which pushes on the electron's charge. The strength of the coupling between an initial state ψ i and a final state ψ f is measured by the transition dipole moment

μ fi = ∫ψ f μ̂ ψ i dτ, where μ̂ = −e r is the electric-dipole operator.

The intensity of a line is proportional to μ fi ². If the integral is exactly zero, the transition is forbidden in the electric-dipole approximation; if it is non-zero, it is allowed . The integral measures how far the mixture of the two states behaves like an oscillating dipole: during an allowed transition the electron cloud sloshes from one side of the nucleus to the other at the frequency of the light.

Splitting the integral. Each hydrogen orbital is a product R nl(r)Y l^mₗ(θ, φ). The components of r are r multiplied by functions of angle, and those angular functions (sin θ cos φ, sin θ sin φ, cos θ) are themselves combinations of the l = 1 spherical harmonics. The dipole integral therefore factorises into a radial integral and an angular integral:

μ fi ∝ [∫R n'l' r R nl r² dr] × [∫Y l'^mₗ' Y 1^m Y l^mₗ sin θ dθ dφ]

The radial integral is almost never zero, so it places no restriction on n. The angular integral of three spherical harmonics is zero unless strict conditions are met.

The rules. For a one-electron atom, electric-dipole transitions are allowed only when

- Δl = ±1 (l must change by exactly one unit), - Δmₗ = 0, ±1 , - Δn = any value , including zero in principle.

Because the electron's spin is untouched by the electric field, the spin state does not change either (Δs = 0).

Consequences for the spectrum. From 1s (l = 0), absorption can only reach np levels. Emission to 1s comes only from np levels, which is why the Lyman series consists of np → 1s lines. The 2s → 1s transition is forbidden; 2s can decay only by a very slow two-photon process, giving it a lifetime of roughly a tenth of a second, compared with about 1.6 ns for 2p. A state trapped like this is called metastable . Because the energy of hydrogen depends only on n, forbidden transitions do not remove any wavelengths from the Balmer series — they reduce the number of ways each line is produced, which matters for intensities and for fine structure.

Step-by-step reasoning

To decide whether a hydrogen transition is dipole-allowed:

1. Write down n, l and mₗ for both states. 2. Check Δl. If it is not +1 or −1, the transition is forbidden — stop. 3. Check Δmₗ. It must be 0, +1 or −1. 4. Check the spin is unchanged. 5. If all conditions are met, the transition is allowed and its energy follows from ΔE = R H hc(1/n f² − 1/n i²).

Visual explanation

Picture a 1s cloud (a sphere) superimposed on a 2pz cloud (two lobes of opposite sign). Their product is positive above the nucleus and negative below, like a small dipole pointing along z, so ∫ψ(2pz) z ψ(1s) dτ survives. The product of 1s and 2s is spherical, with no "up versus down" character, so multiplying by z and integrating gives zero.

Real-world analogy

Think of pushing a child on a swing. A push timed to swing back and forth sets it moving; a push that squeezes the seat symmetrically from both sides does nothing, however strong. Light's field is a back-and-forth push; only pairs of states whose combination can swing from side to side respond to it.

Real-world example

The famous 21 cm radio line of interstellar hydrogen comes from a spin-flip within the 1s level. It is so strongly forbidden that an isolated atom waits about ten million years to emit it. Radio astronomers detect it anyway because space contains enormous numbers of hydrogen atoms, and the line maps the spiral arms of our galaxy.

Why?

Why must l change by exactly one? A photon carries one unit of angular momentum (ħ). When it is absorbed or emitted, the atom's orbital angular momentum must change to conserve the total, and the only way for vectors of length l and 1 to combine into l' is if l' = l − 1, l or l + 1. Parity then removes l' = l: orbitals have parity (−1)^l, the dipole operator is odd, so the two orbitals must have opposite parity.

Common misconception

"Forbidden transitions never happen." They are forbidden only in the electric-dipole approximation. Magnetic-dipole, electric-quadrupole and two-photon processes can still drive them, just millions of times more slowly. In low-density environments such as nebulae, where collisions are rare, forbidden lines are prominent.

Worked example

Question: An electron in hydrogen is in the n = 3 shell. Which of the transitions 3s → 2s, 3s → 2p, 3p → 2s, 3p → 2p, 3d → 2s and 3d → 2p are electric-dipole allowed?

Reasoning: Apply Δl = ±1 to each. 3s → 2s: Δl = 0, forbidden. 3s → 2p: Δl = +1, allowed. 3p → 2s: Δl = −1, allowed. 3p → 2p: Δl = 0, forbidden. 3d → 2s: Δl = −2, forbidden. 3d → 2p: Δl = −1, allowed.

Answer: Only 3s → 2p, 3p → 2s and 3d → 2p are allowed. All three contribute to the red Balmer-α line at 656 nm.

Quick check

1. Is a single-photon transition from 4f to 2p allowed in hydrogen, and why or why not? Answer: No. Here Δl = 3 − 1 = 2, which breaks the rule Δl = ±1, so the transition dipole moment is zero.

Exam focus

Be ready to state Δl = ±1 and Δmₗ = 0, ±1, to write the transition dipole integral, and to justify the rules using photon angular momentum and parity. A common question asks why 2s is metastable; the answer is that the only lower state, 1s, also has l = 0.

Advanced insight

The same symmetry idea generalises to molecules and complexes. The Laporte rule states that in a centrosymmetric environment, transitions between states of the same parity (g → g or u → u) are forbidden. This is why d–d transitions in octahedral complexes are weak (molar absorption coefficients of order 1–100 dm³ mol⁻¹ cm⁻¹) and why tetrahedral complexes, lacking an inversion centre, are often more intensely coloured.

Summary

The intensity of a transition is proportional to the square of the transition dipole moment ∫ψ f μ̂ ψ i dτ. For hydrogen this integral is non-zero only when Δl = ±1 and Δmₗ = 0, ±1, while Δn is unrestricted and spin is unchanged. The rules follow from photon angular momentum and orbital parity. Forbidden transitions such as 2s → 1s are very slow and create metastable states.

Practice questions

1. Write the expression for the transition dipole moment between states ψ i and ψ f and state what its value tells you. Answer: μ fi = ∫ψ f μ̂ ψ i dτ with μ̂ = −e r . If it is zero the transition is dipole-forbidden; if non-zero it is allowed, with intensity proportional to μ fi ². 2. Which final orbitals with n = 3 can be reached from 2p by absorption of one photon? Answer: 3s and 3d, since both have l differing from 1 by one unit; 3p is forbidden because Δl = 0. 3. Explain why the Lyman series contains only transitions ending on 1s that start from p orbitals. Answer: The final state 1s has l = 0, so Δl = ±1 requires the initial state to have l = 1, that is, an np orbital. 4. Use parity to explain why 3d → 1s is forbidden. Answer: Both orbitals have even parity (l = 2 and l = 0), and the dipole operator is odd, so the integrand is odd overall and the integral vanishes. 5. Why is the 2s state of hydrogen described as metastable? Answer: Its only lower level is 1s, and 2s → 1s has Δl = 0, so it cannot decay by one-photon emission and survives for about a tenth of a second.