Hydrogen Atom: Problem-Solving Workshop

Introduction

Knowing the hydrogen atom is one thing; using it fluently is another. Exam and research problems combine several ideas at once: an energy formula, a count of nodes, an integral over a radial wavefunction, a selection rule. This workshop works through a set of representative problems in the order a careful solver would tackle them, highlighting the checks that catch mistakes. Keep the key results in front of you: E n = −13.6 Z²/n² eV, ψ₁ₛ = (Z³/πa₀³)^½ e^(−Zr/a₀), and nodes counted from n and l.

Core explanation

Toolkit. The problems below rely on five results.

- Energy: E n = −R H hc Z²/n² = −13.6 Z²/n² eV, or −Z²/(2n²) in hartree. - Transitions: 1/λ = R H Z²(1/n lower² − 1/n upper²), with R H = 1.097 × 10⁷ m⁻¹. - Nodes: total nodes = n − 1; angular nodes = l; radial nodes = n − l − 1. - Radial distribution: P(r) = r²R(r)². Its maximum is the most probable radius; for 1s, r mp = a₀/Z. - Selection rules: Δl = ±1, Δmₗ = 0, ±1, any Δn.

Problem A — ionisation energy of an excited ion. How much energy is needed to ionise He⁺ from its n = 2 level? With Z = 2 and n = 2, E₂ = −13.6 × 4/4 = −13.6 eV. Ionisation takes the electron to E = 0, so the answer is 13.6 eV — the same as ionising ground-state hydrogen, because Z²/n² is again 1. Check: from n = 1 He⁺ would need 54.4 eV, four times as much, as expected for Z² = 4.

Problem B — a visible line. Find the wavelength of the Balmer-α line (n = 3 → 2) of hydrogen. 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 0.1389 = 1.524 × 10⁶ m⁻¹, so λ = 656 nm, the red line of the Balmer series. For He⁺ the same transition would have 1/λ four times larger and λ = 164 nm, in the ultraviolet.

Problem C — counting nodes. For a 4d orbital, n = 4 and l = 2. Angular nodes = 2 (for 4d xy these are the xz and yz planes). Radial nodes = 4 − 2 − 1 = 1. Total = 3 = n − 1. A radial distribution function for 4d therefore shows two maxima separated by one zero.

Problem D — probability inside a sphere. What is the probability of finding a hydrogen 1s electron within one Bohr radius of the nucleus? Integrate P(r) = 4r²e^(−2r) (atomic units) from 0 to 1. The standard result is P(r < R) = 1 − e^(−2R)(1 + 2R + 2R²). At R = 1: 1 − e⁻²(1 + 2 + 2) = 1 − 5 × 0.1353 = 0.323. So only about 32 % of the time is the electron closer than a₀, even though a₀ is the most probable radius. The distribution has a long tail, and the mean radius is 1.5 a₀.

Problem E — combining ideas. A hydrogen atom in the 3d state emits a photon. Which lower state can it reach, and what is the wavelength? Δl = ±1 forbids 3d → 1s and 3d → 2s; only 3d → 2p is allowed, and the wavelength is 656 nm, as in Problem B.

Step-by-step reasoning

A reliable routine for any hydrogen problem:

1. Identify Z, n and l for every state involved. 2. Write the relevant formula symbolically before substituting numbers. 3. Keep units consistent: eV with 13.6, metres with R H, a₀ for lengths. 4. Check the answer against a limiting case or a scaling law such as Z². 5. For transitions, confirm the selection rules before calculating.

Visual explanation

Sketch P(r) for 1s as a curve rising from zero, peaking at a₀ and decaying slowly. Shade the area from 0 to a₀: it is visibly less than a third of the total. The unshaded tail stretching to large r holds the remaining two-thirds of the probability.

Real-world analogy

Solving hydrogen problems is like following a recipe with a few core techniques — chopping, boiling, seasoning. Once each technique is second nature, any dish becomes a sequence of familiar steps. The five toolkit results play the role of those techniques.

Real-world example

Astronomers identify ionised helium in hot stars and nebulae by its line at 468.6 nm, the He⁺ n = 4 → 3 transition. Its wavelength is predicted precisely by the Rydberg formula with Z = 2, a direct application of the calculations practised on this page.

Why?

Why does the Z²/n² combination appear everywhere? Increasing Z pulls the electron closer (radius ∝ n²/Z) and strengthens the attraction (∝ Z/r). Both effects together give energies proportional to Z²/n², so levels with the same ratio have identical energies in different ions.

Common misconception

"The most probable radius means the electron is usually within that distance." For 1s the most probable radius is a₀, but the probability of being inside it is only 0.32. Most probable, mean and median radii are different quantities and must not be confused.

Worked example

Question: What is the longest wavelength in the Lyman series of Li²⁺ (Z = 3)?

Reasoning: The longest wavelength is the smallest energy gap, n = 2 → 1. 1/λ = 1.097 × 10⁷ × 9 × (1 − 1/4) = 1.097 × 10⁷ × 6.75 = 7.40 × 10⁷ m⁻¹, so λ = 1.35 × 10⁻⁸ m.

Answer: 13.5 nm, in the extreme ultraviolet; it is the hydrogen value of 121.6 nm divided by nine.

Quick check

1. How many radial and angular nodes does a 5f orbital have? Answer: With n = 5 and l = 3 it has 3 angular nodes and 5 − 3 − 1 = 1 radial node.

Exam focus

Many marks are lost through arithmetic slips with 1/n² terms and by forgetting the Z² factor for ions. Show the symbolic formula, then substitute. When asked about probabilities, state whether you are using ψ² (density at a point) or r²R² (probability per unit radius).

Advanced insight

The same toolkit extends to Rydberg atoms, in which one electron is excited to very high n, such as n = 100. Their radius (about n²a₀ ≈ 0.5 μm) and tiny level spacings make them extremely sensitive to electric fields. Such atoms are now used as quantum sensors and as qubits in neutral-atom quantum computers.

Summary

Hydrogen problems rest on a small toolkit: E n = −13.6 Z²/n² eV, the Rydberg formula, node counts l and n − l − 1, the radial distribution function r²R², and the selection rule Δl = ±1. Working symbolically, keeping units consistent and checking scaling with Z² prevents most errors. The 1s electron is inside a₀ only 32 % of the time.

Practice questions

1. Calculate the energy in eV needed to ionise a hydrogen atom from the n = 4 level. Answer: E₄ = −13.6/16 = −0.85 eV, so 0.85 eV is required. 2. How many radial nodes does a 3s orbital have, and at how many radii is its radial distribution function zero, excluding r = 0? Answer: It has 3 − 0 − 1 = 2 radial nodes, so P(r) is zero at two finite radii. 3. Find the wavelength of the hydrogen n = 4 → 2 transition. Answer: 1/λ = 1.097 × 10⁷ × (1/4 − 1/16) = 2.057 × 10⁶ m⁻¹, so λ = 486 nm (blue-green). 4. What is the most probable radius of the 1s electron in Li²⁺? Answer: r mp = a₀/Z = 52.9/3 = 17.6 pm. 5. A hydrogen atom in 4s emits one photon. List the lower states it can reach. Answer: Only p states with lower energy: 3p and 2p, because Δl must be ±1 and s → s is forbidden.