These problems test the quantitative side of atomic spectra: converting between wavelength, frequency and energy, using the Rydberg equation, working with hydrogen’s energy levels and counting photons. They build in difficulty. Full worked solutions follow the questions.
For background, see wavelength, frequency and energy calculations and the Rydberg equation.
Constants and equations
- Planck constant, h = 6.626 × 10⁻³⁴ J s
- Speed of light, c = 2.998 × 10⁸ m s⁻¹
- Avogadro constant, N_A = 6.022 × 10²³ mol⁻¹
- 1 eV = 1.602 × 10⁻¹⁹ J
- Rydberg constant for hydrogen, R_H = 1.097 × 10⁷ m⁻¹
- Handy shortcut: hc ≈ 1240 eV nm, so E (eV) ≈ 1240 ⁄ λ (nm)
Equations:
- c = λν
- E = hν = hc ⁄ λ
- 1 ⁄ λ = R_H (1 ⁄ n₁² − 1 ⁄ n₂²), with n₂ > n₁
- Hydrogen energy levels: Eₙ = −13.6 eV ⁄ n²
- Hydrogen-like ions: Eₙ = −13.6 Z² eV ⁄ n²
Questions
Q1. Calculate the frequency of light with wavelength 500 nm.
Q2. The red Hα line of hydrogen has wavelength 656.3 nm. Calculate the energy of one photon in joules and in electronvolts.
Q3. Sodium street lamps emit strongly at 589 nm. Calculate the energy carried by one mole of these photons, in kJ mol⁻¹.
Q4. Use the Rydberg equation to calculate the wavelength of the photon emitted when a hydrogen electron falls from n = 3 to n = 2. Which colour is it?
Q5. Calculate the wavelength of the transition n = 4 → n = 2 in hydrogen.
Q6. What is the shortest wavelength in the Lyman series (transitions ending at n = 1)? In which region of the spectrum does it lie?
Q7. Using Eₙ = −13.6 eV ⁄ n², calculate the energy and wavelength of the photon absorbed when a hydrogen atom is excited from n = 1 to n = 3.
Q8. Calculate the ionization energy of hydrogen in kJ mol⁻¹.
Q9. A hydrogen atom is excited to n = 4. How many different spectral lines could it produce as it returns to the ground state by all possible routes? List the transitions.
Q10. Which series of hydrogen lines (Lyman, Balmer or Paschen) falls in the visible region? Explain using the energy levels.
Q11. Light of wavelength 400 nm shines on sodium, which has a work function of 2.28 eV. What is the maximum kinetic energy of the emitted electrons in eV? (See the photoelectric effect.)
Q12. He⁺ is a hydrogen-like ion with Z = 2. Calculate the wavelength of the photon emitted in the transition n = 2 → n = 1 for He⁺, and compare it with the same transition in hydrogen.
Q13. A 5.0 mW green laser pointer emits at 532 nm. How many photons does it emit each second?
Worked solutions
A1.
ν = c ⁄ λ = (2.998 × 10⁸ m s⁻¹) ⁄ (500 × 10⁻⁹ m) = 6.00 × 10¹⁴ Hz
A2.
E = hc ⁄ λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ⁄ (656.3 × 10⁻⁹) = (1.9865 × 10⁻²⁵) ⁄ (6.563 × 10⁻⁷) = 3.03 × 10⁻¹⁹ J
In electronvolts: 3.027 × 10⁻¹⁹ ⁄ 1.602 × 10⁻¹⁹ = 1.89 eV. Shortcut check: 1240 ⁄ 656.3 = 1.89 eV. ✓
A3.
Energy of one photon: E = 1.9865 × 10⁻²⁵ ⁄ 5.89 × 10⁻⁷ = 3.373 × 10⁻¹⁹ J Per mole: 3.373 × 10⁻¹⁹ × 6.022 × 10²³ = 2.031 × 10⁵ J mol⁻¹ = 203 kJ mol⁻¹
That’s comparable to the energy of some chemical bonds, which is why visible and ultraviolet light can drive photochemical reactions.
A4.
1 ⁄ λ = 1.097 × 10⁷ × (1 ⁄ 2² − 1 ⁄ 3²) = 1.097 × 10⁷ × (0.2500 − 0.1111) = 1.097 × 10⁷ × 0.13889 = 1.5236 × 10⁶ m⁻¹ λ = 6.563 × 10⁻⁷ m = 656 nm, red. This is the Hα line from Q2.
A5.
1 ⁄ λ = 1.097 × 10⁷ × (1 ⁄ 4 − 1 ⁄ 16) = 1.097 × 10⁷ × 0.1875 = 2.0569 × 10⁶ m⁻¹ λ = 486 nm (blue-green, the Hβ line).
A6.
The shortest wavelength corresponds to the largest energy gap: from n = ∞ down to n = 1 (the series limit).
1 ⁄ λ = R_H × (1 ⁄ 1² − 0) = 1.097 × 10⁷ m⁻¹ λ = 9.12 × 10⁻⁸ m = 91.2 nm, in the ultraviolet.
A7.
E₁ = −13.6 eV; E₃ = −13.6 ⁄ 9 = −1.511 eV ΔE = E₃ − E₁ = −1.511 − (−13.6) = 12.09 eV (absorbed) λ = 1240 ⁄ 12.09 = 102.6 nm (ultraviolet)
Check with Rydberg: 1 ⁄ λ = 1.097 × 10⁷ × (1 − 1⁄9) = 9.751 × 10⁶ m⁻¹ → λ = 102.6 nm. ✓
A8.
Ionization means going from n = 1 to n = ∞: ΔE = 0 − (−13.6 eV) = 13.6 eV per atom. In joules: 13.6 × 1.602 × 10⁻¹⁹ = 2.179 × 10⁻¹⁸ J Per mole: 2.179 × 10⁻¹⁸ × 6.022 × 10²³ = 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹
This matches the measured first ionization energy of hydrogen (13.598 eV). See the ionization energy trend.
A9.
From n = 4 the electron can reach n = 1 by any sequence of downward jumps. The distinct transitions are:
4→3, 4→2, 4→1, 3→2, 3→1, 2→1
That’s 6 lines. The general formula for the number of lines from level n is n(n − 1) ⁄ 2 = 4 × 3 ⁄ 2 = 6. (A single atom emits only one route’s worth of photons at a time, but a sample with many atoms shows all six.)
A10.
The Balmer series (transitions ending at n = 2).
- Lyman lines end at n = 1. The gap from n = 1 to n = 2 alone is 10.2 eV, so every Lyman photon has at least 10.2 eV → wavelengths below 122 nm → ultraviolet.
- Balmer lines end at n = 2. They range from 1.89 eV (656 nm) up to 3.40 eV (365 nm at the series limit), and several lines fall between about 400 and 700 nm → visible.
- Paschen lines end at n = 3, with photons below 1.51 eV → infrared.
See the hydrogen emission spectrum for a full diagram.
A11.
Photon energy: E = 1240 ⁄ 400 = 3.10 eV Maximum kinetic energy = E − work function = 3.10 − 2.28 = 0.82 eV
A12.
For He⁺, Eₙ = −13.6 × Z² ⁄ n² = −54.4 ⁄ n² eV. E₁ = −54.4 eV; E₂ = −13.6 eV ΔE = 40.8 eV → λ = 1240 ⁄ 40.8 = 30.4 nm (extreme ultraviolet)
In hydrogen, the same 2 → 1 transition gives 10.2 eV and 121.6 nm. Because energies scale with Z², the He⁺ photon has four times the energy and one quarter of the wavelength.
A13.
Energy per photon: E = hc ⁄ λ = 1.9865 × 10⁻²⁵ ⁄ 5.32 × 10⁻⁷ = 3.734 × 10⁻¹⁹ J Photons per second = power ⁄ energy per photon = (5.0 × 10⁻³ J s⁻¹) ⁄ (3.734 × 10⁻¹⁹ J) = 1.3 × 10¹⁶ photons per second
Even a weak laser pointer emits tens of thousands of trillions of photons every second, which is why light usually seems continuous. For more on photons, see photons and energy levels.
Common mistakes
- Forgetting to convert nm to m (multiply by 10⁻⁹) before using SI constants.
- Putting n₁ and n₂ the wrong way round in the Rydberg equation, giving a negative wavelength. The lower level is n₁.
- Mixing up “per atom” and “per mole”. Multiply by N_A for molar quantities.
- Using the hydrogen formula for multi-electron atoms. It only works for hydrogen and hydrogen-like ions (one electron), with the Z² factor.
- Confusing emission with absorption. The energies are the same for a given pair of levels; only the direction differs.
Key takeaways
- Use c = λν and E = hc ⁄ λ to move between wavelength, frequency and energy; E (eV) ≈ 1240 ⁄ λ (nm) is a quick check.
- The Rydberg equation gives hydrogen’s line wavelengths; series limits come from n₂ = ∞.
- Hydrogen’s levels are Eₙ = −13.6 eV ⁄ n²; hydrogen-like ions scale with Z².
- Visible hydrogen lines belong to the Balmer series; Lyman is UV, Paschen is IR.
- Each atom’s line spectrum is a fingerprint used in atomic emission spectroscopy and astronomy.
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