UNSW Sydney · FACULTY OF CHEMISTRY

DPST1031 Chap.1 Light, Quantisation and the Hydrogen Atom

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Chapter 1 of 13 · DPST1031

Light, Quantisation and the Hydrogen Atom

This chapter builds the quantum picture that the rest of the course depends on. Light is treated both as a wave, with wavelength, frequency and speed linked by c = λν, and as a stream of photons carrying E = hν. The photoelectric effect and electron diffraction show why both models are needed.

Hydrogen's electron is then restricted to energy levels labelled by n, so emission and absorption happen only at particular wavelengths. The Rydberg equation converts any transition, including ionisation, into a wavelength and a photon energy.

The chapter closes with the quiz-style single-step calculations and the mastery-style chains, such as a series limit followed by the energy left over after ionisation, that the final examination builds on.

In this chapter

What this chapter covers

  • 01

    Wavelength, frequency and the speed of light

  • 02

    Photon energy and Planck's constant

  • 03

    The photoelectric effect and work function

  • 04

    De Broglie wavelength of moving particles

  • 05

    Hydrogen energy levels and the principal quantum number

  • 06

    Emission, absorption and ionisation

  • 07

    Lyman, Balmer and Paschen series

  • 08

    Rydberg equation calculations

Worked example · free

Wavelength and energy of the first Lyman line

Q [4 marks]. An electron in a hydrogen atom falls from n = 2 to n = 1. Use the Rydberg equation to find the wavelength, frequency and energy of the emitted photon. The marks are an AskSia practice weighting, not a University allocation.
Hydrogen energy levels drawn to scale from E = −2.18 × 10⁻¹⁸ J ÷ n²: each downward arrow is an electron falling and emitting a photon whose wavelength and frequHydrogen energy levels drawn to scale from E = −2.18 × 10⁻¹⁸ J ÷ n²: each downward arrow is an electron falling and emitting a photon whose wavelength and frequency match the gap, the Lyman line from n = 2 to n = 1 and the Balmer line from n = 4 to n = 2, while the upward arrow is ionisation to n = ∞.n = 1n = 2n = 3n = 4n = ∞−2.18 × 10⁻¹⁸ J−5.45 × 10⁻¹⁹ JLyman lineBalmer 486 nmionisation
Hydrogen energy levels drawn to scale from E = −2.18 × 10⁻¹⁸ J ÷ n²: each downward arrow is an electron falling and emitting a photon whose wavelength and frequency match the gap, the Lyman line from n = 2 to n = 1 and the Balmer line from n = 4 to n = 2, while the upward arrow is ionisation to n = ∞.
  • +1With n₁ = 1 and n₂ = 2 the bracket is 1 − 1/4 = 0.750.
  • +11/λ = 1.097 × 10⁷ m⁻¹ × 0.750 = 8.228 × 10⁶ m⁻¹, so λ = 1.215 × 10⁻⁷ m, or 122 nm.
  • +1ν = c/λ = 2.998 × 10⁸ m s⁻¹ ÷ 1.215 × 10⁻⁷ m = 2.467 × 10¹⁵ Hz.
  • +1E = hν = 6.626 × 10⁻³⁴ J s × 2.467 × 10¹⁵ s⁻¹ = 1.63 × 10⁻¹⁸ J, three quarters of the ground-state ionisation energy.
The photon has λ = 122 nm, ν = 2.47 × 10¹⁵ Hz and E = 1.63 × 10⁻¹⁸ J, an ultraviolet Lyman line because the electron ends on n = 1.
Sia tip — Put the lower level in n₁ and the upper level in n₂ so the bracket stays positive, then state emission or absorption in words.
Glossary

Key terms

Photon
A quantum of light whose energy is E = hν.
Work function
The minimum energy needed to remove an electron from the surface of a particular metal.
Ground state
The lowest-energy arrangement of an atom's electrons, n = 1 for hydrogen.
Excited state
Any arrangement in which an electron occupies a higher level than in the ground state.
Ionisation energy
The energy required to remove an electron completely, taking it to n = ∞.
Rydberg constant
The constant 1.097 × 10⁷ m⁻¹ that scales the reciprocal wavelengths of hydrogen lines in the Rydberg equation.
FAQ

Light, Quantisation and the Hydrogen Atom FAQ

Why does brighter light not eject electrons below the threshold frequency?

Brightness only increases the number of photons. Each photon still carries E = hν, and one photon gives its energy to one electron, so if hν is below the work function no single electron ever receives enough energy to leave the metal.

What is the difference between emission and absorption lines?

Both correspond to the same energy gap between two levels. In emission an excited electron falls and a photon leaves the atom; in absorption a photon of exactly that energy is taken in and the electron rises, so the same wavelengths appear as bright or dark lines.

How do I find the longest wavelength that ionises hydrogen?

Use the Rydberg equation with the starting level as n₁ and n₂ = ∞. That gives the smallest photon energy able to reach n = ∞, which corresponds to the longest possible wavelength. Starting from a higher level makes that wavelength longer.

Do I need the de Broglie equation as well as the Rydberg equation?

Yes. The syllabus asks you to describe both photons and electrons with wave and particle models. De Broglie's λ = h/(mv) gives the wavelength of a moving particle; for an electron at a few million metres per second it is a fraction of a nanometre, comparable to the spacing of atoms, which is why electrons diffract.

Study strategy

Exam move

Start by making c = λν and E = hν automatic, including conversions between nanometres and metres. Then draw the hydrogen level diagram from memory with the n = 1, 2 and 3 levels and the ionisation limit, and label one emission, one absorption and one ionisation arrow.

Practise the Rydberg routine on at least one line from each of the Lyman, Balmer and Paschen series and check every answer against the region of the spectrum you expect. Finish with photoelectric problems that ask for kinetic energy and threshold wavelength, because they combine both models of light in one question. Before the Topic 1 quiz, time yourself: each Threshold calculation should take under two minutes.

Before the final, practise explaining in two sentences why a line spectrum proves that energy is quantised, since written-response questions often ask for that reasoning as well as the number.

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