Properties of Photons (College Board AP® Chemistry): Revision Note

Martín

Written by: Martín

Reviewed by: Stewart Hird

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Photons

  • One of the main characteristics of light is its wave-particle duality

  • This means that light can behave as a wave, but also as a particle

  • As seen in previous sections, a light wave has frequency, wavelength and energy

  • However, the discovery of the photoelectric effect suggested that light can also behave as a particle

  • The photoelectric effect was observed when some metals emit electrons when light was shone on them

    • The results of this experiment suggested that light must come into particles

  • A light particle is called a photon

Absorption and emission of photons

  • Electrons move around the nucleus of the atoms in energy levels

  • Atoms and molecules can absorb photons, and the electrons will jump to higher energy levels

    • When this occurs, the energy of the atom or molecule increases by an amount equal to the energy of the photon

  • The excited atom or molecule can then emit a photon as the electron falls back to a lower energy level, often all the way to the lowest one, the ground state

    • When this occurs, the energy of the atom or molecule will decrease by an amount equal to the energy of the photon

Absorption and emission diagram

Diagram of an atom with its electron close to the nucleus (ground state) and, after absorption, farther from the nucleus (excited state, higher energy); an emission arrow points back to the ground state.

The difference between absorption and emission depends on whether electrons are jumping from lower to higher energy levels or the other way around

The hydrogen emission spectrum

  • The hydrogen emission spectrum is a set of separate lines; Niels Bohr’s model explained each line as a photon released when an electron falls from a higher energy level to a lower one

  • Bohr suggested that electrons can only exist in fixed energy levels

    • This means that electronic transitions can only occur when photons with a specific amount of energy are absorbed or released

  • Each electronic transition has to do with photons from specific regions of the electromagnetic spectrum

Electron transitions in the hydrogen atom 

Energy-level diagram of the hydrogen atom with levels n = 1 to 8: arrows fall to n = 1 (Lyman series, ultraviolet), to n = 2 (Balmer series, visible), to n = 3 (Paschen, infrared), to n = 4 (Brackett, infrared) and to n = 5 (Pfund, infrared).

Electron jumps in the hydrogen spectrum

  • The table summarizes these jumps:

Electron transitions summary

Jumps

Region

Energy

From n = 4, 5, 6 … to n = 3

Infrared

Lowest

From n = 3, 4, 5 … to n = 2

Visible

Medium

From n = 2, 3, 4 … to n = 1

Ultraviolet

Highest

Energy of Photons

Frequency, wavelength and energy

  • As the electromagnetic spectrum shows, the lower the frequency of an electromagnetic wave, the longer its wavelength

    • Frequency (ν) and wavelength (λ) are inversely proportional

    • The equation that links them is:

 c=λv

  • In this equation, c is the speed of light, 2.998 × 108 m s-1

  • The lower the frequency, the lower the energy of each photon

    • Frequency (ν) and energy (E) are directly proportional

    • The equation that links the energy emitted or absorbed by a photon and its frequency is:
                                                      E=hv

    • In this equation, h is Planck’s constant, 6.626 × 10-34 J s

  • These two equations can be used to calculate frequency, wavelength or energy when one of them is given

Examiner Tips and Tricks

  • E = hν and c = λν, Planck’s constant and the speed of light are all in the equations and constants reference available in the exam

  • E = hν needs the frequency: if you are given a wavelength, convert it to frequency with c = λν first, and convert nm to m (1 nm = 1 × 10-9 m)

  • Check the size of your answer: a visible or ultraviolet photon carries about 10-19 to 10-18 J, so an answer like 10-40 J means the wavelength was used in place of the frequency

Worked Example

Calculate the wavelength of the photon emitted when an electron transition releases 1.632 × 10-18 J

Answer:

Step 1: Calculate the frequency of the photon by using the Planck’s equation

E=hv

v=Eh

v=1.632×10−18 J6.626×10−34 J s

ν = 2.463 × 1015 s-1 

Step 2: Use the frequency calculated in Step 1 to calculate the wavelength of the photon

c=λv

λ=cv

λ=2.998×108 m s−12.463 ×1015 s−1

λ = 1.217 × 10-7 m = 121.7 nm (1 nm = 1 × 10-9 m), which is in the ultraviolet region — the hydrogen n = 2 → n = 1 line

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Martín

Author: Martín

Expertise: Chemistry Content Creator

Martín, a dedicated chemistry teacher and tutor, excels in guiding students through IB, AP, and IGCSE Chemistry. As an IB Chemistry student, he came from hands-on preparation, focusing on practical exam techniques and rigorous practice. While at Universidad San Francisco de Quito, his academic journey sparked a passion for computational and physical chemistry. Martín specializes in chemistry, and he knows that SaveMyExams is the right place if he wants to have a positive impact all around the world.

Stewart Hird

Reviewer: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.