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First teaching 2023

First exams 2025

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Photon Energy (HL IB Physics)

Revision Note

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Katie M

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Katie M

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Physics

Photons & Atomic Transitions

The Photon Model

  • Photons are fundamental particles that make up all forms of electromagnetic radiation
  • A photon is defined as

A massless “packet” or a “quantum” of electromagnetic energy

  • This means that the energy transferred by a photon is not continuous but as discrete packets of energy
    • In other words, each photon carries a specific amount of energy and transfers this energy all in one go
    • This is in contrast to waves which transfer energy continuously

Atomic Energy Levels

  • Electrons in an atom can have only certain specific energies
    • These energies are called electron energy levels

  • They can be represented as a series of stacked horizontal lines increasing in energy
  • Normally, electrons occupy the lowest energy level available, this is known as the ground state
  • Electrons can gain energy and move up the energy levels if they absorb energy via:
    • Collisions with other atoms or electrons
    • Absorbing a photon
    • A physical source, such as heat

  • This is known as excitation, and when electrons move up an energy level, they are said to be in an excited state
  • If the electron gains enough energy to be removed from the atom entirely, this is known as ionisation
  • When an electron returns to a lower energy state from a higher excited state, it releases energy in the form of a photon

Electron energy levels in atomic hydrogen. Photons are emitted when an electron moves from a higher energy state to a lower energy state

Worked example

Explain how atomic spectra provide evidence for the quantisation of energy in atoms.

Answer:

Step 1: Outline the meaning of atomic spectra

  • Atomic spectra show the spectrum of discrete wavelengths emitted or absorbed by a specific atom

Step 2: Describe the relationship between energy and wavelength

  • Photon energy is related to frequency and wavelength
  • Therefore, photons with discrete wavelengths have discrete energies equal to the difference between two energy levels

Step 3: Explain how atomic spectra give evidence for the quantisation of energy 

  • Photons arise from electron transitions between energy levels
  • This happens when an electron is excited, or de-excited, from one energy level to another, by either emitting or absorbing light of a specific wavelength
  • Since atomic spectra are made up of discrete wavelengths, this shows that atoms must contain discrete, or quantised, energy levels 

Worked example

The diagram shows the electron energy levels in an atom of hydrogen.

7-1-2-we-transitions-between-energy-levels-question-image

Determine the number of possible wavelengths that can be produced from transitions between the n = 4 excited state and the n = 1 ground state.

Answer:

  • There are six possible wavelengths that could be produced from the different energy level transitions

5-1-4-calculating-photon-energy-worked-example-ma

  • The possible transitions are:
    • n = 4 to n = 3
    • n = 4 to n = 2
    • n = 4 to n = 1
    • n = 3 to n = 2
    • n = 3 to n = 1
    • n = 2 to n = 1

Exam Tip

Make sure you learn the definition for a photon: discrete quantity / packet / quantum of electromagnetic energy are all acceptable definitions

Calculating Photon Energy

  • Each line of the emission spectrum corresponds to a different energy level transition within the atom
    • Electrons can transition between energy levels absorbing or emitting a discrete amount of energy
    • An excited electron can transition down to the next energy level or move to a further level closer to the ground state
  • For example, if an atom has six energy levels:
    • At low temperatures, most electrons will occupy the ground state n = 1
    • At high temperatures, electrons may be excited to the most excited state n = 6

12-1-1-photons-ib-hl

Energy and frequency of a photon are directly proportional

  • The energy of a photon can be calculated using the formula:

E space equals space h f

  • Using the wave equation, energy can also be equal to:

E space equals space fraction numerator h c over denominator lambda end fraction

  • Where:
    • E = energy of the photon (J)
    • h = Planck's constant (J s)
    • c = the speed of light (m s−1)
    • f = frequency (Hz)
    • λ = wavelength (m)
  • This equation tells us:
    • The higher the frequency of EM radiation, the higher the energy of the photon
    • The energy of a photon is inversely proportional to the wavelength
    • A long-wavelength photon of light has a lower energy than a shorter-wavelength photon

Difference in discrete energy levels

  • The difference between two energy levels is equal to a specific photon energy
  • The energy of the photon is given by:

increment E space equals space h f space equals space E subscript 2 space minus space E subscript 1

  • Where:
    • E1 = energy of the lower level (J)
    • E2 = energy of the higher level (J)
  • Using the wave equation, the wavelength of the emitted, or absorbed, radiation can be related to the energy difference by the equation:

lambda space equals space fraction numerator h c over denominator E subscript 2 space minus space E subscript 1 end fraction

  • This equation shows that:
    • The larger the difference in energy between two levels ΔE, the shorter the wavelength λ and vice versa

Worked example

Some electron energy levels in atomic hydrogen are shown below.

5-1-4-energy-level-transitions-worked-example

The longest wavelength produced as a result of electron transitions between two of the energy levels is 4.0 × 10–6 m.

(a)
Draw an arrow to show the transition that would produce:
  • A photon of wavelength 4.0 × 10–6 m. Mark with the letter L.
  • The photon with the shortest wavelength. Mark with the letter S.
(b)
Calculate the wavelength for the transition giving rise to the shortest wavelength.

 

Answer:

(a)

  • Photon energy and wavelength are inversely proportional
  • Therefore, the largest energy change corresponds to the shortest wavelength (line S)
  • The smallest energy change corresponds to the longest wavelength (line L)

5-1-4-energy-level-transitions-worked-example-ma

(b)

Step 1: Write down the equation linking wavelength and energy

lambda space equals space fraction numerator h c over denominator increment E end fraction space equals space fraction numerator h c over denominator E subscript 2 space minus space E subscript 1 end fraction

Step 2: Identify the energy levels that give rise to the shortest wavelength

  • The shortest wavelength photon will come from a transition between the energy levels that have the largest difference:
    • E subscript 2 space equals space minus 0.54 space eV
    • E subscript 1 space equals space minus 3.4 space eV
  • Therefore, the greatest possible difference in energy is 

increment E space equals space E subscript 2 space minus space E subscript 1 space equals space minus 0.54 space minus space open parentheses negative 3.4 close parentheses space equals space 2.86 space eV

Step 3: Calculate the wavelength

  • To convert from eV to J: multiply by 1.6 × 10−19 J

lambda space equals space fraction numerator open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3.0 cross times 10 to the power of 8 close parentheses over denominator 2.86 cross times open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses end fraction

lambda = 4.347 × 10−7 m = 435 nm

Worked example

Light of wavelength 490 nm is incident normally on a surface, as shown in the diagram.
The power of the light is 3.6 mW. The light is completely absorbed by the surface.

Calculate the number of photons incident on the surface in 30 s.

Answer:

Step 1: Write down the known quantities

  • Wavelength, λ = 490 nm = 490 × 10−9 m
  • Power, P = 3.6 mW = 3.6 × 103 W
  • Time, t = 30 s

Step 2: Write the equation for photon energy and write in terms of wavelength

 E space equals space h f space space space space space rightwards double arrow space space space space space E space equals space fraction numerator h c over denominator lambda end fraction

Step 3: Calculate the energy of one photon

E space equals fraction numerator h c over denominator lambda end fraction space equals space fraction numerator open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3.0 cross times 10 to the power of 8 close parentheses over denominator 490 cross times 10 to the power of negative 9 end exponent end fraction space equals space 4.06 space cross times space 10 to the power of negative 19 end exponent space straight J

Step 4: Calculate the number of photons hitting the surface every second

fraction numerator p o w e r space o f space l i g h t space s o u r c e over denominator e n e r g y space o f space o n e space p h o t o n end fraction space equals space fraction numerator 3.6 cross times 10 to the power of negative 3 end exponent over denominator 4.06 cross times 10 to the power of negative 19 end exponent end fraction space equals space 8.87 space cross times space 10 to the power of 15 space straight s to the power of negative 1 end exponent

Step 5: Calculate the number of photons that hit the surface in 30 s

number of photons in 30 s = (8.87 × 1015) × 30 = 2.7 × 1017

Exam Tip

The values of Planck’s constant and the speed of light will be included in the data booklet, however, it helps to memorise them to speed up calculations!

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.