Photons and Light Energy

Light behaves in ways that cannot be explained completely by thinking of it only as a continuous wave. In quantum physics, light can also be described as travelling in tiny, discrete packets of energy called photons.

Each photon carries a specific amount of energy. The amount depends on the frequency of the electromagnetic radiation.

This idea connects several important concepts:

wavelength ↔ frequency ↔ photon energy ↔ electron transitions

Understanding photons helps explain emission spectra, absorption spectra, lasers, solar cells, the photoelectric effect, and how light interacts with atoms.

What Is a Photon?

A photon is a discrete packet, or quantum, of electromagnetic energy.

Instead of imagining light energy as being transferred in completely arbitrary amounts, quantum physics describes electromagnetic radiation as interacting with matter through photons.

A photon:

• carries energy
• has a frequency
• has a wavelength
• travels at the speed of light in a vacuum
• has no rest mass
• can be emitted or absorbed by matter

Different photons can carry very different amounts of energy.

A radio-wave photon has much less energy than an X-ray photon, even though both are forms of electromagnetic radiation.

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Light as Both a Wave and a Particle

Light demonstrates wave-particle duality.

Some experiments are best understood using a wave model.

Wave behaviour includes:

• diffraction
• interference
• wavelength
• frequency

Other interactions are best understood by treating light as photons.

Photon behaviour is particularly important when studying:

• absorption by atoms
• emission by atoms
• the photoelectric effect
• detectors
• quantum energy transfer

Neither description by itself captures every aspect of light.

The Electromagnetic Spectrum

Visible light is only a small part of the electromagnetic spectrum.

From lower frequency to higher frequency, the major regions are:

Radio → Microwave → Infrared → Visible → Ultraviolet → X-ray → Gamma ray

As we move in this direction:

frequency increases

wavelength decreases

photon energy increases

In the opposite direction:

frequency decreases

wavelength increases

photon energy decreases

Frequency and Wavelength

For electromagnetic radiation travelling in a vacuum:

c = fλ

where:

c = speed of light = 3.00 × 10⁸ m/s

f = frequency in hertz (Hz)

λ = wavelength in metres (m)

Because c is constant in a vacuum, frequency and wavelength are inversely related.

If frequency increases:

wavelength decreases

If frequency decreases:

wavelength increases

For example, blue light has a higher frequency and shorter wavelength than red light.

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Rearranging the Wave Equation

From:

c = fλ

we can calculate frequency:

f = c/λ

or wavelength:

λ = c/f

These relationships are especially important because photon energy depends directly on frequency.

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Worked Example: Finding Frequency

A beam of light has wavelength:

600 nm

First convert nanometres to metres:

600 nm = 600 × 10⁻⁹ m

= 6.00 × 10⁻⁷ m

Use:

f = c/λ

f = (3.00 × 10⁸)/(6.00 × 10⁻⁷)

f = 5.00 × 10¹⁴ Hz

Therefore, light with a wavelength of 600 nm has a frequency of approximately:

5.00 × 10¹⁴ Hz

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Photon Energy

The energy of an individual photon depends on its frequency.

The relationship is:

E = hf

where:

E = photon energy in joules (J)

h = Planck's constant

f = frequency in hertz (Hz)

Planck's constant is approximately:

h = 6.63 × 10⁻³⁴ J·s

This equation is one of the fundamental relationships in quantum physics.

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What Planck's Equation Tells Us

Since:

E = hf

photon energy is directly proportional to frequency.

If frequency doubles:

photon energy doubles

If frequency triples:

photon energy triples

Therefore:

higher frequency → higher photon energy

and:

lower frequency → lower photon energy

This is why ultraviolet photons carry more energy than visible-light photons, and X-ray photons carry much more energy than radio photons.

Connecting Energy and Wavelength

We have:

E = hf

and:

c = fλ

Since:

f = c/λ

we can substitute this into Planck's equation:

E = hc/λ

Therefore, photon energy is inversely proportional to wavelength.

This gives another extremely important relationship:

shorter wavelength → greater photon energy

longer wavelength → smaller photon energy

So the three quantities are connected:

frequency ↑ → wavelength ↓ → photon energy ↑

and:

frequency ↓ → wavelength ↑ → photon energy ↓

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Comparing Red and Violet Light

Visible red light has a relatively long wavelength.

Visible violet light has a relatively short wavelength.

Therefore:

λ_red > λ_violet

Since frequency and wavelength are inversely related:

f_red < f_violet

Since photon energy increases with frequency:

E_red < E_violet

Therefore, an individual violet photon carries more energy than an individual red photon.

Worked Example: Calculating Photon Energy from Frequency

Suppose light has frequency:

f = 6.00 × 10¹⁴ Hz

Use:

E = hf

E = (6.63 × 10⁻³⁴)(6.00 × 10¹⁴)

E = 3.978 × 10⁻¹⁹ J

Therefore:

E ≈ 3.98 × 10⁻¹⁹ J per photon

Notice how small this energy is.

Individual photons carry extremely small amounts of energy when measured in joules.

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Worked Example: Photon Energy from Wavelength

Suppose light has wavelength:

500 nm

Convert to metres:

500 nm = 5.00 × 10⁻⁷ m

Use:

E = hc/λ

E = [(6.63 × 10⁻³⁴)(3.00 × 10⁸)]/(5.00 × 10⁻⁷)

E = 3.98 × 10⁻¹⁹ J

Therefore, a 500 nm photon carries approximately:

3.98 × 10⁻¹⁹ J

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A Reliable Calculation Method

When given a wavelength and asked for photon energy:

Convert the wavelength to metres.

For example:

450 nm = 4.50 × 10⁻⁷ m

Calculate the frequency.

f = c/λ

Calculate the photon energy.

E = hf

Alternatively, calculate directly using:

E = hc/λ

Both approaches give the same result.

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Units Matter

Common wavelength units include:

1 nm = 1 × 10⁻⁹ m

1 μm = 1 × 10⁻⁶ m

Frequency is measured in:

hertz (Hz)

where:

1 Hz = 1 s⁻¹

Photon energy is usually measured in:

joules (J)

At the atomic scale, energy is also commonly measured in:

electron volts (eV).

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Electron Volts

A joule is relatively large compared with typical atomic energies, so physicists often use the electron volt.

The conversion is approximately:

1 eV = 1.60 × 10⁻¹⁹ J

Suppose a photon has energy:

3.20 × 10⁻¹⁹ J

Convert to eV:

E = (3.20 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹)

E = 2.00 eV

Electron volts are especially convenient when discussing atomic energy levels.

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Worked Example: Energy in Electron Volts

A photon has energy:

4.80 × 10⁻¹⁹ J

Convert to electron volts:

E = (4.80 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹)

E = 3.00 eV

Therefore:

4.80 × 10⁻¹⁹ J = 3.00 eV

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Photons and Atoms

Atoms contain electrons that can occupy specific quantized energy levels.

An electron cannot simply have any energy.

Suppose an atom has:

E₁ = −10 eV

E₂ = −6 eV

The energy difference is:

ΔE = −6 − (−10)

ΔE = 4 eV

To move from E₁ to E₂, the electron must gain the required energy.

A photon with the appropriate energy can be absorbed.

So:

photon absorbed → electron gains energy → electron moves to a higher allowed state

Photon Absorption

During absorption, a photon transfers its energy to an atom.

If the photon energy matches an allowed atomic transition:

E_photon = ΔE

the photon can be absorbed and an electron can move to a higher energy state.

For example:

E₁ = −8 eV

E₂ = −5 eV

Required energy:

ΔE = −5 − (−8)

= 3 eV

A photon with the appropriate 3 eV energy can produce that transition.

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Why the Photon Energy Must Match

Suppose an electron requires:

3 eV

to reach the next allowed state.

A 2 eV photon cannot simply move the electron two-thirds of the way upward.

There is no allowed atomic energy state there.

Likewise, in the simplified model, a photon of 2.9 eV cannot produce a transition requiring exactly 3.0 eV.

This is a consequence of quantized energy levels.

The photon must correspond to an allowed energy difference.

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Photon Emission

An electron in an excited state can move to a lower energy state.

When this happens, energy can be released as a photon.

The emitted photon carries an energy corresponding to the energy difference:

E_photon = |E_final − E_initial|

For example:

E₃ = −2 eV

E₁ = −8 eV

An electron falls:

E₃ → E₁

Energy released:

E_photon = |−8 − (−2)|

= 6 eV

So a 6 eV photon is emitted.

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Absorption and Emission

The two processes can be summarized simply:

Absorption

photon in → electron moves up

Emission

electron moves down → photon out

The same atomic energy difference can therefore be associated with absorption or emission depending on the direction of the electron transition.

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From Photons to Spectral Lines

Suppose an atom has several possible electron transitions.

Each transition has a particular energy difference.

Therefore, each emitted photon has a particular energy.

Since:

E = hf

each has a particular frequency.

And because:

c = fλ

each has a particular wavelength.

This produces the characteristic spectral lines of an element.

The chain is:

energy levels → energy gaps → photon energies → frequencies → wavelengths → spectral lines

Photons and Emission Spectra

When excited electrons fall to lower energy states, photons are emitted.

Only particular energy differences are available.

Therefore, only particular photon energies are emitted.

This produces an emission line spectrum.

Different elements have different energy-level structures, so they produce different patterns of spectral lines.

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Photons and Absorption Spectra

When continuous light passes through suitable matter, atoms can absorb photons whose energies correspond to allowed upward electron transitions.

Those wavelengths are reduced in the transmitted light.

This produces absorption lines.

Again:

photon energy = atomic energy difference

This is why emission and absorption spectroscopy provide information about atomic structure.

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Worked Example: Which Photon Can Be Absorbed?

Suppose an atom has:

E₁ = −9 eV

E₂ = −6 eV

E₃ = −2 eV

An electron begins at E₁.

Possible upward energy differences are:

E₁ → E₂:

ΔE = −6 − (−9)

= 3 eV

E₁ → E₃:

ΔE = −2 − (−9)

= 7 eV

Therefore, photons corresponding to 3 eV or 7 eV could produce these transitions in this simplified model.

A 5 eV photon does not correspond to either of the shown transitions.

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Calculating the Frequency of a Photon from Its Energy

Suppose:

E = 4.00 × 10⁻¹⁹ J

Use:

E = hf

Rearrange:

f = E/h

f = (4.00 × 10⁻¹⁹)/(6.63 × 10⁻³⁴)

f ≈ 6.03 × 10¹⁴ Hz

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Calculating Wavelength from Photon Energy

Using:

f = 6.03 × 10¹⁴ Hz

Use:

λ = c/f

λ = (3.00 × 10⁸)/(6.03 × 10¹⁴)

λ ≈ 4.98 × 10⁻⁷ m

Convert to nanometres:

λ ≈ 498 nm

So a photon with energy 4.00 × 10⁻¹⁹ J has a wavelength of approximately 498 nm.

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Combining the Equations

Three extremely useful equations are:

c = fλ

E = hf

E = hc/λ

They allow us to move between:

wavelength ↔ frequency ↔ photon energy

For example:

If wavelength is known:

λ → f → E

If frequency is known:

f → λ and E

If energy is known:

E → f → λ

These relationships are central to photon calculations.

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Comparing Photons

Consider three photons:

The shortest wavelength is C.

Therefore, C has:

• the highest frequency
• the greatest photon energy

The longest wavelength is A.

Therefore, A has:

• the lowest frequency
• the smallest photon energy

So:

Energy: C > B > A

Frequency: C > B > A

Wavelength: A > B > C

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Photon Energy Across the Electromagnetic Spectrum

Moving from radio waves toward gamma rays:

wavelength decreases

frequency increases

photon energy increases

Therefore:

Radio Waves

Very long wavelengths and relatively low photon energies.

Microwaves

Shorter wavelengths and higher photon energies than radio waves.

Infrared

Higher-frequency photons than microwaves.

Visible Light

Photon energies detected by the human visual system.

Ultraviolet

Higher photon energies than visible light.

X-rays

Very high photon energies.

Gamma Rays

Among the highest photon energies in the electromagnetic spectrum.

Brightness and Photon Energy Are Not the Same Thing

A common mistake is to assume that brighter light must consist of higher-energy photons.

Not necessarily.

Photon energy depends on frequency, not simply brightness.

A bright red beam may contain many relatively lower-energy red photons.

A dim violet beam may contain fewer but individually higher-energy violet photons.

So we must distinguish between:

energy per photon

and:

total energy carried by many photons

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Number of Photons and Total Energy

If each photon has energy E_photon and there are N photons, then:

E_total = N × E_photon

Therefore:

N = E_total/E_photon

This allows us to calculate the number of photons carrying a known total amount of energy.

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Worked Example: Number of Photons

Suppose a pulse of light contains:

2.00 × 10⁻³ J

Each photon has energy:

4.00 × 10⁻¹⁹ J

Then:

N = E_total/E_photon

N = (2.00 × 10⁻³)/(4.00 × 10⁻¹⁹)

N = 5.00 × 10¹⁵ photons

Even a small amount of visible light can contain an enormous number of photons.

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Photons and the Photoelectric Effect

Photons also help explain the photoelectric effect.

When light strikes certain materials, electrons can be ejected.

But an individual photon must have enough energy to overcome the required energy barrier.

Since:

E = hf

the light must have a sufficiently high frequency.

This helped demonstrate that light transfers energy in discrete packets rather than simply delivering energy continuously.

Photons in Everyday Technology

Photon interactions are important in many technologies.

Solar Cells

Photons transfer energy to electrons in semiconductor materials, helping produce an electric current.

LEDs

Electrons undergo energy changes in semiconductor materials and photons are emitted.

Lasers

Stimulated emission produces photons with tightly controlled properties.

Digital Cameras

Photons interact with light-sensitive detector pixels.

Fibre-Optic Communication

Information is transmitted using light.

Medical Imaging

High-energy electromagnetic radiation can interact with matter differently depending on photon energy.

Spectroscopy

Photon absorption and emission reveal information about atoms and molecules.

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Why Planck's Constant Matters

Planck's constant:

h = 6.63 × 10⁻³⁴ J·s

is extremely small.

It sets the scale connecting photon frequency with energy.

For example, a frequency of:

1 Hz

corresponds to a photon energy of only:

6.63 × 10⁻³⁴ J

But electromagnetic frequencies can be enormous.

Visible light commonly has frequencies around:

10¹⁴ Hz

so visible photons have energies around:

10⁻¹⁹ J

per photon.

At atomic scales, these energies are significant.

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A Complete Photon Calculation

Suppose light has wavelength:

400 nm

Convert the wavelength

400 nm = 4.00 × 10⁻⁷ m

Calculate frequency

f = c/λ

f = (3.00 × 10⁸)/(4.00 × 10⁻⁷)

f = 7.50 × 10¹⁴ Hz

Calculate photon energy

E = hf

E = (6.63 × 10⁻³⁴)(7.50 × 10¹⁴)

E = 4.97 × 10⁻¹⁹ J

Convert to electron volts

E = (4.97 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹)

E ≈ 3.11 eV

Therefore, a 400 nm photon has approximately:

f = 7.50 × 10¹⁴ Hz

E = 4.97 × 10⁻¹⁹ J

E = 3.11 eV

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Common Mistakes

Thinking all photons have the same energy

Photon energy depends on frequency.

Thinking longer wavelength means higher energy

The opposite is true:

shorter wavelength → higher energy

Thinking higher frequency means longer wavelength

Frequency and wavelength are inversely related.

Using nanometres directly in equations requiring metres

Convert:

nm → m

before substituting into SI equations.

Forgetting Planck's constant

For photon energy calculations:

h = 6.63 × 10⁻³⁴ J·s

Confusing photon energy with total beam energy

E = hf gives the energy of one photon.

A beam can contain enormous numbers of photons.

Thinking brightness determines photon energy

Photon energy is determined by frequency.

Brightness generally involves the amount of light being received, including the number of photons.

Thinking an atom can absorb any photon

Atomic transitions require suitable photon energies.

Thinking electrons can move partway between quantized levels

Bound atomic electrons occupy allowed energy states.

Forgetting the Direction of Absorption

Absorption generally moves an electron to a higher-energy state.

Forgetting the Direction of Emission

Emission generally accompanies a transition to a lower-energy state.

Confusing Joules and Electron Volts

They are both units of energy, but they are different units:

1 eV = 1.60 × 10⁻¹⁹ J

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Key Terms

Photon — A discrete packet of electromagnetic energy.

Quantum — A discrete amount associated with a physical quantity.

Electromagnetic radiation — Energy transmitted through electromagnetic waves and describable in terms of photons.

Wave-particle duality — The idea that quantum objects such as light display both wave-like and particle-like behaviour.

Frequency (f) — The number of wave cycles passing a point each second.

Wavelength (λ) — The distance between equivalent points on consecutive waves.

Speed of light (c) — The speed of electromagnetic radiation in a vacuum, approximately 3.00 × 10⁸ m/s.

Photon energy (E) — The energy carried by an individual photon.

Planck's constant — The constant connecting photon energy and frequency.

Planck relation — The relationship E = hf.

Electron volt (eV) — A unit of energy commonly used in atomic and particle physics.

Energy level — An allowed energy state of an electron in an atom.

Quantized — Restricted to particular discrete values.

Energy gap — The difference in energy between two allowed states.

Ground state — The lowest allowed energy state.

Excited state — An allowed state above the ground state.

Absorption — Transfer of photon energy to matter.

Emission — Release of electromagnetic energy as photons.

Electron transition — A change between allowed electron energy states.

Emission spectrum — A pattern of wavelengths emitted by matter.

Absorption spectrum — A pattern showing wavelengths reduced because they have been absorbed.

Photoelectric effect — The ejection of electrons from a material following interaction with sufficiently energetic electromagnetic radiation.

Electromagnetic spectrum — The complete range of electromagnetic radiation arranged by frequency or wavelength.

Intensity — A measure related to the rate at which electromagnetic energy is delivered per unit area.

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Key Takeaways

• Light can be described as discrete packets of energy called photons.
• Light also displays wave-like behaviour.
• This combination is part of wave-particle duality.
• Every photon carries energy.
• Photon energy depends on frequency.
• The fundamental photon-energy equation is:

E = hf

• Planck's constant is approximately:

h = 6.63 × 10⁻³⁴ J·s

• Electromagnetic frequency and wavelength are related by:

c = fλ

• The speed of light in a vacuum is approximately:

c = 3.00 × 10⁸ m/s

• Photon energy can also be calculated from wavelength:

E = hc/λ

• Higher frequency means higher photon energy.
• Higher frequency means shorter wavelength.
• Shorter wavelength therefore means higher photon energy.
• Violet photons have more energy per photon than red photons.
• Ultraviolet photons have more energy than visible-light photons.
• X-ray and gamma-ray photons have still greater energies.
• Atomic electrons occupy quantized energy levels.
• Atoms can absorb photons corresponding to allowed upward energy transitions.
• Excited atoms can emit photons during downward energy transitions.
• The photon energy corresponds to the energy difference between the states.
• Quantized energy differences produce characteristic spectral lines.
• Photon energy can be measured in joules or electron volts.
• 1 eV is approximately 1.60 × 10⁻¹⁹ J.
• E = hf describes the energy of one photon, not necessarily the total energy of a light beam.
• A beam can contain an enormous number of photons.
• Photon interactions are central to spectroscopy, LEDs, lasers, solar cells, cameras and many other technologies.

The essential relationships are:

c = fλ

E = hf

E = hc/λ

So remember:

frequency ↑ → wavelength ↓ → photon energy ↑

and:

photon absorbed → electron up

electron down → photon emitted

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Check Your Understanding

1. What is a photon?

2. What does it mean to say that light energy is quantized?

3. Explain what is meant by wave-particle duality.

4. State the equation relating frequency and wavelength.

5. What is the speed of light in a vacuum?

6. State Planck's equation.

7. What is the approximate value and unit of Planck's constant?

8. Explain the relationship between photon energy and frequency.

9. Explain the relationship between frequency and wavelength.

10. Explain the relationship between photon energy and wavelength.

11. Which has more energy per photon: red light or violet light? Explain.

12. Arrange these in order of increasing photon energy:

radio, visible, ultraviolet, X-ray

13. Calculate the frequency of light with wavelength 600 nm.

14. Calculate the energy of one photon from Question 13.

15. Calculate the energy of a photon with frequency:

8.00 × 10¹⁴ Hz.

16. Light has wavelength 450 nm. Calculate its photon energy using:

E = hc/λ.

17. Convert an energy of:

3.20 × 10⁻¹⁹ J

into electron volts.

18. Convert:

5.0 eV

into joules.

19. Explain what happens when an atom absorbs a photon with an appropriate energy.

20. Explain what happens when an excited electron moves to a lower energy state.

21. Why can't an electron simply absorb part of the energy needed to move between two allowed atomic states?

22. An atom has energy levels:

E₁ = −10 eV
E₂ = −7 eV
E₃ = −3 eV

What photon energy is required for:

E₁ → E₂?

23. What photon energy is required for:

E₁ → E₃?

24. What photon energy is emitted during:

E₃ → E₂?

25. Which of the transitions in Questions 22–24 involves the highest-frequency photon?

26. Which involves the shortest-wavelength photon?

27. A pulse contains 1.00 × 10⁻³ J of energy. Each photon has energy 5.00 × 10⁻¹⁹ J. Calculate the number of photons in the pulse.

28. Explain why a bright red light does not necessarily contain photons with more energy than a dim violet light.

29. Describe one technology that depends on interactions between photons and matter.

30. Challenge: A hydrogen atom emits a photon with wavelength:

486 nm

a. Convert 486 nm to metres.
b. Calculate the photon's frequency.
c. Calculate the photon's energy in joules.
d. Convert the energy into electron volts.
e. Would a 656 nm photon have greater or smaller energy? Explain without calculating.
f. Which photon has the greater frequency?
g. Which photon has the longer wavelength?
h. Explain why the emitted photon has a particular energy rather than an arbitrary energy.
i. Relate the photon energy to an electron transition in the atom.
j. Explain how many photons of this energy would be required to carry a total energy of 1.00 J.
k. Explain why the same wavelength could appear as a characteristic line in an emission spectrum.
l. Explain how photon absorption could produce a related absorption line.
m. Summarize the relationship between wavelength, frequency, photon energy and atomic energy levels.