Relate electromagnetic radiation, photons and E = hf

Electromagnetic radiation ranges from low-frequency radio waves to high-frequency gamma rays. Spectroscopy uses selected regions because different molecular energy gaps match different photon energies.

GCE A-Level H3 Chemistry 9813-2027

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Learning objectives

Relate electromagnetic radiation, photons and E = hf

Place the radiation on the spectrum

Move across the spectrum from radio → microwave → infrared → visible → ultraviolet → X-ray → gamma. Frequency and photon energy increase in that direction, while wavelength decreases.

Waves, photons and energy

For electromagnetic radiation in vacuum, c = λf, where c is 3.00 × 10⁸ m s⁻¹, λ is wavelength in metres and f is frequency in s⁻¹ or Hz.

One photon carries energy E = hf, where h is 6.63 × 10⁻³⁴ J s. Combining the equations gives E = hc/λ, so shorter wavelength means higher frequency and higher photon energy.

Useful approximate wavelength ranges are infrared 700 nm to 1 mm, visible 400–700 nm and ultraviolet 10–400 nm. Boundaries are conventional, so use ranges supplied in a question when they differ slightly.

Why different spectroscopies use different regions

Radiofrequency photons have relatively low energy and can match small nuclear-spin energy gaps in NMR. Infrared photons can match vibrational gaps, while UV/visible photons can match larger electronic gaps.

A more intense beam contains more photons per unit time or area; it does not make each photon more energetic. The energy of each photon is fixed by frequency.

Always convert wavelength to metres before using SI constants. The prefixes are 1 nm = 10⁻⁹ m and 1 μm = 10⁻⁶ m.

Worked example: energy of a 500 nm photon

Convert first: 500 nm = 5.00 × 10⁻⁷ m. Then E = hc/λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸)/(5.00 × 10⁻⁷).

E = 3.98 × 10⁻¹⁹ J per photon. Its frequency is f = c/λ = 6.00 × 10¹⁴ Hz, which lies in the visible region.

Try this

Calculate the energy of one photon with wavelength 500 nm.

Check your answer

E = hc/λ = 3.98 × 10⁻¹⁹ J after converting 500 nm to 5.00 × 10⁻⁷ m.

Practise comparing two wavelengths

When only a comparison is needed, use E ∝ 1/λ. You do not need to calculate both energies.

A 250 nm photon has half the wavelength of a 500 nm photon, so it has twice the frequency and twice the energy.

Try this

Compare the energy of a 250 nm photon with that of a 500 nm photon.

Check your answer

The 250 nm photon has twice the energy because photon energy is inversely proportional to wavelength.

Calculate from frequency

Frequency in Hz can be substituted directly into E = hf. Check that J s multiplied by s⁻¹ gives joules.

Use a sensible number of significant figures and state that the result is for one photon unless the question asks for a mole of photons.

Try this

Find the energy of a photon with frequency 3.00 × 10¹³ Hz.

Check your answer

E = hf = (6.63 × 10⁻³⁴)(3.00 × 10¹³) = 1.99 × 10⁻²⁰ J.

Common mistake: intensity changes photon energy

Increasing intensity at a fixed frequency increases the number of photons, not the energy carried by each photon.

Another frequent error is to reverse the wavelength trend. Because E = hc/λ, a longer wavelength carries less energy per photon.

Try this

Correct: ‘Brighter red light has more energetic photons than dim ultraviolet light.’

Check your answer

Brightness concerns photon number. Each ultraviolet photon has more energy than each red photon because ultraviolet has higher frequency and shorter wavelength.

Check your understanding

Be ready to order spectrum regions, use c = λf and E = hf, convert prefixes correctly and distinguish photon energy from beam intensity.

Next, compare the electronic, vibrational, rotational and nuclear-spin energy gaps that these different radiation regions can probe.

Try this

State the linked trend among wavelength, frequency and photon energy.

Check your answer

As wavelength decreases, frequency increases and photon energy increases.