The Electromagnetic Spectrum for Chemists

Introduction

Radio waves, microwaves, infrared, visible light, ultraviolet and X-rays are all the same thing: electromagnetic radiation. They differ only in wavelength, frequency and therefore photon energy. Chemists use different parts of this spectrum for different jobs, so fluency with the quantities that describe radiation — and the equations linking them — is the foundation of every spectroscopic technique that follows.

Core explanation

Waves and particles. Electromagnetic radiation behaves both as a wave and as a stream of particles called photons . The wave picture gives us wavelength and frequency; the particle picture gives us the energy carried by each photon.

Wavelength and frequency. All electromagnetic radiation travels through a vacuum at the speed of light, c = 3.00 × 10⁸ m s⁻¹. Wavelength (λ) and frequency (f) are linked by:

c = fλ

Because c is constant, a shorter wavelength means a higher frequency.

Photon energy. The energy of one photon is proportional to frequency (Planck's relation):

E = hf = hc/λ

where h = 6.63 × 10⁻³⁴ J s is the Planck constant. High frequency (short wavelength) radiation carries high-energy photons. To get energy per mole of photons, multiply by the Avogadro constant, 6.02 × 10²³ mol⁻¹.

Wavenumber. IR spectroscopists use wavenumber , the number of waves per centimetre:

wavenumber = 1/λ (in cm⁻¹)

Wavenumber is directly proportional to energy, which makes it convenient: a peak at 3000 cm⁻¹ corresponds to twice the photon energy of a peak at 1500 cm⁻¹.

Regions of the spectrum used by chemists (increasing energy):

Region Typical wavelength Molecular process Technique --- --- --- --- Radio about 1 m Nuclear spin flips in a magnet NMR Microwave about 1 mm – 10 cm Molecular rotation Rotational spectroscopy Infrared about 2.5 – 25 μm Bond vibrations IR Visible 400 – 700 nm Outer-electron transitions Colorimetry Ultraviolet about 200 – 400 nm Electron transitions, bond breaking UV spectroscopy X-ray about 0.01 – 10 nm Inner-electron transitions, diffraction X-ray crystallography

The IR range 2.5 – 25 μm corresponds to wavenumbers of 4000 – 400 cm⁻¹, the range shown on most IR spectra.

Formulae

c = fλ; E = hf = hc/λ; wavenumber = 1/λ. Constants: c = 3.00 × 10⁸ m s⁻¹, h = 6.63 × 10⁻³⁴ J s, N A = 6.02 × 10²³ mol⁻¹.

Step-by-step reasoning

To find the energy per mole of photons from a wavelength:

1. Convert the wavelength into metres. 2. Calculate the energy of one photon: E = hc/λ. 3. Multiply by the Avogadro constant to get J mol⁻¹. 4. Divide by 1000 to express the answer in kJ mol⁻¹.

Visual explanation

Draw a long horizontal arrow. At the left, write "radio — long wavelength, low frequency, low energy"; at the right, "gamma — short wavelength, high frequency, high energy". Mark the narrow visible band near the middle, with red at the longer-wavelength side and violet at the shorter-wavelength side.

Real-world analogy

Think of photons as coins of different values. Radio photons are small coins; ultraviolet photons are large notes. A molecule can only "sell" a change of state for an exact price, so only the right coin will be accepted.

Real-world example

Sunscreens contain compounds that absorb UV-A and UV-B radiation. UV photons carry enough energy to damage DNA, whereas visible photons generally do not, which is why protection targets the ultraviolet region rather than visible light.

Why?

Why does UV radiation cause chemical damage while radio waves pass harmlessly through the body? A UV photon at 250 nm carries about 480 kJ mol⁻¹ — comparable to many covalent bond enthalpies — whereas a radio photon carries a tiny fraction of a joule per mole, far too little to break bonds.

Common misconception

"More intense radiation always means higher-energy photons." Intensity is the number of photons per second; energy per photon depends only on frequency. A bright red lamp emits many low-energy photons, while a weak UV lamp emits fewer but more energetic ones.

Worked example

Question: Calculate the energy, in kJ mol⁻¹, of photons with a wavelength of 500 nm.

Reasoning: λ = 500 × 10⁻⁹ m. E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (5.00 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J per photon. Per mole: 3.98 × 10⁻¹⁹ × 6.02 × 10²³ = 2.40 × 10⁵ J mol⁻¹.

Answer: About 240 kJ mol⁻¹.

Quick check

1. What is the wavenumber, in cm⁻¹, of infrared radiation with a wavelength of 5.0 μm? Answer: 5.0 μm = 5.0 × 10⁻⁴ cm, so wavenumber = 1 ÷ (5.0 × 10⁻⁴) = 2000 cm⁻¹.

Exam focus

Unit conversions cause most lost marks: nanometres to metres (× 10⁻⁹) and micrometres to centimetres (× 10⁻⁴). Always state whether an energy is per photon or per mole, and remember that frequency, wavenumber and energy rise together while wavelength falls.

Advanced insight

Frequency, not wavelength, is the fundamental property of a photon: when light enters glass or water its speed and wavelength decrease, but its frequency and energy stay the same. This is why precise spectroscopic data are often quoted as frequencies or vacuum wavenumbers.

Summary

Electromagnetic radiation is described by wavelength, frequency, wavenumber and photon energy, linked by c = fλ and E = hf = hc/λ. Wavenumber (1/λ, in cm⁻¹) is proportional to energy and is used in IR. Chemists use radio waves for NMR, infrared for vibrations and UV-visible radiation for electronic transitions, in order of increasing photon energy.

Practice questions

1. Calculate the frequency of radiation with a wavelength of 600 nm. Answer: f = c/λ = 3.00 × 10⁸ ÷ 6.00 × 10⁻⁷ = 5.00 × 10¹⁴ Hz. 2. Which carries more energy per photon: IR at 1700 cm⁻¹ or IR at 3400 cm⁻¹? By what factor? Answer: 3400 cm⁻¹, by a factor of two, because energy is proportional to wavenumber. 3. Arrange radio waves, ultraviolet and infrared in order of increasing photon energy. Answer: Radio waves < infrared < ultraviolet. 4. Convert a wavenumber of 1000 cm⁻¹ into a wavelength in micrometres. Answer: λ = 1/1000 cm = 1.0 × 10⁻³ cm = 10 μm.