Light as a Wave: Frequency, Wavelength, Energy
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with basic algebra and unit conversions, and have a general idea of what an atom is from the previous lecture on isotopes.
Why this matters
Understanding light as a wave provides the essential vocabulary for describing how atoms interact with energy.
Before we can explain why electrons jump between energy levels or why specific elements emit specific colours, we must define the properties of the radiation itself. Light is not just "visible"; it is a broad spectrum of electromagnetic waves, each characterised by a specific wavelength, frequency, and energy. These three quantities are mathematically linked, and mastering their relationships is the prerequisite for all subsequent quantum mechanics topics in this unit.
The Wave Nature of Light
Imagine a rope stretched horizontally. If you shake one end up and down, a wave travels along the rope. This wave has a repeating pattern. The distance between two consecutive peaks is called the wavelength, usually denoted by the Greek letter lambda (\(\lambda\)). The number of peaks that pass a fixed point per second is the frequency, denoted by nu (\(\nu\)).
Light behaves similarly, but instead of a physical rope, it is an oscillating electric and magnetic field. Unlike sound, which requires a medium, light can travel through a vacuum. The speed of light in a vacuum, \(c\), is a fundamental constant of nature. It is approximately \(3.00 \times 10^8\ \mathrm{m\,s^{-1}}\). This speed is incredibly fast, which is why we perceive light as instantaneous in everyday life. The relationship between speed, wavelength, and frequency is intuitive: if the waves are long (large \(\lambda\)), fewer of them can pass a point per second (low \(\nu\)). If the waves are short (small \(\lambda\)), more pass per second (high \(\nu\)).
The Wave Equation
The fundamental relationship connecting the speed of light, wavelength, and frequency is given by the wave equation. Since speed is distance per time, and wavelength is the distance of one cycle while frequency is the number of cycles per second, their product gives the speed.
Here, \(c\) is the speed of light (\(3.00 \times 10^8\ \mathrm{m\,s^{-1}}\)), \(\lambda\) is the wavelength in metres (\(\mathrm{m}\)), and \(\nu\) is the frequency in hertz (\(\mathrm{Hz}\) or \(\mathrm{s^{-1}}\)). This equation implies an inverse relationship between wavelength and frequency. If you double the wavelength, the frequency is halved. This is why radio waves, which have very long wavelengths (metres to kilometres), have low frequencies, while X-rays, with extremely short wavelengths (picometres), have very high frequencies.
Energy of a Photon
While the wave equation describes the geometry of light, it does not account for the energy carried by the light. In the early 20th century, Max Planck and Albert Einstein established that light energy is quantised into discrete packets called photons. The energy of a single photon is directly proportional to its frequency.
In this equation, \(E\) is the energy in joules (\(\mathrm{J}\)), and \(h\) is Planck's constant, a fundamental physical constant with a value of \(6.626 \times 10^{-34}\ \mathrm{J\,s}\). Because energy is proportional to frequency, and frequency is inversely proportional to wavelength, we can substitute \(\nu = c/\lambda\) to express energy in terms of wavelength:
This form is often more useful in chemistry because wavelengths are easier to measure than frequencies. It shows that energy is inversely proportional to wavelength. Short-wavelength light (like ultraviolet or X-rays) carries high energy per photon, while long-wavelength light (like radio waves) carries low energy per photon. This distinction is crucial for understanding chemical reactions, as breaking chemical bonds requires specific energy thresholds.
The Electromagnetic Spectrum
The electromagnetic spectrum is the continuous range of all possible wavelengths of electromagnetic radiation. It is divided into regions based on wavelength and frequency, though these boundaries are somewhat arbitrary. From longest wavelength (lowest energy) to shortest wavelength (highest energy), the main regions are:
- Radio waves: \(\lambda > 1\ \mathrm{m}\). Used for communication.
- Microwaves: \(1\ \mathrm{mm} < \lambda < 1\ \mathrm{m}\). Used in radar and cooking.
- Infrared (IR): \(700\ \mathrm{nm} < \lambda < 1\ \mathrm{mm}\). Associated with heat and molecular vibrations.
- Visible light: \(400\ \mathrm{nm} < \lambda < 700\ \mathrm{nm}\). The only part of the spectrum detectable by the human eye.
- Ultraviolet (UV): \(10\ \mathrm{nm} < \lambda < 400\ \mathrm{nm}\). Higher energy than visible light; causes sunburn.
- X-rays: \(0.01\ \mathrm{nm} < \lambda < 10\ \mathrm{nm}\). Penetrating radiation used in medical imaging.
- Gamma rays: \(\lambda < 0.01\ \mathrm{nm}\). Highest energy, originating from nuclear processes.
Note the units: visible light is typically measured in nanometres (\(\mathrm{nm}\), where \(1\ \mathrm{nm} = 10^{-9}\ \mathrm{m}\)), while radio waves are measured in metres. When calculating energy, you must always convert wavelengths to metres to be consistent with SI units.
A common misconception is that "intensity" and "energy per photon" are the same thing. They are not. Intensity depends on the number of photons (and their energy), while the energy of a single photon depends only on its frequency. A dim red laser has low intensity but each photon has low energy; a bright red laser has high intensity, but each photon still has the same low energy. Only by changing the colour (frequency) do you change the energy per photon. This distinction is critical when discussing the photoelectric effect in the next lecture. C
Worked example
Problem. A laser emits light with a wavelength of \(532\ \mathrm{nm}\). Calculate the frequency of this light and the energy of a single photon. Use \(c = 3.00 \times 10^8\ \mathrm{m\,s^{-1}}\) and \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\).
- Convert wavelength to metres: \(532\ \mathrm{nm} = 532 \times 10^{-9}\ \mathrm{m} = 5.32 \times 10^{-7}\ \mathrm{m}\).
- Calculate frequency: Rearrange \(c = \lambda \nu\) to \(\nu = c / \lambda\). \[ \nu = \frac{3.00 \times 10^8\ \mathrm{m\,s^{-1}}}{5.32 \times 10^{-7}\ \mathrm{m}} = 5.64 \times 10^{14}\ \mathrm{s^{-1}} \]
- Calculate energy: Use \(E = h \nu\). \[ E = (6.626 \times 10^{-34}\ \mathrm{J\,s}) \times (5.64 \times 10^{14}\ \mathrm{s^{-1}}) = 3.74 \times 10^{-19}\ \mathrm{J} \]
Units check. For frequency: \(\mathrm{m\,s^{-1}} / \mathrm{m} = \mathrm{s^{-1}}\) (correct). For energy: \(\mathrm{J\,s} \times \mathrm{s^{-1}} = \mathrm{J}\) (correct).
Three quick tests
- Which has higher energy: red light (\(\lambda \approx 700\ \mathrm{nm}\)) or blue light (\(\lambda \approx 450\ \mathrm{nm}\))?
Answer
Blue light. Energy is inversely proportional to wavelength, so shorter wavelength means higher energy. - If the frequency of a wave is doubled, what happens to its wavelength?
Answer
It is halved. Wavelength and frequency are inversely proportional (\(\lambda \propto 1/\nu\)). - Calculate the wavelength of light with a frequency of \(1.00 \times 10^{15}\ \mathrm{Hz}\).
Answer
\(\lambda = c/\nu = (3.00 \times 10^8) / (1.00 \times 10^{15}) = 3.00 \times 10^{-7}\ \mathrm{m} = 300\ \mathrm{nm}\).
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.