Electromagnetic Radiation

Introduction

Atomic spectra are measured as light wavelengths or frequencies, while atomic transitions are described as energy changes. Electromagnetic-radiation equations connect these descriptions. Care with units and the medium of travel is essential: a wavelength in nanometres must become metres before SI constants are applied.

Core explanation

Electromagnetic radiation includes radio waves, infrared, visible light, ultraviolet, X-rays, and gamma rays. In vacuum, every form travels at speed c = 299,792,458 m s⁻¹. Wavelength λ and frequency ν obey c = λν. Wavelength is distance per cycle; frequency is cycles per second. Their product has units of speed. Longer wavelength at fixed propagation speed means lower frequency.

A photon has energy E = hν = hc/λ, where Planck's constant h is exactly 6.62607015 × 10⁻³⁴ J s in SI. The frequency relation makes high-frequency photons more energetic. The inverse wavelength relation makes short-wavelength photons more energetic. This connects a spectral line's measured wavelength to the energy change of the atom that emitted or absorbed the photon.

In a material medium, light's speed and wavelength can differ from their vacuum values while frequency remains set by the source across a boundary. Introductory atomic-spectroscopy calculations commonly use vacuum wavelengths and c. State that context when applying c = λν directly. The photon energy relation E = hν continues to use frequency.

Visible light has wavelengths of a few hundred nanometres. One nanometre equals 10⁻⁹ m, so 500 nm is 5.00 × 10⁻⁷ m. Its frequency in vacuum is about 6.00 × 10¹⁴ s⁻¹. Multiplying by h gives roughly 3.97 × 10⁻¹⁹ J per photon. A mole of such photons would carry far more energy, found by multiplying photon energy by the Avogadro constant.

Do not confuse light intensity with photon energy. A more intense monochromatic beam may have more photons arriving each second without increasing the energy of each photon. Changing frequency changes the energy per photon. This distinction helps explain threshold behavior in the photoelectric effect.

Step-by-step reasoning

1. Convert supplied wavelength to metres or frequency to s⁻¹. 2. Use c = λν to find the missing wave quantity in vacuum. 3. Use E = hν or E = hc/λ for energy per photon. 4. Estimate whether a shorter wavelength gave a higher energy.

Visual explanation

Draw two waves across the same horizontal distance. The shorter-wavelength wave completes more cycles and has higher frequency when both propagate at the same speed.

Real-world analogy

If a conveyor moves at fixed speed, closely spaced marks pass a sensor more often than widely spaced marks. Wavelength is mark spacing and frequency is the passage rate.

Real-world example

An emission spectrometer measures wavelengths from excited atoms. Converting one spectral line's wavelength to photon energy connects the instrument reading with an atomic-state energy difference.

Why?

Why is λν a speed? A wave advances one wavelength per oscillation and undergoes ν oscillations per second, giving wavelength times oscillations per second as distance per second.

Common misconception

“Brighter light always has more energetic photons.” At a fixed frequency, brightness can increase because more photons arrive; each photon still has energy hν.

Worked example

Find ν and E for vacuum light of λ = 600 nm. Convert λ = 6.00 × 10⁻⁷ m. Then ν = c/λ = 2.998 × 10⁸/6.00 × 10⁻⁷ ≈ 5.00 × 10¹⁴ s⁻¹. Photon energy is hν ≈ (6.626 × 10⁻³⁴)(5.00 × 10¹⁴) = 3.31 × 10⁻¹⁹ J.

Quick check

1. At fixed vacuum speed, what happens to frequency when wavelength halves? Answer: Frequency doubles because c = λν.

Exam focus

Convert nm to m and distinguish energy per photon from energy for a mole of photons. State vacuum conditions when using the standard c value.

Advanced insight

Spectral wavenumber, commonly written as inverse wavelength in cm⁻¹, is proportional to photon energy. Spectroscopists use it because differences in wavenumber can be related conveniently to differences between atomic or molecular energy levels.

Summary

In vacuum, wavelength and frequency satisfy c = λν. Photon energy is hν or hc/λ, so shorter wavelengths correspond to higher frequency and higher energy per photon.

Practice questions

1. Convert 450 nm to metres. Answer: 4.50 × 10⁻⁷ m. 2. Which photon has greater energy, 400 nm or 800 nm light? Answer: The 400 nm photon, since E is inversely proportional to wavelength. 3. If frequency is 3.00 × 10¹⁴ s⁻¹, what is vacuum wavelength approximately? Answer: c/ν ≈ 1.00 × 10⁻⁶ m, or 1000 nm.