EM Waves & Atoms

Nature of Electromagnetic Waves

Electromagnetic (EM) waves are transverse waves, meaning their oscillations are perpendicular to the direction the wave travels. They do not require a medium and can travel through a vacuum, such as space.

All EM waves travel at the same speed in a vacuum, which is the speed of light, approximately $3.0 \times 10^8 \text{ m/s}$.

The electromagnetic spectrum consists of different types of EM waves, arranged in order of increasing frequency and decreasing wavelength:

• Radio waves
• Microwaves
• Infrared (IR)
• Visible light
• Ultraviolet (UV)
• X-rays
• Gamma rays

Each type of EM wave has a characteristic range of wavelengths and frequencies, but all travel at the same speed in a vacuum.

Wavelength and Frequency Relationship

The wavelength $\lambda$ and frequency $f$ of an EM wave are related by the equation:

$$c = f \lambda$$

where $c$ is the speed of light ($3.0 \times 10^8 \text{ m/s}$).

This means that waves with a higher frequency have a shorter wavelength, and vice versa.

For instance, visible light has wavelengths roughly between 400 nm (violet) and 700 nm (red), while radio waves have much longer wavelengths, often measured in metres.

Example: Calculate the wavelength of an EM wave with a frequency of $6.0 \times 10^{14} \text{ Hz}$ (visible light).

Using $c = f \lambda$, rearranged to $\lambda = \frac{c}{f}$:

$\lambda = \frac{3.0 \times 10^8}{6.0 \times 10^{14}} = 5.0 \times 10^{-7} \text{ m} = 500 \text{ nm}$

This wavelength corresponds to green light in the visible spectrum.

Interaction of EM Waves with Atoms

Atoms have electrons arranged in specific energy levels or shells around the nucleus. These energy levels are quantised, meaning electrons can only exist at certain fixed energies.

When EM radiation interacts with an atom, electrons can absorb energy and move to a higher energy level. This process is called absorption.

Conversely, when electrons fall back to a lower energy level, they emit EM radiation with an energy equal to the difference between the two levels. This is called emission.

The energy of the absorbed or emitted EM radiation corresponds to a specific frequency or wavelength, which is why atoms produce characteristic spectra.

Example: An electron in a hydrogen atom moves from an energy level of $-3.4 \text{ eV}$ to $-13.6 \text{ eV}$. Calculate the energy of the emitted photon in joules and its frequency.

Energy difference $E = |-13.6| - |-3.4| = 10.2 \text{ eV}$.

Convert eV to joules using $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$:

$$E = 10.2 \times 1.6 \times 10^{-19} = 1.63 \times 10^{-18} \text{ J}$$

Use $E = hf$ to find frequency $f$:

$$f = \frac{E}{h} = \frac{1.63 \times 10^{-18}}{6.63 \times 10^{-34}} = 2.46 \times 10^{15} \text{ Hz}$$

This frequency is in the ultraviolet region of the EM spectrum.

EM Spectrum and Atomic Effects

The EM spectrum is divided into ionising and non-ionising radiation based on the energy of the photons.

• Ionising radiation has enough energy to remove electrons from atoms or molecules, creating ions. This includes ultraviolet (UV), X-rays, and gamma rays.
• Non-ionising radiation does not have enough energy to ionise atoms. This includes radio waves, microwaves, infrared, and visible light.

Ionising radiation can cause damage to atoms and molecules, potentially leading to chemical changes or biological harm, such as mutations in DNA.

Non-ionising radiation mainly causes atoms or molecules to vibrate or heat up but does not remove electrons.

Example: Calculate the energy of a photon of ultraviolet light with a wavelength of 200 nm and determine if it is ionising.

Convert wavelength to metres: $200 \text{ nm} = 200 \times 10^{-9} = 2.0 \times 10^{-7} \text{ m}$.

Calculate frequency:

$$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{2.0 \times 10^{-7}} = 1.5 \times 10^{15} \text{ Hz}$$

Calculate photon energy:

$$E = hf = 6.63 \times 10^{-34} \times 1.5 \times 10^{15} = 9.95 \times 10^{-19} \text{ J}$$

Convert to eV:

$$\frac{9.95 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 6.22 \text{ eV}$$

Since ionisation energies are typically around 5 eV or higher, this UV photon is ionising.