Electromagnetics deals with the interplay between the photons that constitute light waves and electrons, the particles with which these light waves interact. Specifically, light waves have certain universal properties, including a constant speed, and also emit energy, albeit often on a very small scale.

The fundamental unit of energy in physics is the Joule, or Newton-meter. The speed of light in a vaccum is 3 × 10^{8} m/sec, and this speed is a product of any light wave's frequency in Hertz (the number of light waves, or cycles, per second) and the length of its individual waves in meters. This relationship is normally expressed as:

c = ν × λ

Where ν, the Greek letter nu, is frequency and λ, the Greek letter lambda, represents wavelength.

## Sciencing Video Vault

Meanwhile, in 1900, the physicist Max Planck proposed that the energy of a light wave is directly to its frequency:

E = h × ν

Here, h, fittingly, is known as Planck's constant and has a value of 6.626 × 10^{-34} Joule-sec.

Taken together, this information allows for calculating frequency in Hertz when given energy in Joules and conversely.

## Step 1: Solve for Frequency in Terms of Energy

Because c = ν × λ, ν = c/ λ.

But E = h × ν, so

E = h × (c/ λ).

## Step 2: Determine the Frequency

If you get ν explicitly, move on to Step 3. If given the λ , divide c by this value to determine ν.

For example, if λ = 1 × 10^{-6} m (close to the visible light spectrum), ν = 3 × 10^{8}/ 1 × 10^{-6} m = 3 x 10^{14} Hz.

## Step 3: Solve for Energy

Multiply ν Planck's constant, h, by ν to get the value of E.

In this example, E = 6.626 × 10^{-34} Joule-sec × (3 × 10^{14} Hz) = 1.988 x 10^{-19} J.

## Tip

Energy on small scales is often expressed as electron-Volts, or eV, where 1 J = 6.242 × 10^{18} eV. For this problem, then, E = (1.988 × 10^{-19} )(6.242 × 10^{18}) = 1.241 eV.