The energy of a photon

We have seen that the electromagnetic wave cannot have an arbitrarily small strength: it must have a minimum energy \( E \) which corresponds to one photon. But what is the value of \( E \)?

This question as answered by Max Planck. He gave the answer in terms of the frequency of the electromagnetic wave.

What is the frequency of a wave? Imagine standing at one place as the wave moves past you. The frequency is the number of up and down oscillations you see each second. The frequency is usually denoted by the symbol \( \nu.\)

Planck found that the energy of a photon is proportional to the frequency; more precisely

where \( h \) is the Planck constant.

Expressing \( E \) in terms of the wavelength

The frequency \( \nu \) of an electromagnetic wave is related to the wavelength \( \lambda \) in a simple way

where \( c \) is the speed of light

\[\require{color}\colorbox{yellow} {$c ~=~ 3 \times 10^{8} \,\, m/s $}\]

Thus the energy of the photon can be written as

We see that photons with long wavelength have low energy and photons with short wavelength have high energy.

Understanding the value of \( E \)

While the above relation gives the energy of a photon for any wavelength \( \lambda \), we would like to get some feeling for what such energies can do.

1. Visible light has a wavelength \( \lambda \sim 5\times 10^{-7} \, m \). A photon of this wavelength can hit the electron in an atom, and cause this electron to jump to a higher energy level in the atom. Conversely, when electrons jump from higher energy levels to lower energy levels in an atom, they typically emit photons of visible light.
2. X-rays have a wavelength \( \lambda \sim 10^{-9} \, m \). Photons of this wavelength can hit the electron in an atom, and knock it completely out of the atom. For this reason X-rays are capable of damaging human tissue.
3. Electromagnetic waves with wavelengths smaller than about \( \lambda \sim 10^{-9} \, m \) are called gamma rays. Gamma rays from cosmic sources can have wavelengths as small as \( \lambda \sim 10^{-19} \, m \). A single photon of such a cosmic gamma ray can collide with air molecules in the atmosphere and produce a 'shower' of millions of new particles, that spread over several square kilometers.