Energy & Momentum of a Photon: Equation & Calculations

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  • 0:01 Light as a Particle
  • 1:04 Energy of a Photon
  • 2:11 Momentum of a Photon
  • 2:42 Example Calculation
  • 3:55 Lesson Summary
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Lesson Transcript
David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

Expert Contributor
Kathryn Boddie

Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. She has over 10 years of teaching experience at high school and university level.

After watching this lesson, you will be able to explain what wave-particle duality is, provide the equations for the energy and momentum of a photon of light, and use those equations to solve problems. A short quiz will follow.

Light as a Particle

Light is a wave, but that's not all it is. Light acting like a wave was relatively easy to produce. Even in the middle of the 19th century we knew that light could reflect, refract, and diffract. And, as far as we could tell, those effects could only be explained by light being a wave. But then, Albert Einstein noticed something. He showed in his photoelectric effect experiment that light could also behave as a particle.

In that experiment, he shone light onto some metal and found that electrons were ejected. When the light was brighter, more electrons were ejected, but those electrons didn't move any faster - they didn't have any extra energy. However, if he increased the energy of the light, making the light bluer, the electrons did move faster. Suffice it to say, this was not what he expected, and these observations led him to the realization that light must behave both as a wave and a particle; this is called wave-particle duality.

Energy of a Photon

It turns out that light contains energy in discrete packets (or particles) called photons. The amount of energy in those photons is calculated by this equation, E = hf, where E is the energy of the photon in Joules; h is Planck's constant, which is always 6.63 * 10^-34 Joule seconds; and f is the frequency of the light in hertz.

The electrons in the metal were being hit by these photons, which gave them the energy needed to escape. Bluer light has more energy because it has a higher frequency, so the electrons that escaped were moving faster. Brighter light contained more photons, so more electrons left the metal, but those electrons were no faster than with the dimmer light because they could only be hit by one photon of light at a time.

This idea of electrons colliding with photons of light is why these observations can only be explained by treating light as a particle, rather than a wave. A wave can't collide with anything, so if light was purely a wave, then brighter light should lead to higher energy electrons.

Momentum of a Photon

If light contains particles called photons, perhaps they should have momentum like any other particle. In fact, light is both a wave and a particle. So, not only does it have a momentum, it also has a wavelength. We relate these two quantities using something called the de Broglie wavelength: p = h / lambda. This equation says that the momentum of a photon, p, measured in kilogram meters per second, is equal to Planck's constant, h, divided by the de Broglie wavelength of the light, lambda, measured in meters.

Example Calculation

So, how do we use these equations? Let's try an example calculation:

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Additional Activities

Additional Examples of Calculating the Energy and Momentum of a Photon

In the following examples, use the equations for finding the energy and momentum of a photon of light from the lesson to solve. Another equation that may be needed is the equation for the frequency of light given the wavelength: this equation is f=v/lambda, where f is the frequency, v is the velocity of the wave (in other words, the speed of light, which is about 3 x 10^8 m/s) and lambda is the wavelength. After completing the examples, students will be more comfortable using the equations for finding the energy and momentum of a photon of light.


  • Find the energy and momentum of a photon of light in a beam of light with frequency 6.5 x 10^14 Hz.
  • Find the energy and momentum of a photon of light in a beam of light with wavelength 4 x 10^-7 meters.
  • Find the wavelength of the beam and momentum of a photon of light given that the energy of a photon is 3.978 x 10^-19 Joules.


  • To find the energy, we need the formula E=hf, where E is the energy, h is Planck's constant 6.63 x 10^-34 Joule seconds, and f is the frequency. So we have E=(6.63 x 10^-34)(6.5 x 0^14) = 4.3095 x 10^-19 Joules. To find the momentum, we need the formula p= h/lambda where h is Planck's constant and lambda is the wavelength of the light. We do not know the wavelength right now, but we can use the equation f=v/lambda to find it. To find lambda, we have 6.5 x 10^14 = (3 x 10^8)/lambda and so lambda = (3 x 10^8) / (6.5 x 10^14) = 4.62 x 10^-7 meters. Then, the momentum is p=h/lambda = (6.63 x 10^-34) / (4.62 x 10^-7) = 1.435 x 10^-27 kilogram meters per second.
  • To find energy, use E=hf , but first find f using f=v/lambda. f=v/lambda = (3 x 0^8) / (4 x 10^-7) = 7.5 x 10^14. Then the energy is E=(6.63 x 10^-34)(7.5 x 10^14) = 4.9725 x 10^-19 Joules. To find the momentum, use p=h/lambda. p=(6.63 x 10^-34) / (4 x 10^-7) = 1.6575 x 10^-27 kilogram meters per second.
  • To find the wavelength, use E=hf to first find frequency. 3.978 x 10^-19 = (6.63 x 10^-34)f and so f=6 x 10^14 Hz. Then find the wavelength using f=v/lambda. 6 x 10^14=(3 x 10^8)/lambda and so lambda = 5 x 10^-7 meters. To find the momentum, use p=h/lambda. p=(6.63 x 10^-34) / (5 x 10^-7) = 1.326 x 10^-27 kilogram meters per second.


What exactly is wave-particle duality? How does light behave like a wave? How does it behave like a particle?

Guide to Discussion

Students should be able to answer these questions on their own after the lesson - some topics that may come up is how light reflects, refracts, and diffracts like a wave and how light behaves like a particle in how photons collide with electrons.

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