# Deriving photon energy equation using $E = mc^2$ and de Broglie wavelength

So I was learning de Broglie wavelength today in my physics class, and I started playing around with it. I wondered if it was possible to calculate the energy of a light wave given its wavelength and speed. After rearranging a bit, I plugged it into $$E = mc^2$$, and realized I had found the equation for the energy of a photon that I learned at the beginning of my quantum mechanics unit, $$E = hf$$.

$$\lambda = \frac{h}{p}$$

$$p = \frac{h}{\lambda}$$

$$E = mc^2$$

$$E = \frac{p}{c}\cdot c^2$$

$$E = \frac{hc}{\lambda} = hf$$

I have a few questions about this. Firstly, I do not understand how it can make sense to do $$\frac{p}{c}$$ in this context, because, as I understand it, light has no mass. How can I come to $$E = hf$$ using the mass of a massless object?

Secondly, as I was doing those steps, I thought I would be calculating the energy of the entire light ray. I now realize I was finding the energy of a single photon. In hindsight, this makes sense, because the energy of the entire light ray must depend on some length value, correct?

This led me to two other questions: do light rays have some finite length? how do I calculate the energy of a light ray, not just a single photon?

• Jan 29 at 5:41

Firstly, I do not understand how it can make sense to do p/c in this context, because, as I understand it, light has no mass.

That's correct. Light has no mass so using $$m=\frac{p}{c}$$ is technically incorrect.

How can I come to E=hf using the mass of a massless object?

From the relation $$E^2=p^2c^2+m^2c^4$$ given that the mass of a photon is indeed zero, then $$E=pc\rightarrow E=\frac{hc}{\lambda}\ \ \ \text{since }\ \ \ p=\frac{h}{\lambda}$$ You also know that $$c=f\lambda$$ so $$E=hf$$

the energy of the entire light ray must depend on some length value, correct?

Yes, it depends on the wavelength, $$\lambda$$. It's not clear what you mean by "entire". The energy of a light ray is characterized by its frequency, and therefore wavelength. Note that a single photon is a particle, and the "light ray" classical wave character comes from many photons.

• Succinctly answered. I think the OP may be struggling on how the energy of a light ray including many photons may be defined if at all reasonable. Jan 29 at 2:59
• That was my impression too. Thanks Jan 29 at 3:19
• I think I am having difficulty understanding what light is in general :D I find it hard to conceptualize how its energy is only dependent on its frequency. I imagine it as some line/beam, and imagine that the absolute length of that entire line must have some effect on energy, not the wavelength. But, this response answers all my questions, so I marked as accepted :)
– dwib
Jan 29 at 3:26
• @dwib please note that the photon is a point particle , i.e. has no space extention . in the standard model of particle physics. The frequency does not belong to the photon, only the energy, and spin. The energy of the individual photon has been identified with hν. see this simple experiment on how single photons build up the electromagnetic wave. physics.stackexchange.com/questions/388026/… Jan 29 at 5:29