# Energy of Photons

Can the energy of a photon be found out by using the most famous equation in all Physics, $E^2=(mc^2)^2 + (pc)^2$ ? If yes, then what will be the $p$ there?

• Hi, welcome to PSE. Wherever you find that equation, it should tell you what p represents as well, but maybe ask yourself: why is the p sometimes included, and sometimes not? – user171879 Oct 28 '17 at 16:22
• I know that momentum is denoted by "p" whereas p=mv. If the mass of an object (e.g. photon) is zero, then the E becomes zero too. But the energy of a photon can't be zero. What am I doing wrong here? – user173644 Oct 28 '17 at 16:29
• You are not reading the equation correctly. It's split between rest mass, (which is always there, at same value) and kinetic energy, which is related to momentum. Look up Relativistic on Wikipedia – user171879 Oct 28 '17 at 16:35
• Will you please give me a example of finding a photon's energy using that equation? – user173644 Oct 28 '17 at 16:38
• Try using: $p=h/\lambda$. BTW, @user17879 is not correct, the second term does not represent kinetic energy. $p=mv$ is only mechanical momentum. Momentum takes other forms when fields are involved, namely field momentum. But there is a very nice discussion in this S.E. answer. – garyp Oct 28 '17 at 19:31

## 2 Answers

The momentum of a photon is h/ℷ where ℷ is the wavelength and h is Planck's constant. The energy of a photon is hc/ℷ. In the equation you wrote, the rest mass of a photon is zero. Leaving you with E = pc.

$E$ is the total energy of the particle, $m$ is mass, $p$ is momentum and $c$ is the speed of light in vacuum. Photons are massless, so $m=0$, therefore the energy of the photon is: $$E=pc$$

• E=pc=(mv)c assuming m=0, E=(0*v)c=0 . But we all know, E≠0 😲 – user173644 Oct 28 '17 at 16:49
• No, momentum is not just $mv$. Momentum is $p=mv\gamma$, where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ and it is called the Lorentz factor. – Andrei Geanta Oct 28 '17 at 16:53
• Thank you Arthur for your answer. It was giving me a headache since the last few hours. – user173644 Oct 28 '17 at 16:55