# Momentum of a photon

This thing has troubled me for a long time.. but i think this is eventually starting to get clear to me.

When we say a photon has a momentum 'p', the momentum isn't it's mass times the velocity.. but it can be interpreted as: If the photon [having frequency $f_1$] collides with any other particle, and an elastic collision takes place.. and it's frequency now is $f_2$, then the momentum imparted to the particle is: $$\Delta p = \frac{h}{c}(f_2 - f_1)[/itex] am i right? Last edited: ## Answers and Replies Related Other Physics Topics News on Phys.org Astronuc Staff Emeritus Science Advisor For a photon, momentum, p = E/c = h$\nu$/c, and a change would be given by [tex]\Delta p = \frac{h}{c}{(f_2 - f_1)}$$, where h/c is a constant. I used nu for frequency f.

For a photon, momentum, p = E/c = h$\nu$/c, and a change would be given by

$$\Delta p = \frac{h}{c}{(f_2 - f_1)}$$, where h/c is a constant. I used nu for frequency f.
well.. i actually.. said the same thing. I don't get what you were trying to explain. Can u please explain? thanks.

Astronuc
Staff Emeritus