# Momentum of photon proof

1.photon has no mass . so m=0.hence, p=mv=0.by doing some calculations , we can get that
p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?

Doc Al
Mentor
1.photon has no mass . so m=0.hence, p=mv=0.by doing some calculations , we can get that
p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?
p = mv is only useful for massive particles at relatively low speeds. For photons, you'll need the relativistic energy-momentum relation: Energy–momentum relation

UltrafastPED
Gold Member
There is more than one type of momentum - p=mv is the mechanical form.

Momentum for electromagnetic radiation is determined by the momentum that it can impart to a mechanical system. The derivation follows from energy transport properties of radiation derived from Maxwell's equations - look up "Poynting Vector". The electromagnetic radiation momentum is found to be p=E/c; this relation also holds for the photon.

The equation p=h/lambda is the de Broglie relation; it is "derived" from the Planck relation (E=h*f), then divide by c to get E/c=p=h*f/c=h/lambda. Of course this is not a derivation - it merely shows that the two are algebraicly consistent. The experimental proof of the de Broglie relation can be seen experimentally: x-rays and electrons both give diffraction patterns in accordance with the above.

For more detail see http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html

only rest mass of photon is zero

momentum of photon can be calculated by

p = E/c

where E is energy of photon
don't ask me any proof please because i'm an tenthee! studying for iit and was told this on a chemistry lecture and would study more on it guess in my research later
further proofs are encouraged and needed by me as my teacher said it would come later on:tongue:

and a kind request to UltrafastPED pls do no\umerically in separate lines or it's feels scrambled

UltrafastPED