- #1

- 15

- 0

p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the conversation discusses the concept of momentum for photons, which have no rest mass. The equation p = mv is only applicable for massive particles at low speeds, while for photons, their momentum is determined by the energy they can impart to a mechanical system. The de Broglie relation, p=h/lambda, is derived from the Planck relation and is experimentally proven through diffraction patterns. For further understanding, it is encouraged to study electromagnetic field theory.

- #1

- 15

- 0

p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?

Physics news on Phys.org

- #2

Mentor

- 45,530

- 2,126

p = mv is only useful for massive particles at relatively low speeds. For photons, you'll need the relativistic energy-momentum relation: Energy–momentum relationKehsibashok said:

p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?

- #3

Science Advisor

Gold Member

- 1,914

- 216

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

- #4

- 70

- 0

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

- #5

Science Advisor

Gold Member

- 1,914

- 216

I prefer inline formulas, especially when there are chains of implication. This makes for more concise, "unscrambled" chains of logic.

BTW, for the proof of p=E/c see any text on electromagnetic field theory (upper level undergraduate physics); you will arrive at the Poynting vector sometime in the second semester!

The momentum of a photon is given by the formula p = E/c, where E is the energy of the photon and c is the speed of light. This means that photons have momentum because they have energy and travel at the speed of light.

The momentum of a photon can be calculated using the formula p = E/c, where E is the energy of the photon and c is the speed of light. This calculation is based on the fact that photons have energy and travel at the speed of light.

Yes, a photon always has momentum because it has energy and travels at the speed of light. According to the laws of physics, any object with energy and speed must have momentum.

Yes, the momentum of a photon can be changed by changing its energy or direction of travel. This can happen through interactions with other particles or objects, such as when a photon is absorbed or reflected off a surface. However, the speed of light is constant, so the magnitude of the change in momentum will always be equal to the change in energy.

The momentum of a photon is significant because it helps us understand the behavior of light and other electromagnetic radiation. It also plays a role in various phenomena, such as the photoelectric effect and the Compton effect, which have important implications in fields like quantum mechanics and atomic physics.

Share:

- Replies
- 6

- Views
- 315

- Replies
- 5

- Views
- 598

- Replies
- 30

- Views
- 1K

- Replies
- 4

- Views
- 600

- Replies
- 52

- Views
- 2K

- Replies
- 3

- Views
- 900

- Replies
- 1

- Views
- 797

- Replies
- 29

- Views
- 1K

- Replies
- 7

- Views
- 1K

- Replies
- 2

- Views
- 646