How do you derive the units for momentum for photons

In summary, the conversation is discussing how to find the magnitude of momentum for a photon. It is suggested to use the equation p = E/c, but the units do not match up. The conversation then explores different ways to express energy in terms of basic units to simplify the equation. It is ultimately concluded that the equations do have the same units and it is possible to express them in different ways using basic formulas.
  • #1
TheCelt
24
5
Homework Statement
Units for photon momentum confuse me
Relevant Equations
p=hf/c = E/c
So if i have a photon of some energy and i want to find the magnitude of the momentum, i can get the right answer but the units don't make sense.

So i derive p = E/c since i know the energy of the photon and i used f=E/h and substituted this into p=hf/c

This means for units of the equation p = E/c i get:

kg m/s = J / (m/s) = J s/m

This is confusing me since I've always been told to check my units but these equations don't have the same units, so I don't know if i have done something wrong, or momentum has different units applied for photons? Hope some one can explain what's going on here with the units.

Thanks
 
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  • #2
OK, so can you express the Energy units in terms of the basic units kg, m, sec instead of Joules? Then you can simplify the left side of your last equation.

Hint: I thought it was useful to think of Energy=Force*Distance, but there are other ways too. Perhaps the formula for kinetic energy?
 
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Likes hutchphd
  • #3
Oh true i can use kinetic energy to match them! Thanks! :)
 
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Likes DaveE and hutchphd
  • #4
Resolving units like this is a great application for all of those simple formulas that you've memorized. This problem had nothing to do with kinetic energy, but that formula told you the equivalence of some of the units you did have.

So, for example, how can you express Volts in basic SI units? Well, I know that Volts*Amps=Watts=Energy/sec, so Volts=Energy/(Amp*sec)=(kg*m2)/(Coulomb*sec2).
 
  • #5
TheCelt said:
these equations don't have the same units
Yes they do.
The set of standard units in any system is richer than it needs to be. While it is usual to express pressure in Pascals, there's nothing to stop you using the same number of kg m-1 s-2 or J m-3 or any other equivalent.
 
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1. How do you calculate the momentum of a photon?

The momentum of a photon can be calculated by multiplying its velocity by its mass. Since photons have no mass, their momentum is solely determined by their velocity, which is equal to the speed of light (c).

2. What are the units for momentum of a photon?

The units for momentum of a photon are kilogram meters per second (kg*m/s). This is the standard unit for momentum in the SI system of measurement.

3. How is the momentum of a photon related to its wavelength?

The momentum of a photon is directly proportional to its wavelength. This means that as the wavelength of a photon increases, its momentum also increases. This relationship is described by the equation p = h/λ, where p is momentum, h is Planck's constant, and λ is the wavelength of the photon.

4. Why is the momentum of a photon important in physics?

The momentum of a photon is important in physics because it helps us understand the behavior of light and other electromagnetic radiation. It also plays a crucial role in concepts such as radiation pressure and the photoelectric effect.

5. Can the momentum of a photon be measured experimentally?

Yes, the momentum of a photon can be measured experimentally using various techniques such as Compton scattering or the photoelectric effect. These experiments involve measuring the change in momentum of a photon before and after interacting with matter.

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