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Finding the momentum of a photon

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  • #1
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Homework Statement



Calculate the momentum of a 140 eV photon.

Homework Equations



p = E/c

The Attempt at a Solution


[/B]
First, convert eV to Joules

140eV (1.60 x10-19J)

= 2.24 x10-17J



Now for momentum:

p = mv

p = (E/c2)v

p = E/c

In my course, however, it says for photons E = hc/ λ

so p = E/c

is p = hc/λ/c

p = h/λ

My question is is it acceptable to use p = E/c even though this is with a photon, as the question only gives its eV value? or is there some way that I can't think of to determine the wavelength, λ, to then be able to use p = h/λ?
I have written my answer using p = E/c below.


p = E/c

p = 2.24 x10-17J / 3.00 x108m/s

p = 7.47 x10-26kgm/s

The momentum of the photon is .47 x10-26kgm/s.

I would really appreciate any help on this problem.
 

Answers and Replies

  • #2
TSny
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I have written my answer using p = E/c below.


p = E/c

p = 2.24 x10-17J / 3.00 x108m/s

p = 7.47 x10-26kgm/s.
Yes, this is the best way to work this problem and your answer looks correct.

A different approach would be to first find λ from E = hc/ λ, and then find p using p = h/λ. But that would be a more round-about way.
 
  • #3
Orodruin
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Yet another approach would be to use reasonable units suitable for the problem instead of SI units. The typical unit used by particle physicists in this type of problem would be eV/c and the obvious answer in those units based on your p = E/c would be p = 140 eV/c.
 
  • #4
DrClaude
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Now for momentum:

p = mv

p = (E/c2)v

p = E/c
The final result is correct, but the route you used to get there is not. ##p=m/v## is valid in the non-relativistic case, which doesn't apply to photons. In addition, ##m=0## for photons, so that would lead to ##p=0##. You have to start by the relativistic energy
$$
E^2 = p^2 c^2 + m^2 c^4
$$
For a photon, since ##m=0##, you recover ##p=E/c##.
 
  • #5
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The final result is correct, but the route you used to get there is not. ##p=m/v## is valid in the non-relativistic case, which doesn't apply to photons. In addition, ##m=0## for photons, so that would lead to ##p=0##. You have to start by the relativistic energy
$$
E^2 = p^2 c^2 + m^2 c^4
$$
For a photon, since ##m=0##, you recover ##p=E/c##.
That was another part I was unsure of, in my course they usually start with p = mv which seems weird to me, and I wondered how it would work in relativistic problems. If I understand you correctly, the derivative I used, p = E/c is correct, but how I derived it was not?
 
  • #6
DrClaude
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If I understand you correctly, the derivative I used, p = E/c is correct, but how I derived it was not?
Indeed, for a photon ##p = E/c## is correct, but as I wrote the starting point is not ##p=mv##.
 
  • #7
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Indeed, for a photon ##p = E/c## is correct, but as I wrote the starting point is not ##p=mv##.
Ok thank you that makes mcuh more sense.
 
  • #8
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Indeed, for a photon ##p = E/c## is correct, but as I wrote the starting point is not ##p=mv##.
Would this be the correct derivation?

E2=p2c2+m2c4

E2=p2c2 because m = 0 for a photon

p2 = E2/c2

p = √E2 /c2

p = E /c

I just want to make sure I got it right. Thanks
 
  • #9
Orodruin
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Would this be the correct derivation?

...
Yes.
 
  • #10
ehild
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That was another part I was unsure of, in my course they usually start with p = mv which seems weird to me, and I wondered how it would work in relativistic problems. If I understand you correctly, the derivative I used, p = E/c is correct, but how I derived it was not?
p=mv is valid if m means the relativistic mass. Einstein's famous equation E = mc2 is also valid for the relativistic mass. Recently, the term relativistic mass is avoided and "m" is used for the rest mass m0 ("invariant mass). Using this term, the photon has no mass, but according to Einstein's equation, the mass of the photon is E/c2, its momentum is p=mc =E/c. Or E=hf, the mass is hf/c2, the momentum is p=mc = hf/c. c/f= λ, so p=h/λ.
In the equation E2=p2c2+m2c4, m is the invariant mass, and the equation was derived during undergraduate SR General Physics courses (Giancoli, for example) from E=mc2, p=mv, m=m0γ, ##γ=\frac{1}{\sqrt{1-\frac {v^2}{c^2}}}##.
 
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  • #11
Orodruin
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In the equation E2=p2c2+m2c4 m is the invariant mass, and the equation is derived during undergraduate SR courses using E=mc2, p=mv, m=m0γ, ##γ=\frac{1}{\sqrt{1-\frac {v^2}{c^2}}}##.
I certainly do not derive it like this in my SR course. I never mention relativistic mass at all. The more appropriate way of looking at it is as the definition of mass in SR as the magnitude of an object's 4-momentum. The amazing part comes when you show that this definition of mass coincides with the inertia of the object in its rest frame (which of course is the reason to call it mass in the first place even though that is a post hoc naming - the canonical name would be rest energy).
 
  • #12
ehild
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I certainly do not derive it like this in my SR course. I never mention relativistic mass at all. The more appropriate way of looking at it is as the definition of mass in SR as the magnitude of an object's 4-momentum. The amazing part comes when you show that this definition of mass coincides with the inertia of the object in its rest frame (which of course is the reason to call it mass in the first place even though that is a post hoc naming - the canonical name would be rest energy).
Well, I did it long time ago, during courses of General Physics for chemists students, without mentioning 4-momentum at all. I wanted to write "General Physics" instead of SR.
 
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