# Momentum of a photon

1. Feb 7, 2013

### quicksilver123

1. The problem statement, all variables and given/known data
Find the momentum of a 140eV photon.

2. Relevant equations

momentum equations

e=mc^2

3. The attempt at a solution

140eV=2.24*10^-17 J

p=mv
p=(E/c^2)c
p=E/c
p=(2.24*10^-17)/(3*10^8)
p=7.466666*10^-26

Is this correct? (other than significant digits)

2. Feb 8, 2013

### Dick

Your derivation is little suspect since m=0. You should probably really be using E^2=(mc^2)^2+p^2*c^2. But the only problem aside from signficant digits is that you didn't put any units on it.

3. Feb 8, 2013

### Simon Bridge

As it happens - yes. But don't forget to write in the units!
The actual energy-momentum relation is $E^2=m^2c^4+p_2c^2$ and photons have $m=0$.
As Dick says, you got the right answer through the (apparently) wrong method.

Technically you can say that p=140eV/c since "eV/c" are units of momentum :)

4. Feb 8, 2013

### quicksilver123

140eV=2.24*10^-34 J

m=E/c^2
m=2.488888889*10^-34
if a photon has no mass (even though the course asks me to find the mass of a photon several times) then how did i get this answer?

p=h/λ
mv=h/λ
mc=h/λ
λ=.......

never mind. this is getting roundabout.

the course has me calculate the mass value for a photon several times.

when it would ask me for the momentum of a photon, however, it would have me use the formula : p=h/λ
however, it would also supply a frequency so that i could actually calculate wavelength.

i don't know of any way to calculate wavelength in this case, so i just used the standard p=mv for a particle

Last edited: Feb 8, 2013
5. Feb 8, 2013

### Simon Bridge

Nope - m=0 for a photon.
And you left off the units again.

But you are close (see post #3).
Mass is a form of energy so there is a sense in which you can attribute a total mass to a photon. It's just potentially misleading to think of it that way.
i.e. the $m$ in the relation is the rest mass. For a moving mass, the mass-energy relation is $E=\gamma mc^2$.

But you do have the frequency - for a photon: $E=hf=hc/\lambda$

6. Feb 8, 2013

### Dick

You were basically right the first time. E=p/c. Just put units on it! And 140eV is not 2.24*10^-34 J.

7. Feb 8, 2013

### quicksilver123

Ha, yeah. Its getting late. Its 2.24*10^-17

8. Feb 8, 2013

### quicksilver123

ah. haha. thank you mr planck.

intuitively, i know that for photons, m=0
but the course doesn't teach it that way. i don't really know what's what.

E=hf
f=E/h
f=3.378582202*10^16 Hz

λ=v/f
λ=c/f
λ=0.000000009m
λ=9nm

p=h/λ
p=7.366666667*10^-26 Js/m

I don't feel right about the unit.

h is in Js and lambda is in m

9. Feb 8, 2013

### Dick

If you put all that together you just get E=pc again. J*s/m=(kg*m^2/s^2)*(s/m)=kg*m/s. That's a mass times a velocity, so sure it's a good momentum unit.

10. Feb 8, 2013

thanks