# What are the Units for the momentum of a Photon?

1. Apr 16, 2014

### blue_lilly

1. The problem statement, all variables and given/known data
A photon, a packet of electromagnetic radiation, carries both energy and momentum. Consider a photon with a wavelength of 428 nm in vacuum.
A) What is the frequency of the photon? 7.01×1014 Hz CORRECT
B) What is the energy of the photon? 4.64×10-19 J
C) What is the momentum of the photon?

2. Relevant equations
P=m*V
Plancks Constant: (6.6262E-34 J/s)

3. The attempt at a solution
A) (3E8 m/s) / (4.28E-7 m) = 7.009E14 Hz CORRECT
B) (7.009E14 Hz) * (6.6262E-34 J/s) = 4.644E-19 J CORRECT
This is where I need Help
C) What is the momentum of the photon?
[(7.009E14 Hz)*(6.6262E-34 J/s)] / (3E8 m/s) = 1.546E-27

I think I have the number but I can't submit it to check it, until I have the right unit of measurement. I tried (Hz*m/s), (m/s), (nm*m/s).
If I don't have the number right would you let me know as well.

2. Apr 16, 2014

### Staff: Mentor

What are the standard units for momentum? Think mass*speed.

You can figure it out from your calculation by expressing Hz and J in terms of more fundamental units.

3. Apr 16, 2014

### blue_lilly

So Hz is frequency and frequency either be:
1) inverse of wavelength: (1/λ) and in this problem wavelength is measured in nm. So it could be (1/nm)​
2) speed of light divided by wavelength: (c/λ) and c is measured in m/s and wavelength in nm. So it could be [(m*nm)/s]​

Joules is measured as [(kg*m^2)/s^2]. Photons are mass-less so maybe you can remove the kg to get [(m^2)/(s^2)]

So all together I would have: [(m*nm)/s] [(m^2)/(s^2)]

Am I doing this right? I feel like I'm not because I cant cross out any variables that are on the top with ones on the bottom.

4. Apr 16, 2014

### SteamKing

Staff Emeritus
How does nm compare with m?

5. Apr 16, 2014

### blue_lilly

1 meter is 1E9 nano-meters.
1 nano-meter is 1E-9 meters.

6. Apr 16, 2014

### SteamKing

Staff Emeritus
So that there is an equivalent number of meters in 1 nano-meter, right?

7. Apr 16, 2014

### Staff: Mentor

No, frequency is not 1/λ.

Not exactly. Think of c/λ as speed divided by distance. Or distance/time divided by distance, which is 1/time or 1/sec.

No, don't go removing mass. These are units, not formulas.

You messed things up a bit. So you can try again. And don't forget that you are also dividing by speed.

Even simpler is to use standard units for momentum, given by mass*speed. (But it's a good exercise to see if you can arrive at it step by step, as you are attempting to do.)