# Homework Help: Matter & Photon Particle Comparison

1. Jan 14, 2012

### HarleyM

1. The problem statement, all variables and given/known data
Compare and contrast a 2.2 eV photonn with a 2.2 eV electron in terms of energy(J), rest mass (Kg) speed (m/s) wavelength (m) and momentum kgm/s.

2. Relevant equations

For Photon
P=E/C
F=E/h
V=fλ
λ= h/P

For Electron
λ=h/mv
m=E/C2
p=h/λ
Ke=1/2mv2

3. The attempt at a solution

Photon
2.2 ev X 1.6x10-19= 3.52 x10 -19 J

P=E/C
P= 3.52 x10 -19 / 3 x 108
P= 1.17 x10 -27

f= E/h
=3.52 x10 -19 /6.63x10-34
F=5.31x1014

v=fλ
v= 3x108 m/s

λ= h/P
=6.63X10-34/1.17X10-27
=5.66X10-7 m

m=? ( Does a photon have a rest mass?) I don't think a photon has a rest mass.

Electron

2.2 eV x1.6x10-19 = 3.52 x 10-19 J

Should rest mass be the standard 9.11 x10-31 kg

Or should I calculate using E=mc2

m= E/C2
m= 3.52 x 10-19 / (3x108)2
m=3.91 x10 -36 kg

( I use 9.11 x10-31 kg as the mass for the rest of the calculations but I can change it if its incorrect and the calculated mass should be used! )

Ek=1/2mv2
3.52 x 10-19=1/2 (9.11x10-31)v2
v= √[(3.52 x 10-19)(2)/9.11x10-31]
v=8.79x105 m/s ( Unsure whether I can use this equation to determine speed)

λ=h/mv
= 6.63x10-34/(9.11X10-31)(8.79X105)
=8.28X10-10 m

p= h/λ
=6.63x10-34/8.28X10-10
=8x10-25 kgm/s

Parts I have highlighted is where I am unsure! Thanks for all the help!

If I have misused any equations ( i.e used matter equations for light or vice versa can you please alert me! )

2. Jan 14, 2012

### Curious3141

Correct. I didn't check your numbers, but make sure you don't round off your physical constants and intermediate results too early so that you have adequate precision for your final answer.

Correct. You're missing the units though.

Why bother? They didn't ask for the frequency.

Just quote v = c (then the numerical value as you did).

Correct. I didn't check your numerical computation, but make sure you don't have an error in your last few significant digits (because you rounded off the previous answer).

You're right, photons have no rest mass. Answer is 0.

Correct.

No. Rest mass is just the standard constant. It doesn't change.

Unnecessary. You are actually calculating the apparent gain in mass due to kinetic energy. Note that this is small relative to the rest mass (error of 4 parts per million or so).

No, use the rest mass.

You can, since the electron is clearly travelling at speeds significantly less than c. So relativistic effects don't matter. I didn't check your computation, but the principle is right.

Looks right.

You can just do p = mv, which you calculated as the denominator in the previous part.

3. Jan 14, 2012

### HarleyM

Wow very concise answer, thank you very much

WOOO i get physics, its a great feeling