# Photons and matter waves: The photon of quantum light: I need help

1. May 24, 2009

### afcwestwarrior

1. The problem statement, all variables and given/known data
How fast must an electron move to have a Kinetic Energy equal to the photon energy of sodium light at wavelength 590 nm.

2. Relevant equations
photon energ E =hf

h = 6.63 * 10 ^ -34 J * s = 4.14 = 10^-15 eV *s

f = c/ lamda

c = 3 * 10^ 8 m/s

mass of an electron is 9.11 * 10 ^ -31 (kg) or 511 keV

3. The attempt at a solution

Ok I solved for the energy of a photon for sodium light

E = 6.63 * 10 ^ -34 J *s * ( (3* 10 ^ 8 m/s) / 590 * 10 ^ -9 meters) = 1.989 * 10 ^ -25 Joules

Were solving for Velocity. How do I do this problem it's confusing.

Last edited: May 24, 2009
2. May 24, 2009

### Pengwuino

Well you're looking for the kinetic energy (and thus velocity) right? Well $$E_{kinetic} = \frac{{mv^2 }}{2}$$ and you know the energy of the photon so equate them and find the velocity.

3. May 24, 2009

### afcwestwarrior

Sorry about that I'll make my information much clearer next time. So that's the equation I use.

Last edited: May 24, 2009
4. May 24, 2009

### afcwestwarrior

Would it be like this

$$v = \frac{{\sqrt{{2E }} \div {m} }$$

Last edited: May 24, 2009
5. May 24, 2009

### Pengwuino

Yes, where E = hf = hc/lamda

6. May 24, 2009

### afcwestwarrior

So is that E of photon energy of soidum light the same as the kinetic energy

7. May 24, 2009

### Chrisas

Check your math again. I get a different value for energy using the same numbers you gave as input.

8. May 24, 2009

### afcwestwarrior

so then it would be

It would be v = sqrt { 2 (3.37 * 10 ^ -19 joules / 8.19 * 10^ -14 joules}

m of electron in joules is 8.19 * 10 ^ -14 joules

Last edited: May 24, 2009
9. May 24, 2009

### afcwestwarrior

Oh yea the Enery of a photon for sodium light is actually
$$E = 3.371186441 \times 10^-19$$

10. May 24, 2009

### Chrisas

That's what I got.

11. May 24, 2009

### afcwestwarrior

I got 2.86 * 10 ^ -3 m/s

12. May 24, 2009

### Chrisas

That doesn't seem correct. Your equation for v doesn't look right, but I don't know if you are just typing in the tex code wrong. Did you really intend for the mass, m, to be in the numerator of that equation?

13. May 24, 2009

### Chrisas

Also, I'm not sure where you are getting the mass of the electron in joules. If you are using SI units of joules for energy, then the electron mass should be in kg.

14. May 24, 2009

### Pengwuino

Wait yes, your equation for the velocity in terms of the energy changed... its divided by m.

15. May 24, 2009

### Chrisas

Well my last word on this...it looks like you threw another factor of c in there somewhere. If I divide the answer I get in meters/sec by 3*10^8 m/s, then I get the answer you posted above. So your answer would be correct if it was in units of "c".