1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 24, 2009 #1
    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. jcsd
  3. May 24, 2009 #2

    Pengwuino

    User Avatar
    Gold Member

    Well you're looking for the kinetic energy (and thus velocity) right? Well [tex]E_{kinetic} = \frac{{mv^2 }}{2}[/tex] and you know the energy of the photon so equate them and find the velocity.
     
  4. May 24, 2009 #3
    Sorry about that I'll make my information much clearer next time. So that's the equation I use.
     
    Last edited: May 24, 2009
  5. May 24, 2009 #4
    Would it be like this

    [tex]
    v = \frac{{\sqrt{{2E }} \div {m} }
    [/tex]
     
    Last edited: May 24, 2009
  6. May 24, 2009 #5

    Pengwuino

    User Avatar
    Gold Member

    Yes, where E = hf = hc/lamda
     
  7. May 24, 2009 #6
    So is that E of photon energy of soidum light the same as the kinetic energy
     
  8. May 24, 2009 #7
    Check your math again. I get a different value for energy using the same numbers you gave as input.
     
  9. May 24, 2009 #8
    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
  10. May 24, 2009 #9
    Oh yea the Enery of a photon for sodium light is actually
    [tex]
    E = 3.371186441 \times 10^-19
    [/tex]
     
  11. May 24, 2009 #10
    That's what I got.
     
  12. May 24, 2009 #11
    I got 2.86 * 10 ^ -3 m/s
     
  13. May 24, 2009 #12
    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?
     
  14. May 24, 2009 #13
    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.
     
  15. May 24, 2009 #14

    Pengwuino

    User Avatar
    Gold Member

    Wait yes, your equation for the velocity in terms of the energy changed... its divided by m.
     
  16. May 24, 2009 #15
    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".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Photons and matter waves: The photon of quantum light: I need help
Loading...