1. Not finding help here? Sign up for a free 30min 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!

Momentum in a photon -> cathode collision

  1. Mar 14, 2017 #1
    1. The problem statement, all variables and given/known data
    A photon with the length of lambda hits a cathode perpendicularly to its surface. As a result, an electron leaves its surface perpendicularly to the direction of the photon. How much momentum was transfered from the photon to the cathode? The work function of the cathode is W.

    2. The attempt at a solution
    1. I calculate the speed of the electron:
    [tex]v = \sqrt{ \frac{2hc}{\lambda m_e} - \frac{W}{m} }[/tex]

    2. The momentum of the photon:
    [tex]p_{photon} = \frac{h}{\lambda}[/tex]

    3. The momentum of the electron:
    [tex] p_{electron} = m \sqrt{ \frac{2hc}{\lambda m_e} - \frac{W}{m} } [/tex]

    Now, I am stuck and don't know what to do because of this 90 degree angle.
     
  2. jcsd
  3. Mar 14, 2017 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Conserve momentum in two dimensions, perpendicular and parallel to the cathode surface.

    Note: your expression for v is incorrect.
     
  4. Mar 14, 2017 #3
    kuruman,
    Thank you, there is a 2 in front of W in the velocity expression.

    Could you give me some hints how to conserve this momentum? The only thing that comes to my mind is
    [tex]p_{cathode} + p_{electron} = - p_{photon}[/tex]
     
  5. Mar 14, 2017 #4

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    That's it, but you write it as a vector equation. Call the perpendicular direction y and the parallel direction x. What are the x and y components oh the three momentum vectors?

    On edit: Lose the negative sign. The total momentum vector before is equal to the total momentum vector after.
     
  6. Mar 14, 2017 #5
    in x direction:
    [tex]p_{x cathode} = p_{electron}[/tex]

    in y direction:
    [tex] p_{y cathode} = p_{photon}[/tex]

    [tex]p_{cathode} = \sqrt{p_{x cathode}^2 + p_{y cathode}^2}[/tex]
     
  7. Mar 14, 2017 #6

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Something like that. It's not clear if the problem is asking for the vector or its magnitude which is what you found.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Momentum in a photon -> cathode collision
  1. Momentum and collision (Replies: 3)

  2. Momentum and collision (Replies: 1)

  3. Momentum and collision (Replies: 1)

  4. Momentum and collision (Replies: 1)

Loading...