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!

Is this a trick question?

  1. Nov 29, 2011 #1
    The problem statement, all variables and given/known data

    Compare a 2.2eV photon with a 2.2 eV electron in terms of energy, rest mass, speed, wavelength, and momentum

    The attempt at a solution

    So...

    E = (2.2eV) x (1.60 x 10-19 J/eV) = 3.52 x 10-19 J

    Wouldn't 3.52 x 10-19 J be the energy for both the photon and electron? If so, wouldn't that make mass, speed, wavelength, and momentum equal for both particles?
     
  2. jcsd
  3. Nov 29, 2011 #2

    DaveC426913

    User Avatar
    Gold Member

    Well, hopefully it wouldn't make mass the same...
     
  4. Nov 29, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    Tabulate all the values for each particle. What's the rest mass of a photon?
     
  5. Nov 29, 2011 #4
    Isn't it:

    m = E / c2


    ?

    Or is the rest mass of a photon always zero.
     
  6. Nov 29, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    The rest mass of a photon is always zero.
     
  7. Nov 29, 2011 #6
    Okay, so the rest mass for the photon is 0 and for the electron it is 9.11 x 10-31.

    Wavelength would be:

    E = hc / λ, or rather, λ = E / hc

    Momentum would be:

    p = h / λ

    I assume because their energies are the same, the results for wavelength and momentum will be equal for the electron and photon.


    Speed is:

    p = mv, or rather, v = p / m



    In the end: their energies, wavelengths, and momentum are equal while mass and speed are different. Is that right?
     
  8. Nov 29, 2011 #7

    gneill

    User Avatar

    Staff: Mentor

    Wavelength and momentum will not be the same. Look up DeBroglie wavelength, photon momentum.
     
  9. Nov 29, 2011 #8
    Hmm,


    So I first calculate velocity with:

    v = √2eΔV/m

    And then use:

    λ = h / mv

    That gives wavelength


    For momentum I use:

    p = mv


    Because their masses are different, I should get different results.

    So only the energy is the same for both?
     
  10. Nov 29, 2011 #9

    gneill

    User Avatar

    Staff: Mentor

    As I said, you should tabulate all the values for each particle. You will have to be sure to use the appropriate formulas that apply to each. Then compare results.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is this a trick question?
  1. Trick Questions? (Replies: 22)

  2. Trick question? (Replies: 2)

  3. Trick question (Replies: 6)

Loading...