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Stopping Potential

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm doing some practice problems for an introductory quantum mechanics course and am unsure whether or not I'm solving this problem properly - I need confirmation if I'm doing it right and help if I'm doing it wrong! :)

    "What is the stopping potential when 250 nm light strikes a zinc plate?" (Chapter 3, #18 in Modern Physics 2nd ed. by Randy Harris)

    2. Relevant equations

    K = E - [itex]\varphi[/itex] (where K is the kinetic energy, E is the energy of the incident light and [itex]\varphi[/itex] is the work function)

    This can be written as:

    [itex]\frac{mv^2}{2}[/itex] = [itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] (where m is the mass of a scattered electron, v is the speed of this electron, h is Planck's constant and c is the speed of light)

    [itex]\frac{mv^2}{2}[/itex] = qV (where q is the electron charge and V is the stopping potential)

    h = 6.626 x 10^(-34) Js
    c = 3 x 10^8 m/s
    [itex]\varphi[/itex] = 4.3 eV

    3. The attempt at a solution

    If my equations above are correct, I can write:

    [itex]\frac{mv^2}{2}[/itex] = [itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] = qV
    [itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] = qV
    V = [itex]\frac{\frac{hc}{\lambda} - \varphi}{q}[/itex]

    I can then simply plug in my values (remembering to either convert h in eV*s or [itex]\varphi[/itex] into J) and this should give me the stopping potential, correct?
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    Delphi51

    User Avatar
    Homework Helper

    Yes, that all looks good. No quantum mechanics in there, though; its all high school physics.
     
  4. Feb 5, 2012 #3
    Thanks! And you're right, there isn't really any quantum mechanics here; it's just included in part of the course and its part of the introduction leading into the actual quantum mechanics.
     
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