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Checking answer for velocity of photoelectron

  1. Mar 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the speed, v, which an electron, emitted with kinetic energy W = 0.200 eV, arrives at the anode when the accelerating voltage between the anode and the cathode is U = 2.00 V.

    b) The value I of the photoelectric current

    2. Relevant equations
    From what i gather,
    Etotal = Ekinetic + work function = qV + work function

    3. The attempt at a solution

    Im not sure if the first question is poorly worded or I'm just misunderstanding it, but I assume by 'kinetic energy W = 0.200' they mean the work function. I solved it like so

    Etotal = Ekinetic + work function = qV + work function

    ETOT = ((1.6x10-19)x(2) + (0.2 x 1.6x10-19)
    = 3.52x10-19 joules
    3.52x10-19 = 1/2 mv2, solving for v i get 879077m/s.

    Does the electron actually move 879 KILOMETERS a second? Wow. Is this right? Is it true only because it travels a very short distance in total, over a short time? I'm having a hard time conceptualizing this.

    Now for b
    I'm really not sure how to approach this, I know current I = voltage / resistance, but it's not applicable in this situation,

    from the last question we know that the voltage is (3.52x10-19)/(1.6x10-19) which seems to be 2.2 volts? But where do I go from here?


    If possible I need urgent clarification, I need to go sleep soon. Thanks for any help!
     
  2. jcsd
  3. Mar 25, 2015 #2
    Since it was said that the kinetic energy is 0.2eV, you will not need the work functiin of the metal.
     
  4. Mar 25, 2015 #3

    gneill

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    Staff: Mentor

    I think they mean that the electron that escapes the surface is left with a kinetic energy W = 0.200 eV. Your calculation treated it this way, so the result looks fine.
    Yup. Subatomic particles can move really fast with small amounts of energy. They tend to run into obstacles pretty quickly when they're not in a vacuum.
    Current is also the rate of movement of charge: An ampere is a Coulomb per second.

    Is there information missing in the problem statement that would allow you to quantify the rate of electron emission?
     
  5. Mar 25, 2015 #4
    Ah yes! The question I've posted is a sub-question to a larger problem. I was also given this info:
    The power absorbed by the photo-cathode is Pc = 0.540 W, but only 0.2% of this incident energy is effective in producing photoelectrons.
    I have no clue how i didn't realize the rate of emission is right there.
    Current I = power W / voltage V
    So would this calculation be correct? I = (0.540 x 0.2) / 2.2 = 0.049 amps?

    A big thank you for helping me with this!
    :)
     
  6. Mar 25, 2015 #5

    gneill

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    Staff: Mentor

    I would not look to the voltage as a contributor to the current. It doesn't cause the emission of the electrons from the surface of the cathode, it merely accelerates the ones that are emitted.

    Instead, consider the rate at which the absorbed energy is being converted to emitted electrons (you know the energy they are each given).
     
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