Stopping potential and photo-electric effect?

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When they state "just emitted electrons from the surface" they (usually) mean "the voltage is such that a current is just detected in the circuit", the assumption being that there is a 2V potential difference between cathode and anode, with the cathode being more positive.

That being the case, the electrons will need to leave the cathode with some additional energy in order to overcome the (decelerating) potential difference of 2 V.

Now you know the energy of the 700 nm photon - E. Some of this is used to overcome the work function W, and some of it to overcome the potential difference V.

It's easy to get confused with the signs because q of the electron is negative. Expressing everything in eV (a convenient unit of energy for these kinds of problems - with $\rm{hc = 1240 eV\cdot nm}$)), we have E = 1240/700 = 1.77 eV. Since the electron needs to be able to overcome a 2 eV energy barrier, it must be leaving the plate with some excess energy of 0.23 eV (plus the 1.77 eV from the photon = 2 eV).

I am sure you can figure out how that relates to work function.

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Updated on December 18, 2022

Comments

  • John
    John 11 months

    The surface of a photo-electric metal is at a potential of +2.0V. The wavelength of light needed to just emit electrons from the surface is 700 nm. Find the work function of the metal.

    My attempt at a solution:

    By energy conservation,

    E= work function + 1/2 mv^2

    For the electrons to be just emitted, v=0.

    hc/700 x 10^-9 = E

    But the answer I get from solving this is incorrect. The actual answer says: E = work function + qV And then uses this to calculate the work function. But I have a couple of problems with this :

    1) The electrons are just emitted from the surface; how are they passing a potential difference of +2V?

    2) If the electrons did pass a potential difference of +2V, this means that when emitted, they must have had some non-zero kinetic energy. Doesn't this violate the terms of the question? ( just emitted)

    3) By energy conservation, the initial energy of the system = the final energy. qV + hf = work function + 1/2 mv^2 + qVfinal

    Even if the speed is 0, and the final potential is 0 as well, we get E = work function - qV, not +qV.

  • John
    John over 6 years
    Thank you for the answer! Why would 'emitted from the surface' be equivalent to travelling all the way to the anode? And if we aren't told the potential of the surroundings, do we just take it to be 0? Lastly, this is somewhat unrelated, but why does the frequency of the photons not affect the size of the photocurrent? For any given stopping potential, a higher frequency means more electrons will reach the anode, so a larger current should be detected. And I've read that the intensity of photons is proportional to the photocurrent, but if the frequency (and intensity) is doubled, no change.