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Countdown to MCAT: Random Topic #1: Photoelectric Effect

Discussion in 'MCAT Discussions' started by Mudd, Jul 26, 2002.

  1. Mudd

    Mudd Charlatan & Trouble Maker

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    It was suggested in another post that we have an MCAT chat. In addition, I propose a few topic discussions, because I'm sure that some people here have some great perspectives on material that may help. The photoelectric effect was mentioned in another thread, so I'll start this ball rolling:

    There are a few simple rules to follow:

    Start with the name: photoelectric effect. A photon goes in and an electron is ejected. In order to eject the electron, the energy of the incident photon must exceed the energy needed to eject the electron (called the binding energy to a chemist or work function to a physicist). Any energy beyond the threshold amount belongs to the electron in the form of kinetic energy.

    I am going to steal an analogy from the Berkeley Review general chemistry notes, so my apology to the author of their materials... take it as a compliment that your analogy works well.

    Just as money can be added to a vending maching to get a product, energy can be added to a material to get an electron. Vending machines only accept money in $0.05 increments, making the machine quantized (it won't take 42.3 cents, only amounts divisible by 5). From there, consider a $0.75 candy bar. You can add money to achieve one of three states.


    • 1. Not enough money to get the item. The machine will hold the money until the coins are released. This is similar to adding energy to an atom (or molecule), but not enough to ionize it. An electron is excited, but the energy is released when the electron returns to its original ground state.

      2. Just enough energy is added to ionize the material (exactly $0.75 has been added).

      or

      3. More than $0.75 has been added. The machine will release the item and give back change (in a perfect world at least). This is analogous to the photoelectric effect. If the incident photon is of greater energy than the binding energy (work function), then an electron is ejected (ionized) and it gets the excess energy (change if you will) in the form of kinetic energy. With more money in the vending machine, you get more change. With more energy in the incident photon, the ejected electron gets more kinetic energy. Either way though, the candy bar is still $0.75 and the electron only costs a certain amount of energy (the binding energy).

    This works best with metals, because they have low ionization energies, which means they have low binding energies. Hence, shining light of a high enough frequency on the surface of a metal will eject an electon. The surface is better for losing an electron than the core, because an electron on the surface has fewer neighboring atoms and thus is held least tightly.

    This is the principle behind solar cells, where incident sunlight ejects an electron, which then travels from the anode (site of electon loss) to the cathode (sight of electron storage/reduction).

    I hope this perspective helps. Additions are strongly welcomed.
     
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  3. Mr. Z

    Mr. Z Senior Member

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    That's a good analogy.

    So let me see if i have this correct: Increasing the frequency of the incident light on a piece of metal has the effect of ejecting more electrons, but not increasing their kinetic energy. Where as, decreasing the wavelength will result in electrons being ejected at a higher kinetic energies and vice versa, assuming that the light has enough energy to meet the work function.

    Now how does intensity of the light play a role here? Is more intense light at a higher frequency? or is it at a shorter wavelength? what about the amplitude of the wave?

    thanks
     
  4. mpp

    mpp SDN Moderator
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    I think you've got that backwards. Increasing the frequency of the light (frequency is the same as the energy of each individual photon) increases the amount of kinetic energy each ejected electron will have.

    Increasing the intensity of the light will (which is the same as increasing the number of photons, or the amplitude of the wave) just increases the number of electrons that are ejected, but does not change their kinetic energy.

    If you think about it in particle terms (which was how Einstein won his Nobel prize), think that each individual photon has a certain energy (i.e., frequency). If the energy is above the binding energy of any electron on the surface of the metal, that that electron will be knocked off. The difference between the photon's energy and the binding energy gives the amount of kinetic energy the ejected electron will have.
     
  5. mpp

    mpp SDN Moderator
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    To continue the money analogy...

    If the coke machine took dollar bills (and th coke's cost 75 cents), if you put in ten $1 bills (send 10 photons of equal energy/frequency), you'd get 10 cokes (10 ejected electrons) and 25 cents in kinetic energy back for each coke.

    If however the coke machine took ten dollar bills and you put 1 $10 bill in the machine (1 photon with a bucketload of energy) then you'd get a single coke with $9.25 of kinetic energy.
     
  6. Mudd

    Mudd Charlatan & Trouble Maker

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    mpp hit the nail on the head. I think this idea of topical discussions is going to be great. The analogy for intensity being several $1 bills versus a high energy photon being a $10 bill is excellent. I do like that analogy, but again want to give credit where it is due. This is lifted from general chemistry notes I used while tutoring for BR.

    This is more of a question for a moderator. As I mentioned, the analogy I used in a part of a copyright protected set of general chemistry lecture notes from BR. I use these when I tutor students, and BR doesn't seem to mind. I doubt they would mind my commenting here, but if you have any reservations, please accept my apology and lock the thread.
     
  7. limit

    limit Molesting my inner-child

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    Mudd, any tips/tricks for general chem topics, more specifically acids/bases (or any other general chem topic you think tends to stump mcat takers? Consider this a request for any general chem info you are willing to share... Much appreciated.
     
  8. lady bug

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    Keep this coming! I'm learning a lot!
     
  9. Mr. Z

    Mr. Z Senior Member

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    I believe i've got the idea, the kinetic energy of the ejected electrons increases linearly with the frequency of the incident light (assuming you've exceeded the work function).

    Energy = hv

    KE = hv - work function

    notice the above equation fits y=mx-b form, which means the slope of the line will be equal to plancks constant.:p

    So what about a larger amplitude at the same frequency? I would think that the KE of the ejected electrons would remain the same, but we would see more electrons ejected. And this may be a silly question, but how do you increase the amplitude but not the frequency?
     
  10. Mudd

    Mudd Charlatan & Trouble Maker

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    You nailed it perfectly Z!!! :)

    You increase the amplitude by increasing the number of photons, not changing the frequency of the photon (brighter light with the same color if you will). Frequency in this case is a property of the photon, not a description of the rate at which photons strike the surface of the metal.

    You are ready to teach the topic now.
     
  11. Mr. Z

    Mr. Z Senior Member

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    This was the source of my confusion, I wasn't thinking of frequency as a property of the individual photons, but more as a rate of how fast they are striking the metal.

    I now see the light. sorry, cheap pun intended.

    Thanks for your input mudd, appreciate it. As far as teaching it, i wouldn't go that far :rolleyes:

    This is a good thread, perhaps we should start another? a little magnetism anyone?

    Damn, its friday night and i'm on the computer talking physics:eek:
     
  12. neelyboy

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    bump. seems pretty valuable and helpful to me and possibly could be to others.
     

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