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Photoelectric effect

Explaining the photoelectric effect using wave-particle duality, the work function of a metal, and how to calculate the velocity of a photoelectron. Created by Jay.

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  • blobby green style avatar for user TRILOCHAN BHARDWAJ
    If photon is a particle and it passes its energy to an elcetron then where does the photon go after collision when it has transmitted its energy to the electron???
    (146 votes)
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    • orange juice squid orange style avatar for user Joshua
      I believe it gets absorbed into the system and changes its form to either an electron or it gets added to the element. This works better with metals because metal ions are more stable than other ions
      (6 votes)
  • blobby green style avatar for user TRILOCHAN BHARDWAJ
    if you are talking about classiscal physics . as we know that the photon is massless then how the momentum is being conserved?
    (53 votes)
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    • blobby green style avatar for user ldr709
      Although a photon is massless, it still has momentum. Einstein's E = mc^2 formula is actually a special case of the special relativity formula E^2 = p^2c^2 + m^2c^4, where p is momentum, m is rest mass, E is energy, and c is the speed of light. If you substitute 0 for m (because a photon is massless), and E = hv (the formula for the energy of a photon), you get hv=pc. Because wavelength (λ) is c / v, the equation can be simplified to p = h / λ, so a photon has momentum even though it has zero rest mass.
      (105 votes)
  • spunky sam blue style avatar for user Srikriti
    What if the frequency of the incident ray of light is EXACTLY EQUAL to the threshold frequency of the metal? What would happen to the photoelectron in this case?
    (38 votes)
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  • marcimus pink style avatar for user Samantha Jensen
    At you begin to use many different formulas to solve these problems, do you have earlier videos describing these formulas or a list of the specific ones that are related to Bohr's model?
    (24 votes)
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    • leaf grey style avatar for user Josiah
      I also ran into the issue of wanting more explanation.

      The first issue is v (the English alphabet character "ve") to represent velocity and v (the Greek alphabet character "nu") to represent frequency. Both use the same character but represent different things. Google "Greek alphabet" and view the wikipedia article to see what I mean. On paper I always write my v's differently than my v's.

      The second issue is the format layout of how he got to his answer.
      Kinetic Energy (KE) equals ½ * mass of electron (m) * the velocity(v)²

      Kinetic Energy (KE) also equals Epsilon photon minus Epsilon naught (E₀)

      [½ * m * v²] equals Epsilon photon minus E₀ because KE equals KE

      ½ * m * v² equals [Planck's Constant(h) * frequency (v)] minus E₀ because Epsilon photon equals h*v Note the substitution of h*v where Epsilon photon was.

      ½ * m * v² equals h* [Speed of Light (c) / wavelength (λ)] - E₀ because frequency(v) equals velocity of light(c) divided by the wavelength (λ) Note the substitution of c/λ where the frequency once was.

      This leaves us with ½*m*v² equals h*c/λ minus E₀
      now solve for velocity which is the v² in the above equation using the following numbers.

      m or the mass (in this case an electron) is 9.11 * 10⁻³¹ kg
      c or the speed of light is 2.998 * 10⁸ meters / second
      E₀ or Epsilon Naught was given as 3.43 * 10⁻¹⁹ Joules
      λ(pronounced lambda) or the wavelength was given as 525 nanometers or 5.25 * 10⁻⁹ meters
      h or Planck's Constant is 6.26 * 10⁻³⁴ m² kg / sec

      Or something like this.

      Clear as mud?

      EDIT: many thanks to Cipher for the correction!
      (37 votes)
  • duskpin ultimate style avatar for user Sahel
    I have a doubt regarding the photoelectric effect that has been nagging me for some time now, so I'd be really thankful if somebody could clarify my honest (if stupid) question. I understand the part about the photons transferring their energy to the electrons and the electrons gaining a kinetic energy and all, but what I don't understand is why the photoelectric effect doesn't take place with light below a certain frequency. Because even though the photons have lesser energy, the electrons could just absorb the energies of more than one photons so as to reach the work function. Arghh...it just doesn't make any intuitive sense.
    (23 votes)
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    • piceratops ultimate style avatar for user Just Keith
      Because the electrons can only absorb one photon of a given wavelength. Electrons can ONLY absorb photons of very specific wavelengths and no others. And they CANNOT absorb two or three photons that add up to the required wavelength.

      When dealing with things in the quantum realm, our "intuitive sense" will fail us. This is just the way that electrons behave.
      (16 votes)
  • piceratops ultimate style avatar for user Kristjan Turnsek
    Can someone tell me if I am correct here?
    Whenever a photon knocks out an electron, the metal gets a little more positively charged. Does that mean that every next photon will need a little more energy, because of the electrical attraction between positive metal and electron?
    (16 votes)
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    • leaf green style avatar for user Mark Zwald
      That's an excellent question.
      My guess is there are so many electrons in the metal that losing a few trillion from photoelectric effect has a negligible impact, but you are right in principle. If the metal becomes more positively charged, it will take more energy to free an electron from that potential well.
      Also if the metal is in contact with another material which is grounded, then it will make no difference how many electrons are lost as the ground has a virtually unlimited supply of charge.
      (14 votes)
  • piceratops seed style avatar for user Saniya Maheshwari
    According to my understanding, different shells in an atom have different energy levels. Therefore shouldn't the energy required to free an electron, be different for different shells? Then, when the work function of metallic caesium is given as 3.43 x 10^-19 J, is this the energy required to free an electron from the outermost shell or the innermost, or is it some kind of average value?
    (5 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      well... when I was younger, we used to talk about the structure of metals like this: "a metal consists of a regular array of positive ions. (Very small crystals) . the outermost electrons of each atom finds sufficient energy to leave its parent atom and joins other such electrons to 'move around' in a "sea of electrons". This is what carries the electric current and what makes metals such good conductors.

      So, if this is still the current atomic model of a metal, I would say that your incident photon does not really get to see an atom as such. It will smash into this cloud or soup of electrons as they swirl around (or , more likely, wave around...) and give its energy to one of them.

      The work function (I would suggest) is not so much a function of the atom but more of the metal as a whole. And the electrons that are ejected would, indeed, come from the outermost shells but not because the photon would find it easier to remove them but because these are the electrons that are involved in the metallic bonding process and thus most available to the photon and any other such physical process.

      Any thoughts??
      (8 votes)
  • starky tree style avatar for user mehulghosal
    I have a hypothetical. Say a certain wavelength of light were to hit a substance, but it did not have the enough energy to make a photoelectron. What would happen if two photons hit the same electron? Wouldn't the combined energy be enough to create that photoelectron?
    (2 votes)
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    • leafers ultimate style avatar for user Stoltzfus
      This does happen, but it is very rare. It is called two-photon absorption in case you want to read the wiki on it.

      Basically, in order for there to be a measurable chance of two photons being absorbed by the same electron you have to have a bunch of photons in the same place at the same time. This is typically done with very short pulses (10^-15 seconds long) of intense light from specialized lasers that is focused to a small spot. It turns out this is really useful in microscopy for generating high resolution images.
      (12 votes)
  • spunky sam blue style avatar for user Samir Mahajan
    why is photon massless ?

    and if it is massless then how is light energy affected by gravity?
    (5 votes)
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  • blobby green style avatar for user andrewdmusk
    In a photoelectric cell where there is no voltage, what causes the ejected electrons to travel to the anode?
    (3 votes)
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    • male robot hal style avatar for user Andrew M
      Are you talking about the metal in the photoelectric effect? If so, the answer is: Nothing. They won't travel there unless you apply at least a little positive voltage. They might drift there anyway, depending on the direction in which they were ejected.

      If you are talking about a photovoltaic cell (like for solar power), it is a little more complicated. Those are made of semiconductor materials in which one side has a shortage of electrons relative to the other side. It's a little complicated to type about. You can read about how it works here http://pveducation.org/pvcdrom/solar-cell-operation/photovoltaic-effect
      (6 votes)

Video transcript

- Sometimes light seems to act as a wave, and sometimes light seems to act as a particle. And, an example of this, would be the Photoelectric effect, as described by Einstein. So let's say you had a piece of metal, and we know the metal has electrons. I'm gonna go ahead and draw one electron in here, and this electron is bound to the metal because it's attracted to the positive charges in the nucleus. If you shine a light on the metal, so the right kind of light with the right kind of frequency, you can actually knock some of those electrons loose, which causes a current of electrons to flow. So this is kind of like a collision between two particles, if we think about light as being a particle. So I'm gonna draw in a particle of light which we call a photon, so this is massless, and the photon is going to hit this electron, and if the photon has enough energy, it can free the electron, right? So we can knock it loose, and so let me go ahead and show that. So here, we're showing the electron being knocked loose and so the electron's moving in, let's just say, this direction, with some velocity, v, and if the electron has mass, m, we know that there's a kinetic energy. The kinetic energy of the electron would be equal to one half mv squared. This freed electron is usually referred to now as a photoelectron. So one photon creates one photoelectron. So one particle hits another particle. And, if you think about this in terms of classical physics, you could think about energy being conserved. So the energy of the photon, the energy that went in, so let me go ahead and write this here, so the energy of the photon, the energy that went in, what happened to that energy? Some of that energy was needed to free the electron. So the electron was bound, and some of the energy freed the electron. I'm gonna call that E naught, the energy that freed the electron, and then the rest of that energy must have gone into the kinetic energy of the electron, and so we can write here kinetic energy of the photoelectron that was produced. So, kinetic energy of the photoelectron. So let's say you wanted to solve for the kinetic energy of that photoelectron. So that would be very simple, it would just be kinetic energy would be equal to the energy of the photon, energy of the photon, minus the energy that was necessary to free the electron from the metallic surface. And this E naught, here I'm calling it E naught, you might see it written differently, a different symbol, but this is the work function. Let me go ahead and write work function here, and the work function is different for every kind of metal. So, it's the minimum amount of energy that's necessary to free the electron, and so obviously that's going to be different depending on what metal you're talking about. All right, let's do a problem. Now that we understand the general idea of the Photoelectric effect, let's look at what this problem asks us. So the problem says, "If a photon of wavelength "525 nm hits metallic cesium..." And so here's the work function for metallic cesium. "What is the velocity of the photoelectron produced?" So they want to know the velocity of the photoelectron produced, which we know is hiding in the kinetic energy right here, and we also know what the work function is. So we know what E naught is here. What we don't know is the energy of the photon so that's what we need to calculate first. And so the energy of the photon, energy of the photon, is equal to h, which is Planck's constant, times the frequency, which is usually symbolized by nu. So, we got the frequency, but they gave us the wavelength in the problem here. They gave us wavelength, so we need to relate frequency to wavelength, and that's related by c, which is the speed of light, is equal to lambda times nu. So, c is the speed of light, and that's equal to the frequency times the wavelength. So we can substitute n for the frequency, all right, 'cause we just use this equation and say that the frequency is equal to the speed of light divided by the wavelength. The frequency is equal to speed of light over lambda, so we can plug that into here, and so now we have the energy of the photon is equal to hc over lambda, and we can plug in those numbers. h is Planck's constant, which is 6.626 times 10 to the negative 34. So, times 10 to the negative 34 here. c is the speed of light, which is 2.998 times 10 to the 8th meters over seconds, and all over lambda. Lambda is the wavelength. That's 525 nanometers. So 525 times 10 to the negative 9th meters. All right, so let's get out our calculator and calculate the energy of the photon here. So, let's go ahead and do that math, so we have 6.626 times 10 to the negative 34, and we're going to multiply that number by the speed of light, 2.998 times 10 to the 8th, and we get that number. We're gonna divide it by the wavelength, 525 times 10 to the negative 9, and we get 3.78 times 10 to the negative 19. So, let me go ahead and write that down here. 3.78 times 10 to the negative 19, and if you did you units up here, you would get joules, and so let's think about this number for a second, 3.78 times 10 to the negative 19 is the energy of the photon. And that energy of the photon is greater than the work function, which means that that's a high-energy photon. It's able to knock the electron free, 'cause remember, this number right here, is the minimum amount of energy needed to free the electron and so we've exceeded that minimum amount of energy, and so we will produce a photoelectron. So, this photon is high-energy enough to produce a photoelectron. So let's go ahead and find the kinetic energy of the photoelectron that's produced. So we're gonna use this equation right up here. So let me just go and get some more room, and I will rewrite that equation. So we have the kinetic energy of the photoelectron, kinetic energy of the photoelectron, is equal to the energy of the photon, energy of the photon, minus the work function. So let's plug in our numbers. The energy of the photon was 3.78 times 10 the negative 19 joules, and then the work function is right up here again, it's 3.43, so minus 3.43 times 10 to the negative 19 joules. So let's get out the calculator again. So, from that we're going to subtract the work function 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20. So let's go ahead and write that. This is equal to 3.5 times 10 to the negative 20 joules. This is equal to the kinetic energy of the photoelectron, and we know that kinetic energy is equal to one half mv squared. The problem asked us to solve for the velocity of the photoelectron. So all we have to do is plug in the mass of an electron, which is 9.11 times 10 to the negative 31st kilograms, times v squared. This is equal to 3.5 times 10 to the negative 20. So, let's do that math. So we take 3.5 times 10 to the negative 20, we multiply that by 2, and then we divide by the mass of an electron, 9.11 times 10 to the negative 31st, and this gives us that number, which we need to take the square root of. So, square root of our answer gives us the velocity of the electron, 2.8 times 10 to the 5th. So if you look at your decimal place here, this'll be one, two, three, four, five, so 2.8 times 10 to the 5th meters per second. So here's the velocity of the photoelectron produced, 2.8 times 10 to the 5th meters per second, and if you increased the intensity of this light, so you had more photons, they would produce more photoelectrons. So one photon knocks out one photoelectron if it has enough energy to do so. So let's think about this same problem, but let's change the wavelength. So, what if your wavelength changed to 625 nanometers. So what would happen now? Well, to save time, I won't do the calculation, but all we would have to do is plug in 625 up here. So instead of 525, just plug in 625 to calculate your energy, and if you did that, so if you used 625 times 10 to the negative 9 here, I'll go ahead and give you the answer just to save some time, you would get 3.2 times 10 to the negative 19 joules. And that is lower than the work function. So let me go ahead and highlight that here. So this number is not as high as the work function. The work function was how much energy we needed to free that electron, and since this is lower than the work function that means we do not get a photoelectron. So, you have to have a high enough energy photon in order to produce a photoelectron. It wouldn't even matter if we increased the intensity. So if we had more and more and more of these photons at this wavelength, we still wouldn't produce any photoelectrons. And so, this is the idea of the Photoelectric effect, which is best explained by thinking about light as a particle.