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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.

## Want to join the conversation?

• 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??? • if you are talking about classiscal physics . as we know that the photon is massless then how the momentum is being conserved? •   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.
• 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? •  if the frequency is exactly equal to the work function of the metal, then the incident ray will have energy to only free the electron from the metal surface, but the electron wont have any kinetic energy to move with.
• 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? •  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!
• 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. • 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.
• 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? • 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.
• 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? • 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.
• 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? • 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??  