Photoelectric effect and light coming in chunks

I was watching a lecture on introduction to quantum mechanics.While explaining the photoelectric effect,the lecturer mentioned that we must think of light as coming in chunks with each chunk having definite energy.So where did he use the fact of the chunks of light to explain the photoelectric effect?The only thing that i understood is that the energy given to the system by the light is proportional to the frequency and not to the intensity,but i do not understand how the theory that light comes in chunks affects this result of Einstein's conclusions.

This is going to take a while. First, you need to know about the photoelectric effect.

Take a fresh surface (no oxidation, so you really need a good vacuum) of many metals. Shine light on it. You'll get electrons being emitted. OK, this sort of makes sense - you've got surface atoms that get knocked around ("excited") by the energy in the light, and if they get to bouncing around hard enough, electrons will get knocked loose. Makes sense, right?

Well, not when you look at it. Let's take potassium as an example. Shine white light on it and you get a photocurrent - you set up a metal plate near the potassium and collect the emitted electrons. More light, more current. Sounds reasonable?

Now shine only red light on the surface. You don't get any electrons. Orange? Nope. Yellow? Well, about the point where yellow shades to green (wavelength of 555 nm) you start to get electrons. But let's look at what happens with red light. If the energy from the light is what knocks out the electrons, if you add more energy surely you'll start getting some emission. So crank up the intensity. Guess what? No matter how strong the red light (within limits, like an order of magnitude more than the white light delivered) you don't get any electrons at all. Same for orange. If you use very pure light, like a laser, any wavelength under that 555 nm has no effect.

Now let's look at the electrons that do get emitted. How fast are they going/how much energy do they have? You can measure this by putting a voltage between the two plates, applying a reverse voltage and measuring the current. Increase the voltage enough and you can stop all the electrons, and that determines the maximum energy. If the electrons are getting knocked out by the arrival of waves, you expect the velocity to follow something like a bell curve, with a central average, and lesser numbers as you go faster or slower. That's how thermal emission works, for instance.

Is this true for your electrons? Nope. What happens is that you get a maximum velocity. And then you notice something else. As you shorten the wavelength, the maximum energy goes up in synchrony. As a matter of fact, for wavelengths less than the threshold, the maximum energy is exactly proportional to the reciprocal of the wavelength, which for light is equivalent to the frequency with a factor of c thrown in. Turning up the brightness of the source increases the total current, but it does not affect the peak energy.

If the electrons are being emitted by classical light waves exciting the surface atoms, none of this makes sense. You make stronger waves, you get more energy exciting the atoms. But now let's change our mental model. Let's say that the electrons are held to their atoms by bonds of a particular strength. Let's also say that light isn't a wave, it's a bunch of little bullets (called photons), each of which has an energy proportional to its classical wavelength. This energy is specific to a photon, or quantized. It isn't a function of the total wave (in classical terms). Then it gets simple. If a photon has less energy than the electron bonds, it just gets absorbed. If it has more, the bond is broken, and the electron is kicked loose with an energy equal to the difference of the photon and the bond. More photons, more electrons emitted, and the current goes up, but the energy of the emitted electrons never gets bigger than the difference between the photon energy and the bond energy.

And thus quantum is born.

• Can i ask something else just to clarify the things that you say?Well,i have a question about the current and the energy.If we have a battery in a circuit,the higher its potential the greater the current.So in this situation,when we have light of higher frequency then we have more photons giving energy to electrons at a given time(the chunks that i mentioned) and then if the frequency is high enough the electrons get kicked loose.So there is the energy.I do not understand what the intensity(brightness) has to do with the current. Commented May 31, 2015 at 8:45
• i mean,if the energy from the photons is like the energy given by a battery in a circuit,then the energy of the photons is the one that determines the current.I do not understand what the intensity-brightness changes.Can you explain to me by giving an analogy to a circuit with a battery?Thank you Commented May 31, 2015 at 8:47
• Nope. That's the point. The photoelectric effect does not behave like a battery and a resistor. Brightness (intensity) is just what it sounds like. It's measured in power per unit area. Shine your light through an aperture and you get a certain amount of power (energy per second) on the target. Cut the aperture size in half, and you get half the brightness. If the frequency is above threshold, you get half the current. If the frequency is below, no current regardless of brightness... Commented May 31, 2015 at 11:18
• And "when we have light of higher frequency then we have more photons giving energy" is not right. For a given brightness, higher frequency means fewer photons. Brightness is energy per unit time. If each photon has certain amount of energy, increasing the individual energy while keeping the total energy (brightness) the same means fewer photons per second. Commented May 31, 2015 at 11:20
• So,why do we make the assumption that only ONE photon transfers energy to an electron? Commented Jun 3, 2015 at 14:23

When the light shining on the metal was red, no electrons were emitted, no matter how bright to light was. When it was violet light, electrons were emitted, but no matter how bright to light was, there kinetic energy did not increase, just there number.

One way to explain this is the particle view of light. An electron can only absorb one photon at a time. If it is a red photon, it simply does not have enough energy. The additional photons can not change that because only one photon can be absorbed by an electron at a time. The violet photons have enough energy, but having more (i.e. shining the light more brightly) photons will not increase the kinetic energy of the electrons because they ca only absorb one photon at a time.

This was not his only evidence, but it is the most compelling.

• The particular lecture was introductory to explain from where we knew that something was wrong with light.Having this in mind, does your answer explain that the lecturer mentioned the photoelectric effect to point out that we cannot perceive light as a wave but as electrons in order to explain the photoelectric effect? Commented May 29, 2015 at 20:03
• Because before your explanation I can not understand why the photoelectric effect is used as an introduction to quantum mechanics Commented May 29, 2015 at 20:04
• You can use the wave model of light to explain sight, so this is a better example. Commented May 29, 2015 at 20:05
• I dont understand how sight has to do with photoelectric effect Commented May 29, 2015 at 20:32
• @LandosAdam Because it introduces the learner to a fundamental concept of QM: energy is quantitized (comes in chunks). Commented May 29, 2015 at 21:11