The Experimental Facts

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Shine a very bright light of a low frequency on a metal plate. Nothing happens. Not one electron comes out, even if you make the light brighter still. Now switch to a very dim light of a higher frequency, and electrons come out almost at once. So how can a dim light do what a bright light cannot? Today we look at the experimental facts of the photoelectric effect, the facts behind this puzzle. Questions on them come up often in JEE, so let's begin.

0:35The photoelectric effect and its apparatus

So, what is the photoelectric effect? When light of a high enough frequency falls on a metal, electrons come out of it. The Greek letter nu, which looks like a v, is the frequency of the light. With me? Now look at the apparatus on screen. Light falls on the plate marked C, the emitter, and the electrons fly across to the plate marked A, the collector. The battery can make the collector positive or negative, and the V in a circle is the voltmeter. The current is read on the micro ammeter, the mu A, where mu, another Greek letter, means micro. Okay? Now let me ask you this. The figure says the tube is evacuated, with the air pumped out, and the window is made of quartz. Why do we need both of these? Yes, exactly, there are two reasons. In air, the electrons would collide with gas molecules and lose energy on the way. And ordinary glass absorbs ultraviolet light, but quartz does not. Got it?

1:40Photocurrent: intensity and saturation

Now the facts, one by one. Keep the frequency fixed, and above the threshold, a minimum frequency we will meet soon. Then the photocurrent, which is set by the number of electrons emitted per second, is proportional to the intensity. The sign between them, a bit like the Greek letter alpha, means is proportional to. Right? Now I make the collector more and more positive, so it pulls the electrons harder. Does the current keep rising forever? No, it does not keep rising. The current rises at first, and then it stays constant, however positive the collector becomes. Clear? This constant value is the saturation current. It stops rising because every electron emitted by the plate is now being collected. Okay?

2:31The stopping potential

Now, reverse the battery and make the collector negative. It now opposes the electrons, and the current falls. At one value, the current becomes zero. That value is the stopping potential, or cut-off potential, written V naught: a capital V with a small zero below it. With me? So here is the question. Just enough reverse voltage stops every electron. What does V naught tell us about them? Yes, it measures their maximum kinetic energy. K max, that is K with max written small below it, equals e times V naught, where e is the magnitude of the electron's charge. We will use this line in both examples. Got it? Now here is a trap that catches many students. I keep the frequency the same, but make the light much brighter. More energy is falling on the plate, so V naught should go up, right? No, no, wait, the stopping potential does not change at all! It does not depend on the intensity. Brighter light gives more electrons per second, so more current, a bigger saturation current, but the same V naught, because the electrons are not more energetic. That is the classic trap here. Clear?

3:50Frequency: the threshold, and no time lag

Now, back to our puzzle from the start. The light has a low frequency, but I make it brighter and brighter. Will electrons come out at some point? The answer is no, never. For each metal there is a threshold frequency, nu naught, that is nu with a small zero. Below it, no electrons are emitted, however intense the light, and that is the bright lamp in our puzzle. Okay? And above nu naught? As the frequency goes up, the stopping potential, and so K max, rises linearly, along a straight line. So the frequency controls the energy of the electrons. Right? And the last fact: there is no time lag. Emission starts within about ten to the power minus nine second, the s on screen, even when the light is very dim. That is the dim lamp in our puzzle: above the threshold, even a dim light gives electrons almost at once. With me?

4:46Why the wave theory fails

So why were these facts such a shock? On the wave picture, the energy of light is spread over the wavefront, and is proportional to the intensity. Now let us see what it predicts. Okay? First, a more intense beam should give electrons of greater energy. The double arrow means leads to. But the stopping potential does not change with intensity, so that prediction is wrong. Second, light of any frequency should cause emission, if it is intense enough. But below the threshold, nothing comes out, so this prediction is wrong too. And third, with dim light, an electron should need a long time to collect enough energy. But there is no time lag. So all three predictions are wrong. The explanation is Einstein's photoelectric equation, the next lesson in this chapter. Clear?

5:37Example 1: K max in eV and joules when V0 = 1.5 V

Now, Example one, an easy one. The stopping potential in an experiment is one point five volts, the capital V on screen. Find the maximum kinetic energy of the photoelectrons in electron volts, written e V, and in joules. Pause the video and try it first. Done? Okay, let's check. K max equals e times V naught, which is e times one point five volts. And e times one volt is one electron volt, so K max is one point five electron volts. Easy, right? Now the same energy in joules. How many joules make one electron volt? One electron volt is one point six times ten to the power minus nineteen joule, the capital J on screen. So one point five electron volts is one point five times that, which gives two point four times ten to the power minus nineteen joule. Clear?

6:36Example 2: maximum speed of photoelectrons when V0 = 2.0 V

Now Example two, a step harder. The stopping potential for a certain surface is two point zero volts. Find the maximum speed of the photoelectrons. Pause the video and try it first. Done? Okay, let's check. The fastest electron has kinetic energy e times V naught, so one half times m times v squared equals e times V naught. Here m is the mass of the electron, and v is its maximum speed. Right? So v equals the square root of, two times e times V naught, over m. The tick-shaped sign is the square root. With me? Now, e times V naught is two electron volts. So can we just put the number two into this formula? Oh no, careful, we cannot put it straight in! The mass is in kilograms, so the energy must be in joules. So e is one point six times ten to the power minus nineteen coulomb, V naught is two point zero volts, and m is nine point one times ten to the power minus thirty one kilogram. Got it? Inside the root, this works out to about seven point zero three times ten to the power eleven. Take the square root, and v is about eight point four times ten to the power five metres per second. The wavy equals sign means approximately, and m s to the power minus one is metres per second. Okay?

8:00Quick revision: saturation, K max = eV0, threshold, no lag

So, quick revision. The photocurrent is proportional to the intensity, and it saturates when every emitted electron is collected. The stopping potential gives K max equals e times V naught, independent of the intensity. Below the threshold frequency there is no emission; above it, K max rises with frequency, and there is no time lag. Now try the practice problems on this page. You can do this. All the best!