Photocurrent, Efficiency and the Photocell

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Light falls on a photocell, and a current flows. Now you switch to light of half the wavelength, with the same intensity, and both lights are above the threshold. Each photon carries twice the energy, so the current should go up, right? Keep that guess in mind, because by the end of this video we will check it properly, and the answer may surprise you. It comes up often in JEE, so let's begin.

0:29Counting photons, then electrons

So, let us count. Light falls on the emitter with some power, so a fixed amount of energy arrives every second. And each photon brings its own small packet of energy. So how many photons arrive every second? Yes, exactly, just divide the power by the energy of one photon! N, the number of photons per second, equals P, the power, over h nu, where h is Planck's constant and nu, the Greek letter like a v, is the frequency. That is also P times lambda, over h c. Lambda, the Greek letter like an upside down y, is the wavelength, and c is the speed of light. Okay? But not every photon frees an electron that reaches the collector. Only a fraction does, and that fraction is the quantum efficiency, eta, the Greek letter like an n with a long tail. With me? So eta times N electrons are collected every second, each with charge e, and current is charge per second. So the saturation current, i, equals eta times N times e, which is eta times P lambda over h c, times e. It is the saturation current, when every emitted electron is collected. Got it?

1:43Point source: the inverse-square law

Now, often the light comes from a point source, of power P with the small word source below it. Our cathode has area capital A, at a distance r from the source. What part of the source's power falls on the cathode? At distance r, the power spreads over a sphere of area four pi r squared, and the cathode catches capital A out of it. So P equals P source times capital A, over four pi r squared. So i is proportional to one over r squared; the sign like an open alpha means is proportional to. Clear? At double the distance, two squared is four, so we get a quarter of the photons and a quarter of the current. Okay? Now here is a trap that catches many students. The light is weaker now, so does the stopping potential fall too? No, it does not change at all! Each photon still brings the same h nu, so the fastest electrons are just as fast. We get fewer photons, not weaker ones.

2:45Same power, higher frequency

Right? Now, back to our puzzle. Keep the power the same, but raise the frequency. Each photon, with energy h nu, now carries more energy. Okay? The same energy arrives every second, but in bigger packets. So do we get more photons per second, or fewer? Fewer photons, that is right! Fewer photons give fewer electrons, so a smaller saturation current, as long as eta stays the same. Got it? But each electron now gets more energy, so the stopping potential goes up: e times V naught equals h nu minus phi naught. V naught, with the small zero read as naught, is the stopping potential, and phi naught, with the Greek letter phi, is the work function. Clear? So our puzzle is solved. Half the wavelength gives each photon twice the energy, so at the same intensity, only half as many photons arrive. The current is halved, not doubled, as long as both lights are above the threshold and eta is the same. With me?

3:49The photocell and its uses

Now, the device that uses all this: the photocell, or photoemissive cell. It is an evacuated bulb, like the tube in the figure. Okay? Inside are a photosensitive cathode and a collecting anode. In the figure, light enters through the quartz window onto C, the emitter, our cathode, and electrons fly across to A, the collector, our anode. The meter marked mu A, a microammeter, reads the current, while the battery and the voltmeter, V, set and measure the anode's potential. Right? Now make the light on the cathode brighter, at the same frequency. What happens to the current? It goes up, in step with the light! More photons per second give more electrons per second, so the current is proportional to the intensity. Got it? That makes the photocell very useful: in light meters in cameras, in automatic door openers, and in burglar alarms. Okay? It is also used in counting devices, for switching street lamps, and to reproduce sound from the sound track of a cinema film. Clear?

4:55Example 1: saturation current, 1 mW of 400 nm light

Let's try the first example. Light of wavelength four hundred nanometres, written n m, and power one milliwatt, written m W, falls on a photocathode. One photon in a thousand ejects an electron. Find the saturation current. Pause the video and try it first. Done? Okay, let's check. N equals P times lambda, over h c: ten to the power minus three, times four times ten to the power minus seven, over one point nine eight nine times ten to the power minus twenty five, which is h times c in joule metres. That gives two point zero one times ten to the power fifteen, and the s to the power minus one on screen means per second. Okay? Now many students multiply this by e, and write zero point three two milliamperes. Is that the answer? No, no, careful, that forgets the efficiency! Only one photon in a thousand counts, so eta is ten to the power minus three, and eta times N is two point zero one times ten to the power twelve electrons per second. Right? Now multiply by e, one point six times ten to the power minus nineteen coulombs. We get three point two times ten to the power minus seven amperes, which is zero point three two microamperes. Got it?

6:19Example 2: current from a 5 W point source 0.5 m away

Now let us try a harder one. Take a point source that emits five watts of light, with photons of three point one electron volts, written e V on screen. Facing it, zero point five metres away, is a photocell with a cathode area of two square centimetres, and one photon in a thousand ejects an electron. Find the saturation current. Pause the video and try it first. Done? Okay, let's check. Let me ask you this first: do all five watts fall on our cathode? Of course not, only its own share. Two square centimetres is two times ten to the power minus four square metres, so P equals five times that, over four pi times zero point five squared. That denominator is just pi, so P is three point one eight times ten to the power minus four watts. Okay? Next, one photon has three point one times one point six times ten to the power minus nineteen joules. Dividing, we get six point four two times ten to the power fourteen photons per second. With me? Only one photon in a thousand ejects a collected electron, so that gives six point four two times ten to the power eleven electrons per second. Right? Finally, multiply by e, and we get one point zero times ten to the power minus seven amperes, which is zero point one zero microamperes. That is the whole lesson in one problem! Clear?

7:52Quick revision: i = eta N e, frequency and 1/r squared rules

So, quick revision. First, i equals eta times N times e, so count the photons first, then the electrons. Second, the same power at a higher frequency means fewer photons and a smaller current, but a larger stopping potential. And third, in a photocell the current is proportional to the intensity, and for a point source, to one over r squared. Now go and try the practice problems on this page. All the best!