Electron Emission and the Work Function

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Copper is packed with free electrons; that is why it conducts so well. Now shine bright visible light on a clean copper plate. How many electrons come out? Not a single electron comes out. Shine the same light on caesium, and electrons come flying out. Same light, different metal. So what decides it? The answer is one number that every metal carries, its work function. It comes up often in JEE, so let's begin.

0:34What holds an electron inside a metal?

So, inside a metal, the outer electrons are free to wander around. Then let me ask you this: if they are so free, why don't they simply fly out of the surface? Because the metal pulls them back, yes, exactly! The positive ions of the metal attract them and hold them inside. Okay? So, to get an electron out, we must pay some energy. The least energy that takes an electron just out of the surface is the work function of the metal. We write it phi naught: the Greek letter phi, with a small zero below it. Right? It is usually stated in e V, the electron volt. One electron volt equals one point six times ten to the power minus nineteen joules. The cross is the times sign, the small raised minus nineteen is the power of ten, and capital J stands for joule. Clear? And notice, it depends on the metal, and also on the state of its surface; a thin layer of oxide or dirt, for example, can change it.

1:35Three ways to supply the energy

Got it? Now, how do we hand an electron this energy? There are three ways. The first is heating the metal. This is thermionic emission, the way electrons leave the hot filament of a vacuum tube. Okay? The second is a very strong electric field at the surface, about ten to the power eight volts per metre; the V and the m with the power minus one mean volts per metre. This is field emission. With me? And the third is shining light of suitable frequency on the metal. This is photoelectric emission, the heart of this unit, and the electrons it frees are called photoelectrons. Right?

2:13Typical work functions, in eV

Now some real values, in electron volts. C s is caesium, at two point one four, K is potassium, at two point three zero, and N a is sodium, at two point seven five. Okay? Next row: calcium, C a, three point two zero; aluminium, written A L, four point two eight; copper, C u, four point six five; and platinum, P t, five point six five. Right? Now look at the first row: caesium, potassium and sodium. What family do these three belong to? The alkali metals, that is correct! They have the lowest work functions, so even visible light frees their electrons, while most other metals need ultraviolet light. Got it?

3:01The condition, and the threshold

So, how does light free an electron? Light comes in photons, each carrying energy h nu, where h is Planck's constant and nu, the Greek letter that looks like the letter v, is the frequency. One photon frees an electron only if h nu is greater than or equal to phi naught; that is the sign on screen. Okay? Now back to our copper plate. Visible light gives no electrons, so just make the light much brighter, and they will come out, right? No, no, wait, that is the trap! Brighter light means more photons, not more energetic ones. Each visible photon still has h nu less than phi naught for copper, so no electron comes out, however bright the light is. Clear? So, the smallest frequency that works is the threshold frequency, nu naught, with a small zero below the nu. Set h times nu naught equal to phi naught, and nu naught equals phi naught over h. With me? The matching threshold wavelength is lambda naught, with lambda the Greek letter for wavelength. It equals h c over phi naught, where c is the speed of light. Now a topper's trick: h c is about twelve hundred and forty electron volt nanometres. So lambda naught in nanometres is just twelve hundred and forty over phi naught in electron volts, with no powers of ten at all. Okay? Now here is a quick question. The frequency must be at least nu naught, so must lambda be bigger than lambda naught, or smaller? It must be smaller, that is correct! Higher frequency means shorter wavelength. So nu at least nu naught means lambda less than or equal to lambda naught. Students often flip this second sign, so be careful. Got it?

4:52Example 1: threshold frequency and wavelength of caesium

Let's try a problem together. Find the threshold frequency and the threshold wavelength of caesium, whose work function, phi naught, is two point one four electron volts. Pause the video, and try it first. Done? Okay, let's check. We will divide by h, which is in S I units, so first change phi naught into joules. Two point one four times one point six times ten to the power minus nineteen gives three point four two times ten to the power minus nineteen joules. Okay? Now, nu naught equals phi naught over h, that is, three point four two times ten to the power minus nineteen, over six point six three times ten to the power minus thirty four. This gives five point one six times ten to the power fourteen hertz; H z is hertz, which means per second. Right? For the wavelength, use our trick. Twelve hundred and forty over two point one four gives five hundred and seventy nine nanometres. So any wavelength up to that frees electrons from caesium, including plenty of visible light. Clear?

6:04Example 2: which of Na, Cu, Pt responds to visible light

Now a slightly harder one. Find the threshold wavelengths of sodium, at two point seven five electron volts, copper, at four point six five, and platinum, at five point six five. Then tell me: which of them responds to visible light? Pause the video, and try it first. Done? Okay, let's check. For sodium, twelve hundred and forty over two point seven five gives four hundred and fifty one nanometres. Okay? For copper, twelve hundred and forty over four point six five gives two hundred and sixty seven nanometres. Right? And for platinum, twelve hundred and forty over five point six five gives two hundred and nineteen nanometres. So, the bigger the work function, the shorter the threshold wavelength. With me? Now comes the real question. Which of these three metals can visible light free electrons from? To decide, we need the visible range. Visible photons carry about one point eight to three point one electron volts. The most energetic, at three point one, has wavelength twelve hundred and forty over three point one, which is four hundred nanometres. So for visible light, lambda is greater than or equal to four hundred nanometres; it is never shorter. Okay? Emission needs lambda less than or equal to lambda naught. Sodium's threshold, four hundred and fifty one nanometres, is greater than four hundred, the greater-than sign on screen, so some visible light frees its electrons. Copper and platinum need wavelengths below four hundred, that is, ultraviolet. So the answer is only sodium. And our opening puzzle is solved: copper asks for more energy than any visible photon can give. Got it?

7:55Quick revision: work function, h nu vs phi0, 1240 shortcut

So, quick revision. The work function, phi naught, is the least energy needed to free an electron from a metal. Emission happens only if h nu is at least phi naught, and brighter light cannot change that. And lambda naught in nanometres, with the units in brackets, is twelve hundred and forty over phi naught in electron volts, and we need lambda less than or equal to lambda naught. Now go and try the practice problems on this page. All the best!