The Photoelectric Effect

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a plate of a certain metal, and shine a very bright red lamp on it. Really bright. Not one electron comes out. Now switch to a dim violet lamp, quite weak. And electrons fly out at once! How can a weak light do what a strong light cannot? Strange, right? Today, Einstein's one line equation will solve this puzzle. And look, JEE questions on threshold wavelength, stopping potential and graphs come from that one line. So, let's begin.

0:37When do electrons come out?

So, here is the set-up. Light of frequency nu falls on a metal; nu is the Greek letter that looks like a v, remember? Electrons that come out because of light are called photoelectrons. Now tell me. Will light of any frequency knock electrons out? Or is there some condition? Yes, there is a condition! Electrons come out only if nu is greater than or equal to nu naught, the threshold frequency. Nu naught is nu with a small zero below it, and this sign means greater than or equal to, okay? Now, the trap. Suppose nu is below nu naught, and I make the light very, very bright. Now will electrons come out? No, not even one! Below nu naught, no electrons, however intense the light. That is our bright red lamp: its frequency was simply below this metal's threshold. And above nu naught? Emission starts at once, even with a dim lamp. That is our dim violet lamp: violet has a higher frequency than red. So frequency decides, not brightness. That's it, puzzle solved! Now let's see why.

1:46Einstein's equation

So, Einstein's equation. K max equals h nu minus phi naught. K max, K with a small max below it, is the maximum kinetic energy of the photoelectrons. h nu, Planck's constant h times nu, is the energy of one photon. And phi naught, this Greek letter phi, a circle with a line through it, with a small zero, is the work function. So: photon energy, minus the work function, and the rest is kinetic energy. Just like the bond breaking in our last class, right? Now, the work function phi naught equals h nu naught: the photon energy at the threshold frequency. And since c, the speed of light, equals nu lambda, it is also h c over lambda naught, where lambda naught, lambda with a small zero, is the threshold wavelength, okay? Now you try. Put phi naught equals h nu naught into the first line. What do you get? Yes, exactly! K max equals h times, in brackets, nu minus nu naught; we took h common. Now look: if nu is below nu naught, the bracket is negative. Can kinetic energy be negative? Never! So no electrons come out. The threshold comes straight from the equation. Got it? One last piece. We apply a retarding potential, the stopping potential V naught, that just stops the fastest electrons. Then e V naught equals K max, where e is the magnitude of the electron's charge. Clear?

3:18Brightness or frequency? Read the two graphs

Next, the graphs. Keep the frequency the same, but make the beam brighter. Do the electrons come out faster, with more kinetic energy? No! K max depends on the frequency, not on the intensity, the brightness. See the right graph: current I, against voltage V, okay? Blue is the bright beam, orange the dim one, and both drop to zero at the same point, minus V naught. Same stopping potential, so the same K max. Exactly! So what does brightness change? The number of electrons. The flat top of each curve is the saturation current, and it is proportional to the intensity; this curly sign means proportional to. Brighter beam, taller curve, okay? Now the left graph: K max against nu. Look at the equation like y equals m x plus c. What is the slope of this line? Yes, it is h, Planck's constant! K max against nu is a straight line of slope h. And the slope is h for every metal, so the lines for metal one and metal two are parallel. Remember that. Where does it meet the axes? Where K max is zero, nu equals nu naught, so the line meets the frequency axis at nu naught. Extend it backwards, the dashed part, and it meets the energy axis at minus phi naught. Questions often give this graph and ask for the work function, so read the intercepts carefully, okay? And one fact to remember. Alkali metals, especially caesium, have the smallest work functions, the smallest phi naught. Small phi naught means small nu naught, right? So these metals respond even to visible light. You will meet one in example one.

5:09Example 1: threshold wavelength of sodium

Right, example one. The work function of sodium is two point three electron volts, written e V on screen. Find its threshold wavelength. Pause the video, and try it first. Done? Okay, let's check. Phi naught equals h c over lambda naught. The double arrow means this gives: lambda naught equals h c over phi naught. But phi naught is given in electron volts. Which shortcut from Planck's class saves us from h and c? Yes, twelve forty! Lambda naught in nanometres equals twelve hundred and forty over phi naught in electron volts. The brackets give the units. Same shortcut, turned around. So, twelve hundred and forty over two point three: about five hundred and thirty nine nanometres. The wavy equals sign means approximately. That lies between four hundred and seven hundred and fifty nanometres: visible light. And sodium is an alkali metal, the kind that responds to visible light. It all fits, doesn't it?

6:14Example 2: KE, stopping potential and speed for potassium

Now a harder one, JEE level. Light of wavelength three hundred nanometres falls on potassium, with work function phi naught two point two five electron volts. Find the maximum kinetic energy, the stopping potential, and the maximum speed of the photoelectrons. Pause and try. Hint: work in electron volts first. Done? Okay, let's check. First, the photon energy E, with our shortcut: twelve hundred and forty over three hundred. About four point one three electron volts. Now Einstein: K max equals four point one three minus two point two five. One point eight eight electron volts. With me? Next, the stopping potential. e V naught equals K max. So what is V naught, in volts? One point eight eight volts, exactly! Topper's trick: since e V naught equals K max, K max in electron volts and V naught in volts are the same number. And not four point one three: the work function is paid first. Last, the maximum speed. Kinetic energy is half m v squared, where m is the mass of the electron. So v equals the square root of two K max over m. Now, many students put one point eight eight straight into this formula. Will that work? No, no, careful! That is the trap. One point eight eight is in electron volts, but for metres per second we need joules. So multiply by one point six times ten to the power minus nineteen, and divide by the electron's mass, nine point one times ten to the power minus thirty one. About eight point one times ten to the power five, in m s to the minus one: metres per second.

8:04Quick revision: Einstein equation, intensity vs frequency

So, quick revision. Electrons only if nu is at least nu naught; brightness cannot make up for low frequency. K max equals h nu minus phi naught, or h times nu minus nu naught; phi naught is h nu naught, or h c over lambda naught; and e V naught equals K max. Intensity changes the current, not K max, and K max against nu has slope h. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!