Isolated Conductors and Other Applications

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a metal sphere sitting all alone, not connected to anything. Shine ultraviolet light on it, nonstop, and the light keeps knocking electrons out. So, will the sphere keep losing electrons forever, until it has none left? Hold on to your answer. Today we find out when the emission stops, how much charge the sphere collects, and how long it takes. This idea does appear in JEE problems, so let's begin.

0:31An isolated metal body charges itself

So, take an isolated metal body, connected to nothing, no wire, no earth. Light falls on it and knocks out electrons, and electrons are negative. So tell me, what charge is left on the body? It becomes positive, yes, exactly! The light ejects electrons, so the body is left positively charged. Okay? Now here is the real question. The light keeps falling. So will the body keep losing electrons forever? No, and here is why. As the body turns positive, its potential, capital V, rises. And a positive body pulls the negative electrons back. Right? So the emission effectively stops when even the fastest electron cannot escape. The fastest electron has K max, the maximum kinetic energy; the small max below the K means maximum. Then e times V final equals K max, which equals h nu minus phi naught. Here e is the size of the electron's charge, and V final, with the small word final, is the last potential the body reaches. With me? And in h nu minus phi naught, h is Planck's constant, nu, the Greek letter like a v, is the frequency, and phi naught, the Greek phi with a small zero, is the work function. Now, e times V equals K max: where have you seen this before? Yes, you have seen it with the stopping potential! e times V naught equals K max, where V naught, the V with a small zero, is the stopping potential. So the body charges up to exactly the stopping potential, and stops there. Clear?

2:14How much charge does a sphere collect?

Next, the charge it collects. For a sphere of radius r, the charge is q equals four pi epsilon naught, times r, times V final. Here epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Okay? In numbers, one over four pi epsilon naught is nine times ten to the power nine, in S I units. So q is simply r times V final, over nine times ten to the power nine. Simple, right? Each electron that left took away a charge e, one point six times ten to the power minus nineteen coulomb, the capital C on screen. So the number of electrons lost is q over e. Got it? Now, here is a trap that catches many students. Keep the same frequency, but make the light twice as intense. Does the sphere now charge up to a higher potential? No, and this is where students slip! V final comes from K max, which depends on the frequency, not on the intensity. Brighter light only gives more electrons per second, so the sphere reaches the same V final, just sooner. Clear?

3:24Another application: a charged zinc plate

Now, the same idea explains a classic experiment. Take a negatively charged zinc plate connected to an electroscope, whose thin leaves spread apart when it is charged. The plate is lit with ultraviolet light. What happens to the leaves? Yes, the leaves collapse! The light knocks electrons out of the plate, so the negative plate loses its charge, and the leaves fall together. Okay? Now the second part. The plate starts positively charged, under the same light. Will the leaves collapse this time? No, they will not. The positive plate keeps its charge, because any electron that is emitted is pulled straight back. Just like our isolated sphere, right?

4:09Example 1: potential of a sphere lit by 310 nm light

Alright, here is example one. An isolated metal sphere has a work function of two point three electron volts, the e V on screen, and is lit by light of three hundred and ten nanometres, the n m. To what potential does it rise? Pause the video and try it first. Have you done it? Okay, let's check. First, the photon energy, using the shortcut h c equals twelve hundred and forty electron volt nanometre, where c is the speed of light. Twelve hundred and forty over three hundred and ten gives four point zero electron volts. With me? Now subtract the work function. K max equals four point zero minus two point three, which is one point seven electron volts. Right? And now the key step. The sphere stops emitting when e times V final equals K max. So what is V final? It is one point seven volts, yes! e times V final is one point seven electron volts, so V final is one point seven volts, the capital V for volt. The double arrow means which gives. See, the answer in electron volts is the potential in volts. Clear?

5:21Example 2: how long emission lasts from a 1 cm sphere

Now a harder one, at JEE level. Light of power one microwatt, written with the Greek letter mu before the W, mu meaning micro, carries photons of five electron volts. It falls on an isolated sphere of radius one centimetre and work function three electron volts, and one photon in ten to the power four ejects an electron. How long does the emission continue? Pause the video and try it. Here is a hint: find the electrons per second first. Okay, let's go step by step. One photon has five electron volts, and one electron volt is one point six times ten to the power minus nineteen joules. So one photon carries eight times ten to the power minus nineteen joules, the capital J on screen. With me? One microwatt is ten to the power minus six joules per second. So the photons per second are ten to the power minus six over eight times ten to the power minus nineteen, which gives one point two five times ten to the power twelve. Okay? Now remember, only one photon in ten to the power four ejects an electron. So how many electrons come out each second? Yes, only one photon in ten to the power four counts. One point two five times ten to the power twelve, over ten to the power four, gives one point two five times ten to the power eight electrons per second. Right? Next, K max is the photon energy minus the work function, five minus three, which is two electron volts, so V final is two volts. The radius is one centimetre, that is zero point zero one metre. So q equals zero point zero one times two, over nine times ten to the power nine, which is about two point two times ten to the power minus twelve coulomb. Clear? So the number of electrons lost is q over e. Using the unrounded charge, that comes to about one point three nine times ten to the power seven electrons. Got it? Now, we know the electrons to lose and the electrons leaving each second, assuming they leave at a steady rate. So how long does the emission last? Exactly, we just divide! The time is one point three nine times ten to the power seven over one point two five times ten to the power eight, which gives about zero point one one second, the small s. The wavy equals sign means approximately. So, our opening puzzle: the sphere does not lose electrons forever; here it stops after about a tenth of a second!

8:03Quick revision: stopping potential, charge, electrons lost

So, quick revision. An isolated body under light loses electrons and turns positive. It stops when e times V final equals K max, that is, h nu minus phi naught, the stopping potential. For a sphere, q is four pi epsilon naught, times r, times V final, and the electrons lost are q over e. Now go and try the practice problems on this page. All the best!