Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a zinc plate and shine a very bright red lamp on it. Not a single electron comes out. Now switch to a faint ultraviolet lamp, and electrons come out at once! But brighter light carries more energy, right? So how can the dim lamp win? Hmm, keep your guess ready. Today we see how to use this unit's problem bank, settle this puzzle with one of its concept checks, and solve two of its problems. Questions on the photoelectric effect come up often in JEE, so let's begin.

0:39How to use this chapter

So, what is in this chapter? Part A has eighteen solved problems, at JEE Main level. And they mix the whole unit: photons, the photoelectric effect, and matter waves. Right? Then Part B has fourteen harder problems, at JEE Advanced level. With me? And Part C has twelve concept checks. They test the ideas behind the formulae, the why questions, like our zinc puzzle. After that come twelve practice problems, at the end of Part C, for you to solve alone. Got it? Now tell me, when should you read a solution? Straight away, or only after you have tried the problem yourself? Only after you try, yes, exactly! The course says it clearly: attempt each part as a test, and only then read the solutions. Clear?

1:31Part C: why can't bright red light free electrons?

Now a taste of Part C, and our puzzle. Capital E, the energy of one photon, equals h nu, which equals h c over lambda. Here h is Planck's constant, nu, the Greek letter that looks like a v, is the frequency, lambda, the Greek letter like an upside down y, is the wavelength, and c is the speed of light. In the figure, the wavy orange line is a photon of energy h nu. Okay? And one photon gives all of its energy to one electron, the blue arrow in the figure. Right? But the metal holds the electron in. To escape, it must spend at least phi naught, the work function, the gap you see below the surface. Phi is the Greek letter drawn as a circle with a line through it, and the small zero below is read as naught. What is left is K max, the kinetic energy, with the small max below for maximum. So K max equals h nu minus phi naught, which also equals e times V naught. Here e is the charge of the electron, and V naught is the stopping potential. Clear? Let me ask you this. What if one photon brings less energy than phi naught? Can a brighter light still get electrons out? No, there is no emission at all, however intense the light! The sign between h nu and phi naught is the less-than sign. Below this threshold, not a single electron comes out. The frequency at this cut-off is the threshold frequency, nu naught, the nu with a small zero you see in the figure. With me? Then what does brightness do? For the same frequency nu, brighter light has more photons, so more electrons come out every second. But each photon is the same, so K max stays the same. Got it? Now, back to our zinc plate. Why does the faint ultraviolet lamp win over the bright red one? Yes, exactly, it is all about one photon at a time! Each red photon has less energy than the work function of zinc, and brightness only adds more such weak photons. Each photon of this ultraviolet light has a much higher frequency, high enough that h nu is greater than phi naught of zinc; that is the greater-than sign. So frequency decides whether electrons come out, and intensity decides how many. Okay?

3:55Example 1: max kinetic energy for 350 nm light on sodium

Now, Example one, an easy problem from Part A, to warm up. Light of wavelength three hundred and fifty nanometres, written n m, falls on sodium, whose work function phi naught is two point seven five electron volt, written as e V on screen. Find the maximum kinetic energy of the photoelectrons. Pause the video and try it first. Done? Okay, let's check. First, a topper's trick from the course: work in electron volt. h c equals twelve hundred and forty electron volt nanometres. So the photon energy in electron volt is twelve hundred and forty over the wavelength in nanometres, and the powers of ten disappear. Right? Here, twelve hundred and forty over three hundred and fifty gives three point five four electron volt. Okay? Before we use Einstein's equation, one quick question. What must we check first? We must check that emission happens at all, that is correct! The photon energy must be more than the work function. Three point five four is greater than two point seven five, so photoelectrons are emitted. Clear? Now we just subtract: K max equals three point five four minus two point seven five, which is zero point seven nine electron volt. That is the answer, no joules needed. With me? And by the same trick, V naught in volt equals K max in electron volt, so the stopping potential here is zero point seven nine volt. Got it?

5:27Example 2: work function from a 3:1 ratio of max speeds

Now Example two, a harder one from Part B, at JEE Advanced level. Light of wavelengths three hundred nanometres and six hundred nanometres falls in turn on a metal. The maximum speeds of the photoelectrons are in the ratio three is to one. Find the work function. Pause the video and try it first. Done? Okay, let's check. Same trick, photon energies in electron volt. E one, the E with a small one below, is twelve hundred and forty over three hundred, which gives four point one three three electron volt. And E two is twelve hundred and forty over six hundred, which gives two point zero six seven electron volt. Okay? Now, the speeds are in the ratio three is to one. So the kinetic energies are also three is to one, right? No, no, wait, that is the trap in this question! K equals half m v squared, so K is proportional to v squared; that sign, a bit like the Greek letter alpha, means is proportional to. So K one over K two is three squared, which is nine, and not three. With me? Now Einstein's equation for each light: K max is the photon energy minus phi naught. So the ratio, four point one three three minus phi naught, over two point zero six seven minus phi naught, equals nine. Right? Cross multiplying, four point one three three minus phi naught equals nine times two point zero six seven, which is about eighteen point six, minus nine times phi naught. Clear? Now collect the phi naught terms. Eight times phi naught equals eighteen point six minus four point one three three, which is fourteen point four seven. So phi naught is one point eight one electron volt, and that is the answer! Now a quick check: does the six hundred nanometre light free electrons? Its photon energy, two point zero six seven, is greater than one point eight one, so yes, both wavelengths eject electrons. Always finish with a check like this; it keeps you safe.

7:34Quick revision: Einstein equation, intensity vs frequency

Got it? So, quick revision. Part A is at JEE Main level, Part B at JEE Advanced level, and Part C checks the ideas; attempt each as a test first. K max equals h nu minus phi naught, which equals e times V naught, with no emission if h nu is less than phi naught, however intense the light. Intensity controls the number of electrons; frequency controls their maximum kinetic energy. Photon energy in electron volt is twelve hundred and forty over the wavelength in nanometres, and since K goes as v squared, square a speed ratio. Now open Part A, solve it like a test, and keep going right up to the practice problems. All the best!