Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Light falls on a metal, and the stopping potential is one volt. Now we halve the wavelength, so every photon carries double the energy. And the stopping potential jumps to four volts! Four times, when the photon energy only doubled? Strange, right? And from just these two readings, we can find the work function, and even the wavelength. It is one of this chapter's quick JEE Main style problems on dual nature. So, let's begin.

0:34How this chapter works

So, how does this chapter work? There are three timed sets, with fifteen original problems in each, in the style of recent JEE Main papers. Okay? Now, the course gives each set twenty five minutes. So how much time do you get for one problem? About one hundred seconds a problem, that is correct! So set a timer, and only then check the solutions. Right? Now the constants. h, Planck's constant, is six point six three times ten to the power minus thirty four joule second, written J s; the cross means times. And m sub e, the m with a small e below, is the electron's mass, nine point one times ten to the power minus thirty one kilogram, written k g. With me? And the most useful one: h c equals twelve hundred and forty electron volt nanometre, written e V n m, where c is the speed of light. So work in electron volt, and most powers of ten disappear. Clear?

1:37Tool 1: photons and Einstein's equation

First tool, the photon. Its energy, capital E, equals h nu, which equals h c over lambda. Nu, the Greek letter like a v, is the frequency, and lambda, the Greek letter like an upside down y, is the wavelength. So in electron volt, it is twelve hundred and forty over lambda in nanometres, with e V and n m in brackets for the units. Okay? Now look at the figure. The wavy orange line is a photon of energy h nu, and it gives all its energy to one electron, the dark blue dot, one of the most loosely bound electrons in the grey metal. To escape, the electron spends phi naught, the work function, shown as the gap up to the surface; phi is the circle with a line through it, and its small zero is read as naught. Whatever is left is K max, the K with a small max below, the maximum kinetic energy, and the blue arrow carries it away. So the figure writes h nu equals phi naught plus K max. With me? So K max equals h nu minus phi naught. And K max also equals e V naught, where V naught is the stopping potential and e is the electron's charge. The figure's second line also writes it as h times, nu minus nu naught, where nu naught is the threshold frequency. Right? Now let me ask you this. What if h nu is less than phi naught? Will a very bright light still pull out an electron? Not a single electron comes out, however intense the light, that is the key point! So below the threshold frequency, nu naught, equal to phi naught over h, there is no emission at all. Got it? Now a quick one from Set one. The intensity of the light on a photocell is tripled, without changing its frequency. What happens to the saturation current, and to the stopping potential? The current triples, but the stopping potential stays the same, that is correct! Brighter light brings more photons each second, so more electrons, not faster ones, and the same V naught. With me?

3:49Tool 2: the stopping potential graph

Next, the graph. Divide Einstein's equation by e, and V naught equals h over e, times nu, minus phi naught over e. On the left of the figure, metals A and B give parallel lines, because the slope, h over e, is the same for every metal. Okay? Each line cuts the frequency axis at its threshold frequency: nu naught, or for metal B, nu naught prime, with the little dash. Produced back, the dashed parts cut the V naught axis at minus phi over e, with A or B written small below phi. On the right, at a fixed frequency above the threshold, saturation current against intensity is a straight line through the origin; the fish-like sign means proportional to. Clear?

4:35Tool 3: matter waves

Last tool, matter waves. Every moving particle has a de Broglie wavelength: lambda equals h over p, the momentum, which is m times v, mass times speed. With kinetic energy capital K, p is the square root of two m K, so lambda equals h over root of two m K; the tick-shaped sign is the square root. Okay? Now a charge q, accelerated from rest through a potential difference, capital V, gains kinetic energy q V, the charge times the voltage. So lambda equals h over the square root of two m q V, for non-relativistic speeds, well below the speed of light. Right? For an electron, this becomes one point two two seven over root V, in nanometres, with V in volt. But careful, this number is for electrons only. For a proton or an alpha particle the number is different, so use the general form above. Got it? Now, a photon and an electron both have a wavelength. Can we use the same formula for both? No, we cannot, and this is a common trap. For a photon, lambda is h c over capital E, the photon energy. For a particle, it is h over root of two m K, from the momentum. Never swap them. With me?

5:56Example 1: de Broglie wavelength of electron through 225 V

Alright, Example one, from Set one. Find the de Broglie wavelength of an electron accelerated through two hundred and twenty five volts; the capital V after the number is the unit, volt. Pause the video and try it first. Done? Okay, let's check. It is an electron, so our shortcut works: lambda equals one point two two seven over root V, in nanometres. Clear? The square root of two hundred and twenty five is fifteen. So lambda is one point two two seven over fifteen, in nanometres. Right? That gives about zero point zero eight two nanometres; the wavy equals sign means approximately. That took well under a minute.

6:44Example 2: stopping potential 1 V to 4 V, find work function

And now Example two, from Set three, and it is our puzzle. The stopping potential is one point zero volt for light of wavelength lambda, and four point zero volt for wavelength lambda over two. Find the work function, and lambda. Pause the video and try it first. Done? Okay, let's check. Work in electron volt, because the stopping potential in volt equals K max in electron volt. So for the first light, h c over lambda, minus phi naught, equals one. Okay? Now the second light has wavelength lambda over two. What is its photon energy? It doubles, to two h c over lambda, yes, exactly! So two h c over lambda, minus phi naught, equals four. Right? Now subtract the first equation from the second. Phi naught cancels, and h c over lambda equals three electron volt. Put that back in the first equation, and phi naught is three minus one, so the work function is two point zero electron volt. With me? And the wavelength: h c over lambda is three electron volt, so lambda is twelve hundred and forty over three, in nanometres, which gives about four hundred and thirteen nanometres. Clear? So our puzzle is solved. The photon energy went from three to six electron volt; the arrow means goes to. But the two electron volt work function is subtracted first, so K max went from one to four. Got it?

8:16Quick revision: K max in eV, intensity, de Broglie

So, quick revision. Work in electron volt: K max is twelve hundred and forty over lambda in nanometres, minus phi naught, the same number as the stopping potential in volt. Intensity sets the number of electrons, and frequency sets K max. For electrons, lambda is one point two two seven over root V, in nanometres, and for other charges, h over root of two m q V, the general form. Now open Set one, start a twenty five minute timer, and go. All the best!