Evidence for Matter Waves

Transcript

0:00Introduction

Hello everyone! De Broglie made a bold claim: a moving electron is also a wave. But a claim is not a proof, right? So here is a puzzle: how would you prove that something is a wave? What does a wave do that a tiny ball never does? In nineteen twenty seven, the proof came from a crystal of nickel. This experiment does come up in JEE questions, so let's begin.

0:28Davisson–Germer experiment (1927)

So, here is the experiment, by Davisson and Germer in nineteen twenty seven. On the left of the figure, an electron gun fires electrons, accelerated through a potential difference, capital V, okay? The beam strikes a nickel crystal. Then a movable detector measures the intensity scattered in each direction. Theta, the Greek letter on screen, is the scattering angle. With me? Now look at the result. With V equal to fifty four volts, a sharp maximum appears at a scattering angle of fifty degrees; the small circle up top means degrees. It is the bump in the graph on the right of the figure, okay? Now let us go back to our puzzle. What can make such a sharp peak, in just one direction? It is interference, yes, exactly! The crystal has regularly spaced layers of atoms. The electron waves scattered from these layers add up, crest on crest, in that one direction. That is constructive interference, which only waves can do. Clear? From this diffraction, the wavelength comes out as zero point one six five nanometres. Lambda, the Greek letter, stands for wavelength, and n m for nanometre. Okay? Now comes the real test. What wavelength does de Broglie's formula predict for electrons accelerated through fifty four volts? Use the electron shortcut from the de Broglie lesson. The shortcut gives lambda as one point two two seven over the square root of fifty four, in nanometres; that tick with a long roof is the square root sign. That gives zero point one six seven nanometres. Measured, zero point one six five; predicted, zero point one six seven. This confirmed the wave nature of electrons. Got it?

2:20More evidence, and a use

And there was more evidence. G P Thomson passed electrons through thin metal foils, and obtained diffraction rings. Diffraction again, so once more the electrons behave as waves. Okay? Now a use of matter waves. An electron microscope uses fast electrons. So tell me, is their wavelength long, or short? It is short, correct. Lambda equals h over p, with h Planck's constant and p the momentum, and fast electrons have a large momentum. Their wavelength is thousands of times shorter than that of light. With me? And a shorter wavelength can show much finer detail. So an electron microscope has a far higher resolving power than an optical microscope. Clear?

3:10Standing matter waves in an orbit

Now let us look inside the atom. Picture the electron going round a circular orbit of radius r, as a wave wrapped round the circle. In the figure, the left wave joins up neatly, and the right one does not. So tell me, when can the wave join up with itself? Only when a whole number of wavelengths fits round the orbit, exactly! One, two, three, but never three and a half. Only then can the electron form a steady pattern, a standing matter wave. The wave on the left has exactly four wavelengths, two pi r equals four lambda. On the right, the crossed equals sign means not equal, so that wave cannot fit. Right? Now write it as maths. The orbit's length is its circumference, two pi r, with pi the Greek letter. So two pi r equals n lambda, with n a whole number. With lambda equal to h over m v, m the mass and v the speed, this is n h over m v, okay? Now rearrange this for m times v times r. What do you get, and have you met it before? So m times v times r equals n h over two pi; the double arrow means it follows from the line above. And yes, this is Bohr's quantum condition! Bohr's rule comes straight out of the wave picture. Beautiful, isn't it? So, in the n th orbit of hydrogen, the wavelength is lambda n, with a small n below. It equals two pi times r n, over n: the circumference shared among n waves. Keep it ready for Example two. Got it?

4:50Example 1: Bragg's law wavelength for 54 V electrons

Now Example one. In the Davisson Germer experiment, the diffraction peak for fifty four volt electrons corresponds to a glancing angle of sixty five degrees, on nickel planes zero point zero nine one nanometres apart. Find the wavelength from Bragg's law, two d sine theta equals lambda, where d is the gap between the planes, and compare it with the de Broglie value. Pause the video and try it first. Done? Okay, let's check. Earlier, the peak was at fifty degrees. So theta in Bragg's law is fifty degrees, right? No, no, careful, that is the trap! Fifty degrees is the scattering angle. Bragg's law needs the glancing angle on the planes, and the question gives it: sixty five degrees. And d is zero point zero nine one nanometres. Clear? Now put these into Bragg's law. Lambda equals two d sine theta, so two times zero point zero nine one, times sine sixty five degrees. Okay? Two times zero point zero nine one gives zero point one eight two. Sine sixty five degrees is about zero point nine zero six. Multiply them, and lambda comes to about zero point one six five nanometres; the wavy equals sign means approximately. Right? And de Broglie's value, as we found, is one point two two seven over root fifty four, which is zero point one six seven nanometres. With me? So, zero point one six five against zero point one six seven. That is excellent agreement, isn't it? The electrons diffract with just the wavelength de Broglie predicted. Got it?

6:30Example 2: de Broglie wavelength in hydrogen orbits n=1, 3

Now Example two, a little harder. Find the de Broglie wavelength of the electron in the ground state, and in the third orbit of hydrogen. You are given the Bohr radius, the small letter a with a zero below it, equal to zero point zero five two nine nanometres. Pause the video and try it first. Done? Okay, let's check. Start from our standing wave: lambda n equals two pi times r n, over n, as we found. And from Bohr's model, the radius of the n th orbit is n squared times the Bohr radius; the small two up top means squared. Okay? Now, the radius grows as n squared. So the wavelength must also grow as n squared, right? No, no, wait, there is an n below the line too! Two pi times n squared times the Bohr radius, over n, leaves two pi times the Bohr radius, times n, nothing more. So the wavelength grows only as n, not as n squared. Clear? Ground state means n equals one. So lambda one is two pi times zero point zero five two nine, which is about zero point three three nanometres. With me? Third orbit means n equals three, so lambda three is three times lambda one. Three times zero point three three is about one point zero nanometres. Right? So, a bigger orbit holds a longer wave: lambda n is proportional to n; this sign, like a stretched alpha, means proportional to. Got it?

8:01Quick revision: Davisson-Germer result and Bohr's condition

So, quick revision. Davisson and Germer showed that a nickel crystal diffracts electrons: electrons behave as waves. For fifty four volt electrons, the measured wavelength is zero point one six five nanometres, and de Broglie predicts zero point one six seven. And a standing wave fits an orbit only if two pi r equals n lambda, which gives Bohr's condition, m v r equals n h over two pi. Now go and try the practice problems on this page. All the best!