The de Broglie Wavelength

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In the last chapters, light behaved like a stream of particles, the photons. In nineteen twenty four, de Broglie turned the idea around. If a wave can act like a particle, can a particle act like a wave? He said yes, every moving particle, even a cricket ball. So a ball bowled at thirty metres per second is a wave? Then why has no fielder ever seen it spread out, the way light spreads through a narrow slit? Keep that question in mind; one line of maths will answer it. Questions on this come up often in JEE, so let's begin.

0:42Matter waves: the de Broglie wavelength

So, here is the big idea, from de Broglie in nineteen twenty four. Just as light has a particle side, every moving particle has a wave tied to it. We call it a matter wave. Okay? Its wavelength is called the de Broglie wavelength. Lambda, the Greek letter you see on screen, equals h over p. Here h is Planck's constant, the same h we used for photons, and p is the momentum of the particle. Momentum is mass times speed, so lambda also equals h over m times v, with small m the mass and small v the speed. Everything today comes from this one line. Clear? Now let me ask you this. If the momentum gets bigger, does the wave get longer, or shorter? It gets shorter, yes, exactly! The momentum p sits in the denominator, so lambda is proportional to one over p. This sign, like a stretched alpha, means proportional to. So a larger momentum means a shorter wave. Right? Now a quick trap. An electron and a proton move with the same momentum. The proton is much heavier, so its wave must be shorter, isn't it? So tell me, which one has the longer wave? The answer is neither of them; the two wavelengths are exactly equal! Lambda depends only on the momentum. Same p, same lambda, whatever the charge or the type of particle. The mass matters only through the momentum. Got it?

2:14Useful forms: from energy and from voltage

Alright, often you are given the kinetic energy, capital K, and not the speed. So tell me, if you know K, how do you get the momentum? Here it is: p equals the square root of two times m times capital K; that tick with the long roof is the square root sign. So lambda equals h over the square root of two m K, okay? And see the word non-relativistic: this holds only when the particle moves much slower than light. With me? Now take a charge, small q, that starts from rest and is accelerated through capital V, a potential difference. It gains kinetic energy q times V, so K equals q V, as the line shows. Clear? Put that into our formula. Lambda equals h over the square root of two m q V, as you see on screen. Remember this one; questions on accelerated charges come up often. Right? And here is a topper's shortcut. For an electron, put in its mass and charge once, and you get lambda equals one point two two seven over root V, in nanometres, with V in volts. The n m on screen is the nanometre, ten to the power minus nine metre. Clear? In the same way, for a proton the number is zero point zero two eight seven, for a deuteron, zero point zero two zero three, and for an alpha particle, with alpha the Greek letter, zero point zero one zero one. Each is over root V, in nanometres. Notice that all three numbers are much smaller than the electron's one point two two seven, so at the same voltage, these particles have much shorter waves than the electron. Okay?

3:52Two more forms, and the everyday objects

Now two more forms. For a gas molecule or a neutron at absolute temperature, capital T, moving with its r m s speed, the root mean square speed, lambda equals h over the square root of three m k T, where small k is Boltzmann's constant. Got it? And for a charge moving in a circle of radius r in a magnetic field, capital B, the momentum is q times B times r. So lambda equals h over q B r. Keep this one ready for circular-motion questions. Right? And one last point. For everyday objects, this wavelength is far too small to be observed; it matters only for particles of atomic size. Remember our cricket ball? Let us check it with real numbers.

4:34Example 1: wavelength of a 150 g cricket ball at 30 m/s

Now, back to our puzzle. Find the de Broglie wavelength of a cricket ball of mass one hundred and fifty grams, moving at thirty metres per second; the m s to the power minus one on screen means metres per second. And why are its wave properties never seen? Pause the video and try it first. Done? Okay, let's check. Lambda equals h over m times v. So the mass m is one hundred and fifty, right? No, no, careful, that number is in grams, and Planck's constant needs S I units! So the mass is one hundred and fifty grams, which is zero point one five kilograms. With me? Now, Planck's constant is six point six three times ten to the power minus thirty four, in joule seconds. So lambda equals that, over zero point one five times thirty. And zero point one five times thirty gives four point five. Okay? Six point six three over four point five gives about one point four seven. So lambda is about one point four seven times ten to the power minus thirty four, which rounds to one point five times ten to the power minus thirty four metres; the wavy equals sign means approximately. That is unbelievably small. Clear? So here is the answer to our puzzle. This wavelength is far smaller than any object, or any opening the ball could pass through, so no diffraction can be detected. The wave is there, but we can never see it. That is why matter waves matter only for tiny particles, like electrons. Got it?

6:10Example 2: electron beam voltage raised from 25 V to 100 V

Now a slightly harder one, JEE style. The accelerating voltage of an electron beam is raised from twenty five volts to one hundred volts. By what factor does the de Broglie wavelength change? Pause the video and try it first. Done? Okay, let's check. The voltage becomes four times bigger, and the wavelength falls as the voltage rises. So the new wavelength is one quarter of the old one, right? Oh no, careful, that is exactly the trap here! Look at the formula: V sits inside a square root. For the same electron, h, m and q do not change, so, as the double arrow shows, lambda is proportional to one over root V, and not to one over V, okay? So take the ratio. Lambda two over lambda one, with the small two and one below for the new and the old values, equals the square root of V one over V two. That is the square root of twenty five over one hundred. Twenty five over one hundred is one quarter, and the square root of one quarter is one half. With me? So the wavelength is halved, not quartered. Four times the voltage gives only half the wavelength. Students fall into this square root trap often, so watch for it. Clear?

7:30Quick revision: lambda = h/p, h/root 2mqV, 1.227/root V nm

So, quick revision. Lambda equals h over p, which is h over m times v, or h over the square root of two m K; and same momentum means same wavelength. For a charge accelerated through a voltage V, lambda equals h over the square root of two m q V, and for an electron, it is one point two two seven over root V, in nanometres. And for everyday objects, the wavelength is far too small to observe. Now go and try the practice problems on this page. You will find them easy. All the best!