AIEEE and Earlier JEE Main Questions: Matter Waves

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. One particle hits another particle of half its mass, at rest. The collision is head-on and elastic, and after it, both particles move. Now, which of the two has the longer matter wave, the heavy one or the light one? Most students say the lighter one. Keep your answer ready; we will check it at the end. Today we solve older A I triple E and JEE Main questions on matter waves, and questions like these do appear in the exam. So, let's begin.

0:34The toolkit: one formula, three forms

So, every question today starts from one line. Lambda, the Greek letter you see on screen, is the de Broglie wavelength, and it equals h over p. Here h is Planck's constant, 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. Okay? Now, many questions give you the kinetic energy, capital K, and not the speed. So tell me, how do you get the momentum from capital K? The momentum is 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, as the second line shows. And see the word non-relativistic: this holds only when the particle moves much slower than light. Right? Next, a charge, small q, starts from rest and is accelerated through capital V, a potential difference. It gains kinetic energy q times V, so K equals q V, and lambda equals h over the square root of two m q V, okay? And for an electron, there is a shortcut: 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. Got it?

1:58Topper's trick: work in ratios

Now a topper's trick. Many of these questions ask for a ratio, not a number. Since h is a constant, lambda is proportional to one over p for any particle, where this sign, like a stretched alpha, means proportional to. And for the same particle, m does not change either, so lambda is also proportional to one over root K, capital K the kinetic energy. Write the ratio, and every constant cancels. With me? Now let me ask you this. Take two different particles: is the lighter one's wave always the longer one? The answer is no, not always! Lambda depends only on the momentum, so equal momenta give equal wavelengths, whatever the masses. The mass matters only through the momentum. Hold on to this; our puzzle depends on it. Okay?

2:49Davisson–Germer, and a JEE Main 2015 match

Next, the experiment behind matter waves: the Davisson Germer experiment of nineteen twenty seven. In the figure, on the left, an electron gun fires a beam, accelerated through a voltage V, at a nickel crystal. The movable detector measures how many electrons are scattered at each angle theta, the Greek letter on the figure. Okay? On the right is the intensity against the scattering angle. For fifty four volts, there is a sharp peak at fifty degrees; the small circle means degrees. It comes from constructive interference of the electron waves scattered from the regularly spaced layers of atoms. The wavelength found from this diffraction is zero point one six five nanometres. Clear? Now use our shortcut. One point two two seven over root fifty four gives zero point one six seven nanometres. That is almost exactly the same value, isn't it? Now a quick JEE Main question from twenty fifteen: match each experiment with what it established. First, what did this experiment establish? Yes, exactly, the wave nature of electrons! Electrons were diffracted by a crystal, just like waves, with de Broglie's wavelength. With me? Next, the photoelectric experiment, from the earlier chapters, showed light arriving as photons, so it established the particle nature of light. Right? And the last one, the Franck Hertz experiment, established the discrete energy levels of atoms. Do not swap them: the particle nature is of light, and the wave nature is of electrons, not the other way round. Got it?

4:28Example 1: AIEEE 2005, electron KE doubled, new wavelength

Now, Example one, from A I triple E two thousand and five. The kinetic energy of a free electron doubles. By what factor does its de Broglie wavelength change? Pause the video and try it first. Done? Okay, let's check. The energy doubles, and the wavelength falls as the energy rises. So the wavelength simply halves, right? Oh no, careful, that is the trap! Look at the formula: capital K sits inside a square root. It is the same electron, so h and m are fixed, and lambda is proportional to one over root K, as the double arrow shows. Clear? So take the ratio. Lambda two over lambda one, with the small two and one below for after and before, equals the square root of K one over K two. That is the square root of K over two K, okay? Capital K over two K is one half, and the square root of one half is one over root two. So the wavelength becomes one over root two times the old one, not half of it. The square root makes all the difference. Got it?

5:38Example 2: JEE Main 2017, elastic collision wavelength ratio

Now Example two, our puzzle from the start, from JEE Main twenty seventeen. Particle A, of mass m and initial velocity v, collides head-on and elastically with particle B, of mass m over two, at rest. Find the ratio of the de Broglie wavelengths, lambda A over lambda B, after the collision; the small A and B below name the particles. Pause the video and try it first. Done? Okay, let's check. Particle B is the lighter one, so it must have the longer wave, right? No, no, wait, remember our trick: the momentum decides, not the mass. So first, the velocities after the collision. For a head-on elastic collision with B at rest, the velocity of A becomes m A minus m B, over m A plus m B, times v. And the velocity of B becomes two m A, over m A plus m B, times v. With me? Now put m for A, and m over two for B, the lighter particle. For A, the top is m over two and the bottom is three m over two, so its velocity is v over three. For B, two m, divided by three m over two, gives four over three, so its velocity is four v over three, okay? Now the momenta. The momentum of A is m times v over three, which is m v over three. The momentum of B is m over two, times four v over three, which is two m v over three. Clear? Let us check quickly: m v over three, plus two m v over three, gives m v, the momentum before the collision. Momentum is conserved, so our velocities are correct. With me? Finally, lambda equals h over p, so lambda A over lambda B equals p B over p A, which is two. So the heavier particle, A, has the wave twice as long, because it moves so much slower. That solves our puzzle: the momentum decides, not the mass. Got it?

7:32Quick revision: lambda = h/p, ratios and Davisson-Germer

So, quick revision. Lambda equals h over p, or h over the square root of two m K, and for the same particle, it goes as one over root K, okay? The momentum decides the wavelength, not the mass. And Davisson and Germer established the wave nature of electrons. Now try the other questions on this page. You can do this. All the best!