de Broglie Wavelength

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. In nineteen twenty four, de Broglie made a bold claim: every moving particle has a wavelength. Not just light. Electrons, protons, everything that moves. So when you throw a ball to your friend, that ball is also a wave? Then why has nobody, ever, seen a ball behave like a wave? Hmm, strange, right? Today we settle this, with one small formula. And look, in JEE, de Broglie wavelength questions are quick, sure marks, if you know which form of the formula to pick. So, let's begin.

0:39de Broglie: every moving particle has a wavelength

So, here is de Broglie's idea from nineteen twenty four, in one line. Lambda, this Greek letter, is the wavelength. It equals h over p, where p is the momentum of the particle. And h is Planck's constant: six point six two six, times ten to the power minus thirty four, joule second, the J s on screen. Clear? Now, momentum is mass times velocity: p equals m v, okay? So lambda equals h over m v. Same formula, just written with mass and speed. Now tell me. Look where m and v sit: in the bottom of the fraction. So if a particle is heavier, or faster, is its wavelength longer, or shorter? Shorter, exactly! Bigger bottom, smaller fraction. The double arrow on screen means so. Bigger m or v, so a shorter lambda. Now, keep this in mind. Our ball is much heavier than an electron, right? And one more thing. Even a photon of wavelength lambda has momentum h over lambda, exactly the same as any particle of that wavelength. So same wavelength, same momentum, whether it is light or matter. Got it?

1:52Wavelength from kinetic energy

Alright, but what if the question gives you the kinetic energy, K, and not the speed? How do you get the momentum p from K? Yes, use this: K equals p squared, over two m. Turn it around, and p equals the square root of two m K, and this tick shaped sign is the square root, okay? Now put that into de Broglie's formula. Lambda equals h over p, which is h over root of two m K, okay? So you now have three forms: with momentum, with mass and speed, and with kinetic energy. Just pick the one that matches the data. Clear?

2:28Charge accelerated through a potential difference V

Now, here is a setting you will meet again and again. Take a charge q, starting from rest. It is accelerated through a potential difference V, measured in volts. What kinetic energy does it pick up? Yes, q times V! From rest, it gains kinetic energy K equals q V, okay? Now take h over root of two m K, and in place of K, put q V, okay? So lambda equals h over the square root of two m q V, the new line on screen. This one works for any charged particle: proton, alpha, anything. With me? And for an electron, there is a ready made shortcut. Lambda equals twelve point two seven over root V, in angstrom. Angstrom is the capital A with a small ring on top, and one angstrom is ten to the power minus ten metre. Just put V in volts, and the answer comes straight out in angstrom. Now careful. Suppose a proton is accelerated through some voltage. Can I use the same twelve point two seven over root V for it? No, no! This is a classic JEE trap. That twelve point two seven already has the electron's mass and charge packed inside it. For a proton, or an alpha particle, go back to the general formula, h over root of two m q V, the line lit up again. Got it? Now a trick that toppers use. When two particles go through the same V, then h, the two, and V are common to both. In a ratio, they cancel! So lambda is proportional to one over root of m q. This curly sign means proportional to. Just mass times charge, nothing else. Keep it ready; we need it very soon.

4:14Example 1: de Broglie wavelength of a moving ball

Now, remember our puzzle from the start? Let's finish it. Find the de Broglie wavelength of a ball of mass zero point one kilogram, the k g on screen, moving at ten metres per second, the m s to the power minus one. And why is this wave never observed? Pause the video, and try it first. Done? Okay, let's check. We have the mass and the speed, so which form? h over m v, right? So, six point six two six times ten to the power minus thirty four, over zero point one times ten. Zero point one times ten is just one. So lambda is six point six two six times ten to the power minus thirty four, over one. That is about six point six times ten to the power minus thirty four metre. The wavy equals sign means approximately. Now the real question. The ball does have a wavelength. So why has nobody ever seen it behave like a wave? Look at that number: ten to the power minus thirty four metre! That is far smaller than any object or slit, so no wave effects can be seen. And there you go, puzzle solved! The wave is there, it is just too tiny to show. Remember, bigger mass, shorter wavelength. Nice and clear, right?

5:32Example 2: proton to alpha wavelength ratio, same V

Now a harder one, JEE level. Read it with me. Take a proton and an alpha particle, the Greek letter alpha on screen. Both are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. Pause and try. Hint: use the topper's trick. Done? Okay, let's check. Same V, so the trick works: lambda is proportional to one over root of m q. So we only need the masses and the charges. Alright, the alpha particle is four times heavier than the proton. Root of four is two. So lambda p, the proton's wavelength, to lambda alpha, the alpha's wavelength, is two is to one, right? The small p and alpha below each lambda tell you which particle. No, no, careful! This is the JEE trap. You forgot the charge. The alpha particle is a helium nucleus, so it has four times the mass of a proton, and two times the charge. So m alpha equals four m p, and q alpha equals two q p, okay? Now the ratio. Lambda goes as one over root m q, so the alpha's values go on top. Lambda p over lambda alpha equals root of m alpha q alpha, over m p q p, okay? That is root of four times two, over one times one: root eight, which is two root two. So lambda p is to lambda alpha is two root two, is to one. The proton, lighter and with less charge, gets the longer wavelength. Makes sense, right? Absolutely!

7:08Quick revision: wavelength from p, K and voltage V

So, quick revision. Lambda equals h over p, equals h over m v, equals h over root of two m K: three forms, one idea. For a charge through a potential difference V, lambda is h over root of two m q V; and only for an electron, twelve point two seven over root V, in angstrom. And for ratios at the same V, lambda is proportional to one over root of m q. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!