Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Shine a very bright red lamp on a zinc plate, and not a single electron comes out. Now switch to a faint ultraviolet lamp, and electrons come out. How can faint light do what bright light cannot? Keep your answer in mind; Example one settles it. Today we do quick concept checks on photons and matter waves. Why-questions like these do come up in JEE, so let's begin.

0:30Check 1: does this prove light is not a wave?

So, check one, and many students get this wrong. Does the photoelectric effect prove that light is not a wave? No, it does not prove that. What it shows is that light delivers its energy in quanta, small packets, each of energy h nu. Here h is Planck's constant, and nu, the Greek letter that looks like a v, is the frequency. Okay? But light still shows interference and diffraction, and those need the wave picture; particles alone cannot explain them. So both pictures are needed, and light has a dual nature, which is exactly the name of this unit. Clear?

1:10Check 2: intensity, and one photon at a time

Now check two: take light of a single frequency, a stream of identical photons. In this photon picture, what does the intensity of the light measure? It measures the number of photons crossing unit area per second, times the energy of each photon, h nu. The capital I on screen is the intensity, and the cross means times. With me? So a brighter beam of the same light has more photons, not more energetic ones. Keep this in mind; it is the key to our puzzle. Okay? Now look at the figure. The wavy orange line is a photon of energy h nu, and the blue arrow is the electron it frees from the grey metal. Next, can the photon give part of its energy to the electron, and keep the rest? No, it cannot do that. The photon is absorbed completely, or not at all. In the figure, the electron spends phi naught, the work function, shown as the gap below the surface, to get out of the metal; phi is the circle with a line through it, and the small zero below is read naught. The rest is K max, K with a small max below, the maximum kinetic energy. So h nu equals phi naught plus K max, the first line on the right. The line below it links K max to the stopping potential, and that is our next check. Got it?

2:32Check 3: does the stopping potential depend on distance?

Now check three: we move the collector farther away from the emitter. Does the stopping potential change? Write the energy equation first. e V naught equals K max, equals h nu minus phi naught. Here e is the size of the electron's charge, and V naught, V with a small zero below, is the stopping potential. Right? Now look at the right side: only nu and phi naught, the frequency and the work function. The distance appears nowhere. So the stopping potential does not depend on the distance between the emitter and the collector. Clear?

3:09Check 4: are matter waves electromagnetic?

Now let us go from light to matter. De Broglie said that every moving particle has a wave with it, its matter wave. So tell me, are matter waves electromagnetic waves? No, and here is why. Matter waves go with every moving particle, charged or not. Okay? And they are not oscillations of electric and magnetic fields, the way light is. So matter waves are not electromagnetic waves. With me? Now, if every moving particle has a wave, then a moving car has one too. So why is the wave nature of a car never noticed? Because its momentum, p, is huge. Lambda, the de Broglie wavelength, equals h over p, so it is of the order of ten to the power minus thirty eight metres; the wavy sign means of the order of. That is far too small to produce any observable diffraction. Got it?

4:06Check 5: matter waves in a Bohr orbit

Now check five takes us inside the atom. In the figure, an electron wave is wrapped round a circular orbit, the dashed circle. Why does it matter that a whole number of de Broglie wavelengths fits round the orbit? Look at the dark blue wave on the left: it joins up with itself, with exactly four wavelengths round the circle, so it forms a steady, standing wave. The orange wave on the right does not join up, so that orbit is not allowed; the equals sign with a slash through it, under that wave, means not equal to. Okay? Now, the length of the orbit is two pi r, with pi the Greek letter on screen and r the radius. Set it equal to n lambda, with n a whole number, and put lambda equals h over m v, with m the mass and v the speed. Right? Rearrange, and the double arrow means this gives: m v r equals n h over two pi. That is exactly Bohr's rule for the angular momentum. Clear?

5:06Example 1: why bright red light frees no electrons from zinc

Now, Example one, and it is our puzzle from the start. Red light, however bright, ejects no electrons from zinc, while faint ultraviolet light does. Why? Pause the video and try it first. Are you done? Okay, let's check. Look at the figure again: one photon gives its energy to one electron. So it frees the electron only if h nu is greater than or equal to phi naught; that is the sign on screen. Okay? Each red photon has less energy than the work function of zinc; the sign here means less than. So no red photon can free an electron. Right? But what if we make the red light extremely bright? Surely so much light will free some electrons? No, no, careful, that is the trap! Red light is below the threshold frequency of zinc, the lowest frequency that can eject an electron from it, and brightness only means more such photons, each one still with too little energy. One photon acts on one electron, so more weak photons still free nothing. With me? But each photon of this ultraviolet light has energy greater than phi naught of zinc. Careful: not every ultraviolet wavelength can do this for zinc; the photon energy must still reach phi naught. So even faint ultraviolet light of this kind, with only a few photons, ejects electrons. And that settles our puzzle!

6:34Example 2: electron vs proton, same KE, momentum and lambda

Now Example two, a little harder. An electron and a proton have the same kinetic energy. Which has the larger momentum, and which the longer wavelength? Pause the video and try it first. Are you done? Okay, let's check. Many students say that the same kinetic energy means the same momentum, and the same wavelength. Is that right? No, careful, it is not right. In the non-relativistic case, momentum p equals the square root of two times m times K, where m is the mass and K the kinetic energy; the tick with a long roof is the square root sign. With the same K, p is proportional to root m; the curly sign means proportional to. Okay? Now, m p and m e, with small p and e below, are the masses of the proton and the electron. The proton is about one thousand eight hundred and thirty six times heavier; the wavy equals sign means approximately. So the proton has the larger momentum. Right? Now let us look at the wavelength. Which one has the longer de Broglie wavelength? It is the electron, yes, exactly! Lambda equals h over p, so the smaller momentum gives the longer wavelength. The lighter particle has the longer wavelength. Got it?

7:55Quick revision: photons, intensity, V0 and matter waves

So, quick revision. Each photon gives all of h nu to one electron, or nothing, and emission needs h nu at least equal to phi naught. Intensity counts photons, and the stopping potential depends only on the frequency and the work function. Matter waves have lambda equals h over p, they are not electromagnetic, and a whole number of wavelengths fits round a Bohr orbit. You handled some tricky questions today, so well done. Now try the practice problems on this page. All the best!