Set 1

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Light falls on a photocell, and a current flows. Now make the light three times as intense, at the same frequency. Then the electrons come out with three times the energy, and the stopping potential triples too, right? Hold that thought; problem seven of today's set answers it. This is our first mixed set on dual nature, fifteen problems in the style of JEE Main. So, let's pick up the tools and the traps.

0:31Tool 1: the photon, in electron volt

Tool number one: the photon. Its energy, capital E, equals h nu, which is h c over lambda. Here h is the Planck constant, nu, the Greek letter like a v, is the frequency, c is the speed of light, and lambda is the wavelength. Its momentum, small p, is h over lambda, or capital E over c. Okay? Joules are awkward, so we use the electron volt, written small e, capital V, on screen. h times c is twelve hundred and forty electron volt nanometres, n m being the nanometre. One electron volt is one point six times ten to the power minus nineteen joule, written J; the cross means times, the raised number is the power of ten. Right? So the energy in electron volt is twelve hundred and forty over the wavelength in nanometres; the brackets show the units, so never put the wavelength in metres. Clear? And a source of power capital P sends out capital N photons per second, where N equals P lambda over h c. Now let me ask you this: two lamps of the same power, one red and one blue. Which sends more photons per second? The red one, yes, exactly! Red light has the longer wavelength, so each photon carries less energy, and you need more of them for the same power. With me?

1:58Tool 2: Einstein's photoelectric equation

Tool number two is Einstein's equation; look at the figure. The wavy orange line is a photon of energy h nu, giving all its energy to one electron, one of the most loosely bound electrons on the dashed line, and it leaves along the dark blue arrow. To escape, the electron spends phi naught, the work function; phi is the Greek letter drawn as a circle with a line through it, and the small zero below is read as naught. The two equations on the right are exactly what we build now. Okay? What is left is K max, the maximum kinetic energy, K with a small max below, marked above the surface. So K max equals h nu minus phi naught, which equals e V naught; e is the electron's charge, and V naught the stopping potential. Now tell me, if h nu is less than phi naught, but the light is very intense, will any electron come out? Not a single electron, very good if you said that! One photon frees one electron, so if one photon is not enough, more photons do not help. No emission if h nu is less than phi naught, however intense the light; the pointed sign on screen means less than. Clear? Where h nu just equals phi naught, we get the threshold frequency, nu naught, equal to phi naught over h, and the threshold wavelength, lambda naught, is twelve hundred and forty over phi naught in electron volt, in nanometres. With me? And the topper's shortcut: work in electron volt throughout, and K max in electron volt is the stopping potential in volt, the same number.

3:36Problem 7: three times the intensity

Got it? Now, back to our puzzle, problem seven of the set. The intensity is tripled at the same frequency. Each photon still carries the same energy, so three times the intensity means three times the photons every second. Right? So, three times the electrons too. The saturation current, small i, equals eta N e: N is the photons per second, e the electron's charge, and eta, the Greek letter like an n with a long tail, is the fraction of photons that eject electrons. So the current triples. Now, does the stopping potential triple as well? No, no, wait, that is exactly the trap, because K max equals h nu minus phi naught, and there is no intensity in it. Okay? So the stopping potential is unchanged. Puzzle solved: the current triples, but the stopping potential stays exactly the same. Got it?

4:31The graphs say the same

Now see the course's graphs of photocurrent against collector potential. On the left the frequency is fixed, and the intensities I one, I two and I three, a capital I with small numbers below, go up in that order; the pointed signs read greater than. Three different saturation currents, but all three curves meet the axis at the same point, minus V naught. So, same frequency, same stopping potential, and the saturation current is proportional to the intensity; that open, sideways alpha sign is read proportional to. Okay? On the right it is the other way round: the same intensity, but the frequencies nu one, nu two and nu three go up. Now the stopping potentials, V naught one, V naught two and V naught three, move apart; higher frequency, larger stopping potential. Right? But careful here: the course draws all three curves flattening at the same saturation current, the usual textbook picture. At the same intensity, though, a higher frequency means more energy per photon, so fewer photons every second; with equal efficiency, the saturation current is actually smaller, as the course shows in its graphs chapter. So the frequency sets the stopping potential, and, at a fixed frequency, the intensity sets the saturation current. With me?

5:54Tool 3: the de Broglie wavelength

Tool number three is for matter waves. Every moving particle of momentum p has a de Broglie wavelength, lambda equals h over p. For a charge q of mass m, accelerated from rest through a potential difference capital V, it becomes h over the square root of two m q V, for non-relativistic motion, far slower than light. Okay? For an electron only, it becomes one point two two seven over root V, in nanometres, V in volt. Problem twelve: an electron is accelerated through two hundred and twenty five volts. Quick, what is the square root of two hundred and twenty five? It is fifteen, very good! So lambda is one point two two seven over fifteen, about zero point zero eight two nanometres; the wavy equals sign means approximately. Right?

6:48Example 1: 200 nm light on a 4.2 eV metal, find K max and V0

Now, example one, problem five of the set. Light of two hundred nanometres falls on a metal of work function four point two electron volt. Find the maximum kinetic energy of the photoelectrons, and the stopping potential. Pause the video and try it first. Done? Okay, let's check. By the shortcut, the photon energy is twelve hundred and forty over two hundred, which gives six point two electron volt. Okay? Six point two is greater than four point two, the sign you see, so electrons do come out. Right? So K max is six point two minus four point two, which equals two point zero electron volt. And the stopping potential, is it two point zero electron volt as well? Careful, a potential is measured in volt. Since e V naught equals K max, V naught is two point zero volt; the capital V after the number is the unit. Same number, different unit. Clear?

7:50Example 2: K max ratio at 2 and 3 times threshold frequency

Now a harder one: example two, problem nine of the set. Light of frequencies two nu naught and three nu naught falls, in turn, on a metal of threshold frequency nu naught. Find the ratio of the maximum kinetic energies. Pause the video and try it first. Done? Okay, let's check. The frequencies are in the ratio two to three, so the kinetic energies are also two to three, right? Oh no, careful, that is the classic trap: K max is not proportional to the frequency, because the work function is subtracted first. And phi naught equals h times nu naught. Okay? For the first light, K one, K with a small one below, equals two h nu naught minus h nu naught, which is just h nu naught. Right? For the second light, K two equals three h nu naught minus h nu naught, which is two h nu naught. With me? So K one to K two is h nu naught to two h nu naught, which is one to two; the colon means is to. The frequency went up by only half, yet the energy doubled, because the work function is subtracted first. Got it?

9:10Quick revision: photon energy, Einstein equation, de Broglie

So, quick revision. Photon energy in electron volt is twelve hundred and forty over lambda in nanometres, and photons per second are P lambda over h c. K max equals h nu minus phi naught, which equals e V naught; frequency sets the energy, intensity the current. And for an electron, lambda is one point two two seven over root V, in nanometres. Now try all fifteen problems on this page in twenty five minutes, then check the solutions. All the best!