Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a lab puzzle. Light falls on a metal, and the stopping potential is one volt. Now we halve the wavelength, so every photon carries twice the energy, and the stopping potential jumps to four volts. Four times, from photons only twice as energetic? Is the reading wrong? Example two will settle it. Today it is Set three of the mixed problems on dual nature, in the style of JEE Main. Three quick tools first, then two full problems. Let's go.

0:33Tool 1: work function from the threshold

So, here is tool one. The threshold frequency of a metal, nu naught, is eight times ten to the power fourteen hertz. Nu is the Greek letter that looks like a v, the small zero below it is read as naught, and H z is hertz. Find the work function, phi naught; phi is the circle with a line through it. Okay? Let me ask you this: if the light is just below the threshold frequency, but extremely intense, will any electrons come out? Not a single one, that is correct. Emission happens only if h nu, the photon energy, with h the Planck constant, is greater than or equal to phi naught; that is the greater-than sign with a line under it. And this holds however intense the light is. Right? At the threshold, h nu naught just equals phi naught. So phi naught is six point six three times ten to the power minus thirty four, times eight times ten to the power fourteen, in joule, the J on screen. With me? That gives about five point three times ten to the power minus nineteen joule. One electron volt is one point six times ten to the power minus nineteen joule, so divide, and phi naught is about three point three electron volt, the e V on screen. The wavy equals sign means approximately. Clear?

1:57Tool 2: intensity and the saturation current

Now tool two, an assertion and reason question. The capital A on screen stands for assertion: the saturation photocurrent is proportional to the intensity of the light. The sign like a small fish means proportional to. Okay? And the R on screen stands for reason: a more intense beam, of the same frequency nu, brings more photons per second. Is the reason true, and does it explain the assertion? Both are true, and yes, it explains it. Look at the left graph: three blue curves for one frequency, at intensities capital I one, I two and I three. More photons per second means more electrons per second; the double arrow means so. So the flat part, the saturation current, rises with intensity. Got it? But then brighter light must also give faster electrons, and a bigger stopping potential, right? No, no, careful, that is the trap! All three blue curves meet the axis at the same point, minus V naught; V naught, the capital V with a small zero, is the stopping potential. Same frequency, same V naught, and since K max equals e V naught, with e the electron's charge, the same maximum kinetic energy; K max is the capital K with the small word max. With me? So both are true, and the reason explains the assertion. Intensity sets the number of electrons, and frequency sets their energy.

3:30Tool 3: the same de Broglie wavelength

Okay? Now tool three, from matter waves. An electron and a proton have the same de Broglie wavelength, lambda, the Greek letter like an upside down y. Find the ratio of their speeds, v e over v p; the small e and p below mean electron and proton. Right? Same wavelength, so they must have the same speed, isn't it? Oh no, careful, look at the formula! Lambda equals h over p, the de Broglie wavelength, where p is the momentum. So the same lambda means the same momentum, not the same speed. Clear? Momentum is mass times speed, with m for the mass and v for the speed. So the electron's mass times its speed equals the proton's mass times its speed, which gives v e over v p equals m p over m e. With me? The proton's mass is about eighteen hundred and thirty six times the electron's, so the ratio is eighteen hundred and thirty six is to one; the colon is read as is to. Got it?

4:35Example 1: new stopping potential when 500 nm becomes 400 nm

Now, Example one. The stopping potential is zero point five volt with light of five hundred nanometres; the V on screen is volt, and n m is nanometres. Find it with light of four hundred nanometres on the same metal. Pause the video and try it first. Done? Okay, let's check. Work in electron volt. Since h c is twelve hundred and forty electron volt nanometre, V naught in volt equals twelve hundred and forty over lambda in nanometres, minus phi naught. Okay? Now, the wavelength goes from five hundred down to four hundred. V naught goes as one over lambda, so the new value is zero point five times five hundred over four hundred, which is zero point six two five volt, right? No, no, wait, that is the trap! V naught is not proportional to one over lambda, because the work function is subtracted from the photon energy. But on the same metal phi naught is the same, so it cancels when we subtract the two equations. With me? V naught prime, with the little dash, is the new stopping potential. V naught prime minus zero point five equals twelve hundred and forty times, one over four hundred, minus one over five hundred. That bracket is one over two thousand, so we get twelve hundred and forty over two thousand, which is zero point six two. Clear? So V naught prime is zero point five plus zero point six two, which gives one point one two volt. More than double, though the wavelength fell by only a fifth. Got it?

6:12Example 2: work function from 1 V at lambda, 4 V at lambda/2

And now Example two, our lab puzzle from the start. The stopping potential is one volt for light of wavelength lambda, and four volts for wavelength lambda over two. Find the work function and lambda. Pause the video and try it first. Done? Okay, let's check. In the figure, the wavy orange line is a photon of energy h nu, which is h c over lambda, with c the speed of light. The electron spends phi naught to get out, and what is left is K max. So, in electron volt, h c over lambda, minus phi naught, equals one. Okay? Now the second light, of wavelength lambda over two. What is its photon energy? Yes, exactly, it doubles, to two h c over lambda. So two h c over lambda, minus phi naught, equals four. Right? Now subtract the first equation from the second. Phi naught cancels, and h c over lambda equals three electron volt. With me? Back in the first equation, phi naught equals three minus one, which is two point zero electron volt. And there is our puzzle, solved! The photon energy went from three to six electron volt, just double, but the same two electron volt of work function is paid both times. So the check on screen reads six minus two equals four, so the stopping potential went from one to four, and the reading was right. Clear? Finally, h c over lambda is three electron volt, so lambda is twelve hundred and forty over three nanometres, which gives about four hundred and thirteen nanometres. Got it?

7:56Revision: threshold, intensity vs frequency, de Broglie

So, quick revision. At the threshold, phi naught equals h nu naught, and below nu naught there is no emission, however intense the light. Intensity sets the saturation current, and frequency sets V naught and K max. With two wavelengths on one metal, subtract, and phi naught cancels; and the same de Broglie wavelength means the same momentum. Now go and try the other problems of this set on this page. All the best!