The Graphs of the Experiment

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two friends test two different metals, and each plots the stopping potential against the frequency of the light. Both get straight lines. The second metal holds its electrons more tightly, so its line should be steeper, right? Hmm, are you sure about that? Keep your guess in mind; the graph will answer it. Graph questions on this come up often in JEE, so let's begin.

0:30Photocurrent against collector potential

So, the first graph, on the left of the figure: photocurrent up the side, collector potential along the bottom. The three blue curves are for one frequency but three intensities, capital I one, I two and I three, with I three the brightest. With the collector negative, the current falls to zero at the stopping potential. Okay? Now let me ask you this. If the light is brighter, do we need a bigger stopping potential? No, we do not; all three curves meet the axis at one point, minus V naught. V naught, the capital V with a small zero read as naught, is the stopping potential. So the line reads: same nu, the Greek letter for frequency, I three greater than I two greater than I one, same V naught. Right? But the flat parts are different. With the collector positive enough, every emitted electron is collected, and that is the saturation current. Brighter light brings more photons, so more electrons every second. The sign like a small fish means proportional to: saturation current is proportional to intensity. Clear? Now the orange curves on the right: the same intensity, but three frequencies, with nu three the highest. So tell me, which part of the curve moves now? It is the stopping potential, yes, exactly! Higher frequency gives faster electrons, so minus V naught three, for nu three, sits farthest to the left. Same intensity, higher nu, larger V naught. Here the flat parts are drawn at one height, which means the same number of electrons every second. So intensity moves the flat part, frequency moves the starting point. Got it?

2:14Stopping potential against frequency

Now the key graph, on the left of this figure: stopping potential against frequency. Start from Einstein's equation: e V naught equals h nu minus phi naught. Here e is the electron's charge, h is Planck's constant, and phi naught, the circle with a line through it, is the work function. Okay? Divide everything by e. V naught equals h over e, times nu, minus phi naught over e. That is a straight line in nu. So, what is its slope? Yes, exactly, the slope is h over e! Neither h nor e depends on the metal, so the slope is the same for every metal. And there is our puzzle, solved: metal A in blue and metal B in orange give parallel lines. So a tighter hold does not make the line steeper. Right? Now, where does this line cut the frequency axis? Think about what a zero stopping potential means. At the threshold frequency, nu naught, that is correct. There V naught is zero, so h nu naught equals phi naught, and below nu naught no electrons come out, however intense the light. For metal B, it is nu naught prime; the little dash is read as prime. Clear? Produce the line back, along the dashed part, and it cuts the V naught axis at minus phi naught over e; the figure shows minus phi A over e and minus phi B over e, with A and B written small. Metal B has the larger work function, so its line sits lower and starts farther right. Got it? And the graph on the right is saturation current against intensity, at a fixed frequency above the threshold. The current is proportional to the intensity, so it is a straight line through the origin. Okay?

4:06More graphs from the same equation

Now, suppose we plot K max against nu. K max, the capital K with the small word max written at its foot, is the maximum kinetic energy, and it equals e V naught. So what is the slope now? It is just h, well done, because multiplying by e cancels the e in h over e. And the line cuts the energy axis at minus phi naught. With me? Next, V naught against one over lambda; lambda, the Greek letter like an upside down y, is the wavelength. Since nu is c over lambda, with c the speed of light, the slope is h c over e, and the line cuts the axis at one over lambda naught, where lambda naught is the threshold wavelength. Okay? Now, careful. What if we plot V naught against lambda itself, and not one over lambda? Is that a straight line too? No, it is not, and this is a common trap. Longer wavelength means less energy in each photon, so V naught falls along a curve, and reaches zero at lambda naught. So K max or V naught against wavelength is a falling curve. Right? And for a fixed frequency, K max and V naught do not change with intensity at all, just as the blue curves showed us, so against intensity both are horizontal lines. Got it?

5:30Example 1: work function from the V0 vs frequency graph

Now, Example one. The V naught against nu line for a metal cuts the frequency axis at four point six times ten to the power fourteen hertz. The dash means against, and the unit H z is hertz. Find the work function. Pause the video and try it first. Done? Okay, let's check. Where the line cuts the frequency axis, the stopping potential is zero, so that is the threshold frequency. So nu naught is four point six times ten to the power fourteen hertz. Right? At the threshold, phi naught equals h nu naught: six point six three times ten to the power minus thirty four, times four point six times ten to the power fourteen, in joules. Divide by one point six times ten to the power minus nineteen to get electron volts, the e V on screen. Okay? Work it out, and phi naught comes to about one point nine electron volts; the wavy equals sign means approximately. So you need just one reading from the graph, and that is really all there is to it.

6:39Example 2: stopping potential ratio for two metals

And now Example two, the harder one, built on our puzzle. The V naught against nu lines of metals A and B have threshold frequencies four times ten to the power fourteen hertz, and six times ten to the power fourteen hertz. For light of frequency eight times ten to the power fourteen hertz, find the ratio of the stopping potentials. Pause the video and try it first. Done? Okay, let's check. Remember, both lines have the same slope, h over e. So how can we write V naught using only the threshold frequency? Put phi naught equals h nu naught into our line. Then V naught equals h over e, times, nu minus nu naught, and h over e is the same for both metals. Right? So the ratio is just the ratio of the brackets. V naught A over V naught B, with A and B written small, equals eight minus four, over eight minus six, because the ten to the power fourteen cancels. That is four over two. Clear? So the ratio is two is to one; the colon is read as is to. And look, we never needed h, e or the work functions, because the same slope did all the work. Got it?

7:57Quick revision: slope h/e, intercepts, saturation current

So, quick revision. Brighter light gives a higher saturation current, but the same V naught. V naught against nu has slope h over e for every metal, and cuts the nu axis at nu naught. K max against nu has slope h, and against wavelength it is a falling curve. Now try the practice problems on this page. All the best!