Einstein's Photoelectric Equation

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. You shine light on a metal and measure the stopping potential; call it capital V, okay? Now you use light of half the wavelength, so every photon carries double the energy. So the stopping potential should also double, right? But for a certain metal, it comes out as three times V, not double, but triple! Where does the extra come from? Einstein's equation explains it, and it comes up often in JEE, so let's begin.

0:32Einstein's picture: one photon, one electron

So, what did Einstein say? Light of frequency nu is a stream of photons, each of energy h nu. Nu, the Greek letter that looks like a v, is the frequency, and h is Planck's constant. Okay? Let me ask you this. When one photon reaches the metal, does it share its energy among many electrons, or give it to just one? Just one, that is right. One photon gives all its energy to one electron. In the figure, the wavy orange line is the photon, and the blue arrow is the electron it frees. With me? But the metal holds the electron. To escape, it needs at least phi naught, the work function. Phi is the Greek letter drawn as a circle with a line through it, and the small zero below it is read as naught. The figure shows phi naught as a gap below the surface. Clear? So the photon brings h nu, and the electron spends phi naught to get out. How much energy is left for its motion? Yes, exactly, h nu minus phi naught! So K max equals h nu minus phi naught. K is the kinetic energy, and the small max below means maximum, for the most loosely bound electrons, which need only phi naught. This is Einstein's photoelectric equation, and the figure on the right writes the same thing as h nu equals phi naught plus K max. Got it?

1:56Threshold, and other forms of the equation

Now, a question for you. What if the photon's energy, h nu, is less than the work function? Then no electron comes out at all: no emission if h nu is less than phi naught, and the sign on screen means less than. And that holds however intense the light is. Okay? The frequency where h nu just equals phi naught is the threshold frequency, nu naught, equal to phi naught over h. Right? Now put phi naught equals h nu naught into the equation. K max equals h times the bracket, nu minus nu naught. And since nu is c over lambda, it also equals h c times the bracket, one over lambda, minus one over lambda naught. Lambda is the wavelength, lambda naught the threshold wavelength, and c the speed of light. With me? And we measure K max with the stopping potential, V naught. K max equals half m v max squared, which equals e V naught. Here m is the electron's mass, v max its maximum speed, and e its charge. Clear?

3:00What intensity does, and what it does not

Now, a classic question. Keep the frequency fixed, and make the light ten times brighter. Do the electrons come out faster? No, they do not, and this is a common trap. Our equation, h nu minus phi naught, has no intensity in it. So K max depends on the frequency, not on the intensity. Right? So brighter light means more photons every second, so more electrons every second, not faster ones. Got it? And one more point. Absorbing a photon is a single, sudden event, so there is no time lag. Okay?

3:38Topper's trick: work in electron volt

Now, a trick that saves a lot of time: work in electron volt. K max in electron volt equals twelve hundred and forty over lambda in nanometres, which is the photon energy, minus the work function in electron volt. On screen, e V in brackets means in electron volt, and n m in brackets means in nanometres. Okay? And the best part: the answer in electron volt is the stopping potential in volt, the same number, because K max equals e V naught. Clear?

4:08The trap: double the frequency, double K max?

Now, the big trap. Suppose we double the frequency of the light. The photon energy doubles, so K max also doubles, right? No, no, careful, the work function is subtracted first! Let K be the old K max, and K prime, with the little dash, the new one. K prime equals two h nu minus phi naught, which equals two K plus phi naught, so more than double. Okay? Same with wavelength: halving lambda doubles the photon energy, but K max does not double either. Right? So what is simple? Only the difference. Between two frequencies, delta K max equals h times delta nu, where delta, the Greek letter shaped like a triangle, means the change in. Got it?

4:51Example 1: 400 nm light on caesium, find K max, V0, v max

Now, Example one. Light of wavelength four hundred nanometres falls on caesium, whose work function is two point one four electron volt. Find the maximum kinetic energy, the stopping potential, and the maximum speed of the photoelectrons. Pause the video and try it first. Done? Okay, let's check. Using our trick, the photon energy is twelve hundred and forty over four hundred, which gives three point one zero electron volt. Okay? Subtract the work function: three point one zero minus two point one four, equals zero point nine six electron volt, and that is K max. So the stopping potential is zero point nine six volt, the same number; the capital V after it is the unit, volt. Right? Now the maximum speed, from half m v max squared. Be careful here, K max must be in joule, so it gets a factor of one point six times ten to the power minus nineteen. So v max is the square root of, two times zero point nine six times one point six times ten to the power minus nineteen, over nine point one times ten to the power minus thirty one, the electron's mass in kilogram. With me? Work it out, and v max comes to about five point eight times ten to the power five metres per second. The m s to the power minus one is metre per second, and the wavy equals sign means approximately. That's it!

6:18Example 2: threshold wavelength when half lambda gives 3V

And now Example two, our puzzle from the start. With light of wavelength lambda, the stopping potential is capital V, and with wavelength lambda over two, it is three times V, okay? Find the threshold wavelength. Pause the video and try it first. Done? Okay, let's check. For the first light: h c over lambda, minus phi naught, equals e V, where e is the electron's charge, with me? 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 three e V, right? Now subtract the first equation from the second. Phi naught cancels, and we get h c over lambda equals two e V, okay? See, only the difference is simple. Now back to the first equation. Phi naught equals h c over lambda, minus e V, that is two e V minus e V, just e V, which is half of h c over lambda. So phi naught equals h c over, two lambda. Clear? Finally, the work function is h c over the threshold wavelength, lambda naught. Compare the two, and, as the double arrow on screen says, it follows that lambda naught equals two lambda. And that solves our puzzle: halving the wavelength gave three times V, because the work function is subtracted first. Okay?

7:50Quick revision: K max = h nu - phi0, frequency vs intensity

So, quick revision. K max equals h nu minus phi naught, which equals e V naught, with no emission below the threshold frequency, nu naught, equal to phi naught over h. Frequency sets K max, and intensity sets the number of electrons. And the shortcut: twelve hundred and forty over lambda in nanometres, minus phi naught, in electron volt, gives the stopping potential in volt. Now go and try the practice problems on this page. All the best!