AIEEE and Earlier JEE Main Questions: Photoelectric Equation

Transcript

0:00Introduction

Hello everyone! Let us start with a quick question. Light falls on a metal and knocks out electrons. Now I double the frequency of that light. Does the fastest electron now come out with double the kinetic energy? Many students say yes. Hold on to that answer, because an A I triple E question asked exactly this. Today we solve such A I triple E and earlier JEE Main questions on the photoelectric equation, so let's begin.

0:31Einstein's equation: one photon, one electron

So, first, a quick recap. Each photon of light carries energy E equals h nu, which is also h c over lambda. Here h is Planck's constant, nu, the Greek letter that looks like a v, is the frequency, lambda is the wavelength, and c is the speed of light. Okay? Now look at the figure on the right. The wavy orange arrow is the photon, with energy h nu, and it hits one of the most loosely bound electrons, on the dashed line inside the grey metal. One photon gives all its energy to that one electron. So, reading the figure, does that electron fly out with all of h nu? No, it has to pay first. Leaving the metal costs phi naught, the work function, the gap from the dashed line up to the surface; that is the Greek letter phi with a small zero. What is left is K max, the maximum kinetic energy, shown by the arrow above the surface; the little max written after the K means maximum. So h nu equals phi naught plus K max, Einstein's photoelectric equation. Got it? And K max equals half times m times v max squared, with m the electron's mass and v max its top speed. It also equals e times V naught, with e the electron's charge, and V naught, the V with a small zero, the stopping potential. Clear? Now suppose h nu is less than phi naught, but the light is extremely intense. Will any electron come out? No, not a single one, however intense the light, because one photon cannot pay the work function. The frequency where h nu just equals phi naught is the threshold frequency, nu naught, written as nu with a small zero, and it equals phi naught over h. That is why the figure also writes K max as h times, nu minus nu naught. Right? So what does intensity do, at the same frequency? More intense light means more photons per second, so more electrons per second. But each electron still gets one photon, so K max stays the same.

2:38Topper's trick: work in electron volts

Okay? Now a time-saving trick: skip joules, and work in electron volts, the e V on screen. h c is twelve hundred and forty electron volt nanometre, so the photon energy in electron volts is twelve hundred and forty over lambda in nanometres, the n m in brackets. With me? For example, take light of four hundred nanometres. Twelve hundred and forty over four hundred gives three point one zero electron volts. Simple, right? Then subtract the work function, also in electron volts, and you have K max in electron volts. Okay? And the best part: the stopping potential in volts is the same number as K max in electron volts. Clear?

3:20The trap: linear is not proportional

Now, back to our opening question, from A I triple E two thousand eleven. Statement one: light of frequency nu, above the threshold nu naught, falls on a metal, and if the frequency is doubled, K max and the stopping potential are both doubled. Statement two: K max and V naught depend linearly on the frequency. So, is statement one true? Let us check with the equation. Nu changes to two nu; the arrow on screen just means changes to. The new kinetic energy, K prime, the K with a small dash, is two h nu minus phi naught. That is two times, h nu minus phi naught, plus phi naught. With me? And h nu minus phi naught is just K max. So K prime equals two K max plus phi naught, which is more than double. So statement one is false, and V naught is not doubled either. Right? But statement two is true. K max equals h nu minus phi naught, and that is linear in nu. Linear is not the same as proportional, because the work function is subtracted first. Okay? Only the change in K max between two frequencies is simple. Delta K max equals h times delta nu, where delta, the little triangle, means change in. Got it?

4:40Example 1: spectrum region for 6.2 eV and 5 V (AIEEE 2006)

Now, Example one, an A I triple E question from two thousand six. The threshold frequency of a metal corresponds to an energy of six point two electron volts, and the stopping potential for a certain radiation is five volts. In which region of the spectrum does the radiation lie? Pause the video and try it first. Done? Okay, let's check. The threshold energy is the work function, so phi naught equals h nu naught, six point two electron volts. And a stopping potential of five volts means K max equals e times V naught, five electron volts. Okay? Now, many students subtract here. Six point two minus five gives one point two electron volts. Is that the photon energy? No, no, careful, that is the trap! The photon pays the work function and supplies the kinetic energy, so we add. h nu equals phi naught plus K max, six point two plus five, which gives eleven point two electron volts. Right? Now use our trick. Lambda equals twelve hundred and forty over eleven point two, which gives about one hundred and eleven nanometres. The wavy equals sign means approximately. Clear? On the electromagnetic spectrum, ultraviolet runs from about four hundred nanometres down to one nanometre. So one hundred and eleven nanometres is ultraviolet, and that is our answer. Well done if you added instead of subtracting!

6:12Example 2: wavelength made 3/4, speed of fastest electron

And now Example two, the harder one, from JEE Main two thousand sixteen. Radiation of wavelength lambda falls on a photocell, and the fastest electron emitted has speed v. The wavelength is changed to three lambda over four. What can be said about the speed of the fastest electron now? Pause the video and try it first. Done? Okay, let's check. Before the change, Einstein's equation gives half times m times v squared equals h c over lambda, minus phi naught. Okay? Now the wavelength becomes three quarters of lambda. What happens to the photon energy, h c over lambda? Yes, exactly, it becomes four thirds of h c over lambda, since lambda is in the denominator. So half times m times v prime squared equals four h c over three lambda, minus phi naught, where v prime, the v with a small dash, is the new speed. With me? Now the tempting shortcut. Four thirds the photon energy, so four thirds the kinetic energy, so v prime equals v times root four thirds, right? Oh no, careful, it is the same trap as our opening question! The work function is not multiplied by four thirds. So write the right side as four thirds times, h c over lambda minus phi naught, plus phi naught over three. Clear? Now, the bracket is just the old kinetic energy, half times m times v squared, and phi naught over three is positive. So the new kinetic energy is more than four thirds of the old one; the dot means times. Got it? Cancel the half and the m. So v prime squared is greater than four thirds of v squared, and the double arrow, meaning so, gives v prime greater than v times root four thirds. The speed is more than v root four thirds, not equal to it. Okay?

8:05Quick revision: photoelectric equation and 1240 eV nm

So, quick revision. h nu equals phi naught plus K max, and K max equals e times V naught; below the threshold frequency, no emission. Work in electron volts: twelve hundred and forty over lambda in nanometres, then subtract the work function. And K max is linear in frequency, not proportional, so doubling nu more than doubles it. That answers our opening question. Now go and try the practice problems on this page. All the best!