JEE Advanced Questions (2011–2018): Graphs and Applications

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Silver has a work function of four point six electron volts, and sodium has two point three, exactly half. We plot the stopping potential against frequency for each metal. Will sodium's line be twice as steep? Hold that thought. JEE Advanced asked exactly this in two thousand thirteen, and graphs like these come up often. So let us solve the JEE Advanced questions on these graphs and their uses, one by one. Let's begin.

0:33The stopping-potential line

So, first, the key equation: e times V naught equals h nu minus phi naught. Here e is the electron's charge, V naught, the V with a small zero, is the stopping potential, and h is Planck's constant. Nu, the Greek letter like a v, is the frequency, and phi naught, phi with a small zero, is the work function. Okay? Divide by e, and V naught equals h over e, times nu, minus phi naught over e. So V naught against nu is a straight line, as in the left graph. With me? Now tell me, the slope of this line is h over e. Does that slope depend on which metal you use? If you said no, that is correct: Planck's constant and the electron's charge are the same for every metal. So the slope is always h over e, and the lines for metal A and metal B are parallel. Clear? The line cuts the frequency axis at nu naught, the threshold frequency; below it there is no emission at all, however intense the light. So the dashed part is just the line produced back, and it meets the V naught axis at minus phi over e: minus phi A over e for metal A, and minus phi B over e for metal B, okay? Metal B has the larger work function, so its threshold, nu naught prime, with a small dash, sits further to the right on the axis. So, our puzzle: silver at four point six electron volts, the e V on screen, and sodium at two point three. Both lines have slope h over e, so the slope ratio is one, not two. Sodium's line is not steeper at all; the work function only shifts the line sideways. Got it?

2:20Other axes: wavelength and one over wavelength

Now, JEE Advanced two thousand fifteen. Light of wavelength lambda, the Greek letter, gives a stopping potential V naught. How does V naught vary with lambda, and with one over lambda? Is V naught against lambda a straight line too? Let us check. Nu equals c over lambda, with c the speed of light, so V naught equals h c over e, times one over lambda, minus phi naught over e; the dot means times. Okay? So against one over lambda, a straight line of slope h c over e, starting from one over lambda naught; lambda naught, lambda with a small zero, is the threshold wavelength. Right? So, if you said a straight line against lambda too, careful, that is the trap! Against lambda itself it is not a straight line, because lambda sits in the denominator. V naught falls along a curve and reaches zero at lambda naught. Clear?

3:17Applications: counting the electrons

Now the applications. Light of power P brings photons of energy h nu each, so capital N, the photons per second, is P over h nu. If a fraction eta of them eject electrons that are collected, the saturation current i equals eta times N times e; eta, the Greek letter like an n, is the quantum efficiency. Brighter light, more current: that is the straight line through the origin on the right. At a fixed frequency above threshold, the current is proportional to the intensity; that curly sign means proportional to. Got it? And if a voltage V accelerates an electron from rest, it gains kinetic energy e times V, so its momentum p is the square root of two times m times e times V, with m the electron's mass. Right? If absorbed at the anode, each electron hands over its momentum p, so the force, the momentum delivered per second, is electrons per second times p. With me?

4:19Example 1: Planck constant from stopping potentials (2016)

Now, Example one, from JEE Advanced two thousand sixteen. To find Planck's constant, a metal was lit with different wavelengths and the stopping potential measured. Point three micrometre gave two point zero volts, point four gave one point zero volt, and point five gave point four volt; the mu m on screen means micrometre. Find h, with c, three times ten to the power eight metres per second, and e, one point six times ten to the power minus nineteen coulomb. Pause the video and try it first. Done? Okay, let's check. For each reading, e times V naught equals h c over lambda, minus phi naught. Okay? Now, can we get h from just one reading? No, because the work function is not given. Subtract two readings, and phi naught cancels: h c times, one over lambda one minus one over lambda two, equals e times, V one minus V two; the small one and two number the readings. With me? Take the first two readings, in metres: point three and point four times ten to the power minus six; do not forget this change. And V one minus V two is one point zero volt. Clear? One over point three micrometre, minus one over point four micrometre, gives eight point three three times ten to the power five per metre. Right? So h equals one point six times ten to the power minus nineteen, over three times ten to the power eight, times eight point three three times ten to the power five. That gives six point four times ten to the power minus thirty four joule second, the J s on screen. Got it?

6:08Example 2: force of photoelectrons on the anode (2018)

And now Example two, from JEE Advanced two thousand eighteen. Light of power two hundred watts, just above the threshold, falls on a perfectly absorbing cathode of work function six point two five electron volts, so the photoelectrons have negligible kinetic energy, with one hundred percent efficiency. Five hundred volts is applied, and the electrons strike the anode normally and are absorbed. The force on the anode is n times ten to the power minus four newton; find n, given m with a small e, the electron's mass, nine times ten to the power minus thirty one kilogram. Pause the video and try it first. Done? Okay, let's check. Just above the threshold, each photon carries about the work function, six point two five electron volts; the wavy equals sign means approximately. Okay? The efficiency is one hundred percent, so how many electrons leave the cathode every second? Yes, exactly, one electron per photon. So divide two hundred watts by six point two five times one point six times ten to the power minus nineteen joules. Six point two five times one point six is ten, giving two times ten to the power twenty electrons per second. Right? Now, a tempting thought: they leave with negligible kinetic energy, so they hit the anode gently, and the force is almost zero, right? No, no, wait, that is the trap: the five hundred volts accelerate every electron. So p is the square root of two times m times e times V, with the values on screen. With me? Multiply the numbers under the root, and collect the powers of ten. So p is the root of one point four four times ten to the power minus forty six, that is, one point two times ten to the power minus twenty three kilogram metre per second. Clear? Absorbed electrons hand over all their momentum, so the anode feels a push. Force equals two times ten to the power twenty, times one point two times ten to the power minus twenty three, which is twenty four times ten to the power minus four newton. So n equals twenty four. Very good if you got it!

8:28Quick revision: V0 vs frequency and 1/lambda, force on anode

So, quick revision. V naught equals h over e times nu, minus phi naught over e, with slope h over e for every metal. Against one over lambda, a straight line; against lambda, a falling curve to zero at lambda naught. Electrons per second are eta times P over h nu; force is electrons per second times p. Now try the other JEE Advanced questions on this page. All the best!