IIT-JEE Questions (1989–2009): Graphs and Applications

Transcript

0:00Introduction

Hello everyone! Picture a small lamp, twenty centimetres from a photocell. The current saturates at eighteen milliamperes, and the cut-off voltage is zero point six volt. Now we move the lamp three times farther away. Weaker light, so both numbers must fall, right? Maybe the current drops to one third? Hmm, I am not so sure about that. Today we solve past IITJEE questions on graphs and the photocell, and one of them settles this puzzle. Questions like these come up often in JEE, so let's begin.

0:35Frequency and intensity: two different jobs

In this figure, the photocurrent is up the side, and the collector potential, the collecting plate's voltage, is along the bottom. The dark blue curves are for one frequency at three intensities, capital I one, I two and I three; the orange curves are for one intensity at three frequencies. With me? Behind every curve is one equation: e V naught equals K max, which equals h nu minus phi naught. Here e is the electron's charge, V naught, with the small zero read as naught, is the stopping potential, and K max, with the small word max, is the maximum kinetic energy. h is Planck's constant, nu, the Greek letter like a v, is the frequency, and phi naught, the circle with a line through it, is the work function. Okay? And one condition: if nu is less than nu naught, the threshold frequency, no electrons come out, however intense the light. Right? Let me ask you this. Same frequency, but much brighter light. Does the stopping potential go up? No, it does not, because all three dark blue curves start from the same point, minus V naught: K max depends on the frequency, not the intensity. Same nu, brighter light, the same V naught. Clear? So what does brightness change? More photons every second, so more electrons, and a higher flat part, the saturation current. The sign on screen, like an open alpha, means proportional to, so the saturation current is proportional to intensity. Got it? And a higher frequency means faster electrons and a larger stopping potential, so on the right, as nu goes from nu one up to nu three, the orange curves start farther to the left, at minus V naught one, two and three. Remember the pair: frequency sets where the curve starts, intensity sets how high it goes. Okay?

2:35Counting electrons: the saturation current

Next, the saturation current. Say N photons fall on the emitter every second, and a fraction eta of them eject electrons; eta, the Greek letter like an n with a long tail, is the quantum efficiency. Each electron carries charge e, so the current, small i, equals eta times N times e. With me? And N is the light power on the emitter, capital P, divided by the energy of one photon, h c over lambda, with lambda the wavelength and c the speed of light. So N equals P lambda over h c. Right? Now, a small lamp is a point source, sending light equally in all directions. Move the photocell twice as far away: how much power does it catch now? It catches one quarter, yes, exactly! The light spreads over a sphere of area four pi r squared, r being the distance, and the cell, of area capital A, catches only its share. So P equals P source, the lamp's power, times A, over four pi r squared, and the current goes as one over r squared. Clear?

3:45Topper's shortcut: work in electron volt

Now a topper's shortcut: work in electron volt. Since h c is twelve hundred and forty electron volt nanometre, the photon energy in electron volt is twelve hundred and forty over lambda in nanometres. With me? Subtract the work function to get K max in electron volt, and that same number is the stopping potential in volt. No joules, no powers of ten. Got it?

4:10Example 1: lamp 3 times farther, find V0 and current

Now, Example one, our puzzle, from IITJEE nineteen ninety two. When a monochromatic point source, light of one frequency, is zero point two metre from a photocell, the cut-off voltage and saturation current are zero point six volt and eighteen point zero milliampere; m A is milliampere. Find them when the source is zero point six metre away. Pause the video and try it first. Done? Okay, let's check. The cut-off voltage is just the stopping potential. The lamp is farther away now, so does it change? No, it does not, and that is the first trap. Same source, so the same frequency, and every photon carries the same energy. Right? So K max is the same, and the cut-off voltage stays at zero point six volt. Moving the lamp changes the number of photons, not their energy. Clear? Now the current. From zero point two metre to zero point six metre is three times farther. So the saturation current falls to one third, right? No, no, wait, that is the trap from our puzzle! The distance r changes from zero point two metre to zero point six metre, and the arrow on screen means changes to. That is three times farther, and intensity goes as one over r squared, so the intensity becomes one ninth. Okay? So the saturation current, proportional to intensity, is eighteen point zero over nine, which gives two point zero milliampere. Puzzle solved: the cut-off voltage stays zero point six volt, and the current falls to two point zero milliampere. With me?

5:53Example 2: I-V graphs of 3 plates lit by 3 wavelengths

And now Example two, the harder one, from IITJEE two thousand nine. Plates p, q and r have work functions two point zero, two point five and three point zero electron volt, the e V on screen. Each is lit by a beam with wavelengths five hundred and fifty, four hundred and fifty, and three hundred and fifty nanometres, the n m, in equal intensities. Describe the current voltage graphs, taking h c as twelve hundred and forty electron volt nanometre. Pause the video and try it first. Done? Okay, let's check with our shortcut. Twelve hundred and forty over five hundred and fifty gives two point two five. Over four hundred and fifty, it gives two point seven six. And over three hundred and fifty, it gives three point five four, all in electron volt. Right? This beam has three kinds of photons. To stop every electron, which photons decide the stopping potential? It is the most energetic ones, yes, exactly! The stopping potential must stop the fastest electrons, and they come from the three point five four electron volt photons. Okay? For p, three point five four minus two point zero gives one point five four volt. For q, three point five four minus two point five gives one point zero four volt. And for r, three point five four minus three point zero gives zero point five four volt. Clear? Now the saturation currents. Every plate gets the same beam, with equal intensities. So all three plates give the same saturation current, right? Not quite, so check each photon against each work function. For p, with two point zero electron volt, all three kinds work. For q, with a work function of two point five, the photons of two point two five electron volt fail, so only two kinds work. And for r, with a work function of three point zero, only the photons of three point five four work. Got it? The smaller the work function, the faster the electrons, so the stopping potentials go p greater than q greater than r. And more useful photons means more electrons every second, so the saturation currents go p greater than q greater than r too. So p's curve starts farthest left and rises highest. With me?

8:26Quick revision: frequency sets V0, intensity sets current

So, quick revision. Frequency sets the stopping potential, and intensity sets the saturation current. For a point source, intensity and current go as one over r squared. And with mixed light, the most energetic photons fix the stopping potential. Now try the other IITJEE questions on this page. You can do this. All the best!