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Integrated ConceptsA laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.71 eV? (c) Calculate the current of ejected electrons. (d) If the photoelectric material is electrically insulated and acts like a 2.00-pF capacitor, how long will current flow before the capacitor voltage stops it?

a) $\left(\frac{\mathrm{N}}{t}\right)=4.03 \times 10^{15} / \mathrm{s}$b) $\mathrm{P}=2.55 \times 10^{-4} \mathrm{W}$c) $I=6.45 \times 10^{-4} \mathrm{A}$d) $1.23 \times 10^{-9} \mathrm{s}$

Physics 103

Chapter 29

Introduction to Quantum Physics

Quantum Physics

Simon Fraser University

University of Winnipeg

McMaster University

Lectures

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So in this problem. Initially, we have a laser with power. People too. Maybe what's? And we know that the wave lands off. This laser is 400 uh, Nana meters. Okay, so in part A, we want to know how many photos will be ejected each second. So the energy of a single photon is h time. Seaver. Nanda. So you can see that this energy of three points want why? Electro volts and ah, so you can see that the total number of electrons that will be released in each second is the total power times one second. So here. Tease the peace. One second divide by the energy of signal photo. Okay, so this gives you a full 0.2 times 10 to the 15th photons. Okay. Okay. So in part B, I suppose so. Want to know? Uh, how many energies? Ah, that Ah, that are carried away by this? Ah, by the electrons. So that means when we need to find out of the convent energy of the electoral. So according to the electric photo, in fact, the total energy off the photo on equal to the funding energy plus kind energy off the electoral, right? And according to the problem, the binding energy is a 2.71 electro volts. And so that ah and E is the one that would have on here. Right? So we wait to see that, uh, the kind of energy ihsaa 0.398 Electric bolts. Okay, So the energy that are cared away by the electron equal to the total energy times K, you are by the pool and by the by the energy of the photo. Right. So the models like this. So this is the Adam, and this is the photo with energy. And this is the electron was my neck energy key. Okay, so then the energy carried away by the electron is Katie. So Katie, But I'II come to you. This is the energy carried away by the by the electrons and a you equal P times t ana Here, tease. One second. Okay, The times, Katie. Bye bye. Eat. So P is the power of the laser, which is a two million me waltz and tease One second and Kay is the one that we found here. On e is the energy of a single photo which is this 13.11. Okay. So you can find out the video of this. Ah, um, energy is ah, 0.2 55. Many Jos and in part in Parsi. Ah, we want to know the currents off this. Ah ah. I want in the current formed by the ejected electrons in part A. We already find out to the total number four towns each second so and is full 0.2 times 10 to the 15th photons. Angel second. Right. So we can see that to the charge, Uh, for each second, equal to cube equal. Negative in time z right. Which we remove this one? Yeah, negative end times easy. So the charge is Ah, negative. Six point. They're 43 times 10 to the native, fourth columns. So this is This means that each second they're totally negative. 6.43 times 10 to the fourth columns charge are released from this Ah, material. Okay, so the current will be go to here. We just We just need to focus on the magnitude. So let's say this is Q the magnitude you're bi t. So here t steal. It's one second. Okay, so this gives you four point. Ah, 6.43 times 10 to the navy forthem peers. So this is the current that are formed by the release electrons and in part D Ah. Suppose Ah, The material itself is an insulator and ate. It contains some, uh, passes and the sea is two people. Faraday's. So we want to know how long that we need. Thio, get this. Ah, but you were charged, so come back to the part B. So the energy of a photo on equal to the binding energy plus kind Any energy of this electron, right? So the binding energy according to this problem, is to 0.71 electoral votes and the energy of the full times 3.1 Wine electro volts. Right. So s So this is the material, right, Because a ce electric a cz the electron rejected the charge will start accumulated on the surface of the off the material and ah ah, this If the potential difference between the two surface ah could overcome with the kind netting energy of this electron, then the electron would stop going out right? So that means if kay equal, let's see v times Hey, so we're V is the potential difference off this off this material. Okay, so every time, see there the electron will stop going out the electron. We were just to stay on the surface. Okay, so in that case, we can see that. Ah, because, Kay, as we found, it's, ah, point 398 electoral votes. So we're gonna see that to the potential difference. The maximum potential different should be called 2.398 votes. All right, And then, uh uh, because we already know that it sees two people Faraday's and the potential difference is a 0.398 votes. So we're gonna see that the maxim charge accumulated on the material is Ah ah. See Humpty. Right. So this gives you 7.96 times 10 to the NATO 12 cools. All right. Ah, And remember that Ah, in each second. There, there. Ah, number off. 6.43 times turned to the native. Fourth columns comes out from the material, as we found in part. Ah, part seat. Right. So this is, uh, on per second and times teeth and times T. So this is the This is the charge speed that are released from the material. And this is the total time. So we're gonna so for the tea, right? So, t as you guys just solve this equation, you contend that equal 1.24 times 10 to the *** nine seconds. So that means after 1.24 times 10 to 189 seconds, then the material will stop releasing the electrons.

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