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A photo sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase(a) the voltage applied to t...

Question

A photo sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase(a) the voltage applied to the light source(b) the intensity of light(c) the wavelength of light(d) the frequency of light

A photo sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase (a) the voltage applied to the light source (b) the intensity of light (c) the wavelength of light (d) the frequency of light



Answers

A photo sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase (a) the voltage applied to the light source (b) the intensity of light (c) the wavelength of light (d) the frequency of light

Okay, so at the top here, I've written to broke the relation and the relationship between kinetic energy and momentum. If we can combine these two equations, we get the relationship between the kinetic energy and the broccoli wavelength of the particle. So in part A, we're told that in electron has the wavelength of five times er sorry. Five nano meters. So it's energy will be 6.63 times tends the negative 34 squared over two times the mass of an electron times five nano meters squared. This comes out 9.71 times 10 to the negative. 21 Jules. To convert that TV, we divide by 1.6 times 10 to the negative 19 and that comes out to 0.6 TV, then, by definition of electron vole being the energy that's gained by an electron, Um, when it's passed through 14 it follows that potential needed is 0.6 words in part B were asked what energy excuse yet what energy of photons would have had the same wavelength of five nano meters. So Lando's equal Teo HC, over G okay. And to calculate this, you can either use H in jewels dot seconds, which is 6.63 times sends the negative 34 with the expressed in tools 9.71 times 10 to the negative. 21. Yeah, Or you can use on h expressen tv dot seconds. And he expressed in electron volts on in part C were asked what wavelength the photon would have if it had the same. Oh, excuse me. Sorry, I I got these two swat, so let's back up a little. Well, we want to use for this one is e is equal to H C over Lambda Where? Lambda Sze five Nana meters. Um, now And see, we want to know what the wavelength would be if we were using the same energy as the electron Hot in part A. And now we just rearrange this equation. Tau Lambda equals HC over

We are given data based on a photo electric experiment, wavelengths and nanometers and stuffing potential and vaults. The question asks us to plot the frequency on the X axis and the starting potential on the why access and then from there, enter and give the frequent the special frequency, the threshold leveling the work function of the material and plants constant, given the value of E, which is the charge of Trump. So our first job is to convert wavelength into frequency to plot on the X axis. And for that, we use this equation which is see the speed of light legal to frequency times. Wavelength frequency is equal to the speed of light divided, by the way. Thanks. And from there we can get our data that is updated. We can use this frequency data to plot frequency on the X axis and stuffing potential on the Y axis. If we would apply on it, we get a linear graph, a linear function, actually concede There data points here on this graph, which are the red dots, and you get this linear fit from the spot from these data points. What this is is a, um an approximation based off of the data. The slope here is, um, 4.112 times 10 to the negative 15 and the Y intercept of this linear fit is minus 1.894 So in this chapter, you're given equations before electric experiment equation 38.4. And what this is is that it tells you that the energy of the electron after its ejected is he going through the energy of the light that's shown onto the surface where the electron was living, essentially minus the work function, which is the minimum energy needed to reject the electron from that surface. If we rearrange this into point Slope formula, where V is the why and the frequency is the X, we got this equation so you can see the slope is age over E. And the intercept is minus work function over here. So now we can answer some of these questions here. Professional wavelength are sorry that special frequency and the threshold wavelength occurs at a special point. It occurs at a plant where the current just begins to occur and we're just where the stopping bolted just needed to stop the flow of current. That's where the stopping voltage is equal to zero or were V is equal to zero. So in our equation, if we set vehicle to zero, we can move the wire sets onto that side by adding work function over E. And we get this equation here down below. So we get the work function over equal to age over e times the frequency. Well, if we use our substitution ins, we actually get the frequency equal to minus B over em. And we know what these values are from Margraff above. It's one ar minus. B is 1.892 a. M is 4.112 times 10 to the negative 15. This gives us a value of 4.59 times 10 to the 14 hurts all from the values that we had above. So now we have to find threshold wavelength, which is easily solved by using this equation that we used from before. It's the way we're flying. Sorry. The speed of light equal to the frequency times away flying thing. From here we can see that a wavelength physical to speed of light over the frequency which is it gives us 653 nanometers. So these two the threshold frequency and verse one wavelike. The next question is to solve the work function, but we have the work function here. It's actually in a relation with the Y intercept. We know what the y intercept, IHS or you know what he is. However, the question asked us to give us the work function Evie's So we have the work function equal to minus B times E. B is already involved. So we get 1.894 votes times he which is 1.9 TV's. So that's simple. Our last question is to find a church, which is Plank's constant Given e. We have another relation for that. Here it's h over equals. Tow em on the slope. You know what the slope is, and we know what is. Slope is 4.112 times 10 to the negative 15 and is 1.602 times 10 to the negative 19. So this gives us an end result of 6.5 million times 10 to the negative 34 Joel seconds, which is in S I units. Thank you for watching this video. I hoped it helped. I'll see you in another one

So here we're going to solve problem 30 31. The question asks us to plot frequency versus stopping potential. So in order to do that, we're gonna use an equation. But first, let's write down the steps world, um, convert our wavelengths wavelengths. I'll put Omega, as is the symbol for wave links to our frequencies, which is F and we're gonna use an equation that's given tow us. It's assumed that we can use this and that. We know this sees equal to Omega F. I'm sure you know this equation for C is the speed of light, which is three times 10 to the eight meters per second. Um, to get our frequencies, we just rearrange like so f equals to see over mega. Now what? It just so happens that our frequencies, our frequencies, are going to be in the magnitude of the hundreds of trillions hundreds of trillions, which is around, um, the 10 to the 12. 10 to the 14 range. So what I'll do is I'll write the frequencies in, um times 10 to the 12. So whatever number right down, I'll just multiply this. You should assume that it's multiplied by 10 to the 12th. Okay, So in other words, it's gonna be in terror. It's so our frequencies that we get from using the equation from below air and terror hurts. 820 uh, 740 688 6 10 5 50 and 5 18 So now that we've gotten that done, our second step is to graph frequency in the X axis. Monograph are frequency on the X axis and our stopping potential on the Y axis. Okay, so let's draw a graph here. These are axes R F R Frequencies on to be on the X axis are stopping potential on the line. Um, in our data does not start at zero. So all kind of right this mark here to indicate that our frequencies start around 500 By the way, this is five. Whatever number is going to be multiplied by 10 to the 12 hurts, just like how I wrote about. So it's around 500. Terror hurts all the way up to over 800 terrorists. So we'll just go to this for good measure. I stopped in potential goes all the way up to about 1.5 volts. So anything above one point for, well, just plan above there. Okay, so that's plot our data. And now a plot it and I'll speak afterward so that we can go by. I'm fast. But for the 1st 1 actually, I'll just say we have 518. Terror hurts, of course. Mining 2.24 votes for our stopping potential. It's about there Texas 5 50 and points three. Sex. Can we have 6 10 in 100.62 and I'll just go ahead and plot the rest? Right? Okay, so now that we have our Dina plotted, I want you to notice that it's posits sort of positively correlated, and it's linear. So we're gonna use point Slope form, which is our former left for mind. And we're gonna use that with another equation that would skip that's given to us and solve some of the questions. So our third step here is to use an equation that's given to us, and that equation is 38. Question 38.4. That is given. Ah, and of course, Chapter 38. And we're gonna use that with point slope formula. Sorry, guys. so formula. And we all know what point that formula is. It's why equals two M x plus B and equation 38.4, which is given to us, is equal to this. It's e v or E vo, which this is stopping potential equal to the energy of light, which is plank's constant times the frequency of light minus the work function of the material. Now we're gonna have to rearrange this to, um, make this look like point slope form. Um, And to do that, we just divide by e on both sides, which is the charge of an electron, by the way, cancels on that side and you're left with vo you gonna each over e v or sorry h over e times f minus the work function over G. Okay, so you can see that we have are starting potential, which is why F for frequency, which is X and have our slope which we now know our slope, which is M, is equal to H over E. And our intercepts for y intercept is the work function over. Remind us the work function over eat? No, um in order to get the slope all you have to do is pick two points on this graph here and use that equation that you learned back in algebra, which is the difference and why between two points is divided by the difference in X and two points. I won't do that here, but you should know that the value would get for Slope is 4.112 times 10 to the negative. 15 folds over hurts. It's such a small number because our hurt value is so big that's on the bottom now, for be all you have to do is use point slope form. Since you know what I m is, you can pick any point on the graph at which gives you a why and the next value we can you can get be. So be is negative. 1.894 Thanks. Okay, these are the two values that we can use to solver our problem. The first problem is to solve for the threshold frequency. Now, with their equation here, the threshold frequency is I'll write it again. I'll write the equation again. E v o equals to age or things constant times the frequency minus the work function theme. The value of the first hole frequency is when the light has an energy that's exactly equal to the work function. That means the elect the electron on the metal plate of the photo electric experiment just rejects. It doesn't have any remaining energy. This is the energy of the electron show. In other words, were saying that this sign is equal to zero. But if we don't, If we divide uhm zero, which is gonna be this value here if we divided by E. And we get this equation, this is on the B zero videos on a vehicle to zero. So this is what I'm saying here. We're going to say that zero is equal to H over E times frequency. When this work function over E if you multiply both sides by e, you would get, um, you get zero equals stay, Jeff minus the work function, which is the threshold. That's a special frequency with a point and what you get the factual frequency. So, um, how do you fix that? Our question, which asks us about the special frequency, could be solved easily by just writing, like so, If is equal to the work function over e divided by each over e. And we know what these values are from up here. Um, the negative of the work function over he is Be So we're gonna get negative, be here divided by the slope, and they're gonna be over. The slope is equal to 1.894 bolts over 4.112 times tend to be negative. 15 Goencz hurts, and this value is 4.59 times 10 to be negative. 14 her. I'm sorry. Uh, it's positive. 14 because through our hearts is going to be in the hundreds of trillions to very large number. So our special frequency is 4.59 times 10 to the 14 hurts. I'll just erase that. So it's clear. Um, our next question is the threshold freaked out. Wavelength and weaken. Solve that by using the equation that we used above and Step one, which was Sequels to Omega f. Not going to write that again. Um, I'll just write it like this are special frequency and, um, Harold label this thresh just to make it clear that we're doing the threshold. It's just see over if their shoulder frequency. Um, in that value, um, I don't write it all out. Um is 650 53 nano meters as the answer for this question. So next we're supposed to solve for the work function in terms of E V E's. So we know the work function minus the work function over E. This is going to be and B is equal to negative 1.89 four volts. So simply all we have to do is multiply both sides by negative one that cancels the negative and multiply both sides by E. And that cancels the E on the left side. And we're let work we're left with. The work function equals to 1.8 nine uh, TVs. So that one was the simplest one so far. And the last question is fine. The value of plank's constant given the value of E. So our slope, which is M, is equal to H over e. So our H value is equal to I m come e. We know what our slope is. And we know what is the charge of the oven electron. So we write everything all out. We get 4.112 times 10 of the 15 volts over hurts times 1.602 times. Tend to be negative. 19 corms. Yeah, and we get approximate value for Plank's constant, which is six point 59 times 10 to the negative. 34 Jewell seconds. Jewell seconds. Because cool arms can be rearranged. Um, um, to get to me when multiply will buy votes over Hurts to get thes units here. So this values kind of close to the true value. But, um, this is the approximate value that you get when using this equation. So thank you for watching this video. I hope to help. I'll see you in the next one.

Part, eh? A maximum. Ah, kinetic energy Often rejected Electron remains the same. Maximum kinetic energy or and ejected electron remains the same. Remains the same art be, uh, the minimum de broccoli. Bad land off in electron also demands the same. So the minimum the minimum, not de Brantley Fab Lind, Often electron. Um oh, and electron remains the same. Remains the same. Artsy, the number of rejected the number of electrons ejected per second in crazies. But the number the number Oh, electrons ejected 1st 2nd burger. Second increases in crazies and body on the electric current I in the photo tube, uh, increases increases.


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