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Which metals will show photoelectric emission

Light of wavelength 430 nm is incident on each of the metals given in the table. We will explore how to predict which of the metals will show photoelectric emission. Created by Mahesh Shenoy.

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Video transcript

let's solve a problem on photoelectric effect we're given light of wavelength 430 nanometer is incident on all the metals in the table which metal would show photoelectric emission so light of this much wavelength is is is incident on all these metals and we have to predict which of these will show photoelectric effect in which of these cases electrons are going to come out so where do we begin now there are so many different kinds of numericals that can be asked on photo on photoelectric effect that i used to think that oh my god for every numerical now i have to remember how to solve it but no for any numerical all you need is einstein's photoelectric equation and it's not enough if you just remember the equation you have to understand it so that's where we'll start let's start by quickly recalling what the whole einstein's photoelectric equation was the whole idea is if you have an electron trapped inside the metal and you want to release it it needs energy and the energy is coming from light which is made of photons particle nature of light and each photon has some energy which is given by the planck's equation h times f f let's use green for f and when that photon when that electron absorbs the photon it uses its energy for two things it will use it a part of it to escape from the metal so part of it is used to escape from the metal and that energy is often called the work function of the metal and that's what's given over here how much energy is needed how much minimum energy is needed to escape from the metal and the rest of the energy will go out as kinetic energy and therefore the photoelectric equation from energy conservation we can say is the energy of the photon is the work function plus the kinetic energy and of course some electron most electrons will not be lucky they might also lose energy internally and so only those few lucky electrons will come out with the maximum kinetic energy and so that's why we say this is k max so this is the first step for any numerical on photoelectric effect this is the heart of solving any problem on photoelectric effect all right so where do we go from here so we're asked to predict which of these will show photoelectric effect so i'm thinking why why won't some of them will not show photoelectric effect why well think about it if the energy of the photon is less than the work function itself then there will be no photoelectric effect because work function is the minimum energy needed so for this to show photoelectric effect the photon needs to have at least 2.14 electron volts of energy for this at least 2.3 and so on so imagine if my photons energy was i don't know maybe let's say 2 electron volt then i know none of them will show photoelectric effect if my photons energy was 5 electron volt as an example then i know all of them will show photoelectric effect so the so you you see what we have to do we have to figure out that means what is the energy of the photon and then see if it is more than the work function if it is photoelectric effect will happen and how do we calculate the energy of the photon e is equal to h times f so now would be a great idea to pause the video and see if you can continue and try and solving and try and solve the answer try and solve the problem all right let's do this so the first thing the main thing we have to find is what's the energy of the photon and we know how to calculate that that's going to be the planck's constant h times the frequency of light but here's the thing frequency is not given to us so here's the frequency frequency is not given to us we are given the wavelength of light how do you calculate frequency from wavelength what's the connection between them now again you can remember a formula but i don't like to i like to think of this logically so again if you have forgotten it don't worry go back to your waves again very quickly let me show you how i do it what i do is i take example so let's say frequency of a wave was 5 hertz as an example and let's say it was traveling at some speed i don't know maybe 100 meters per second or something 100 meters per second then the way i think about this is here's the source if i wait for one second then i know in one second five waves will come out so it'll be one wave we have number two three four five and this would be the first wave this would be second third fourth and fifth this the wave is going to the right and that first wave in that one second would have traveled a hundred meters right because it's going hundred meters per second and so i now know five waves occupy a space of 100 meters so one wave will occupy a space of 100 divided by 5 and that's our wavelength so we're going to be 100 divided by 5 but what are we doing in general we are dividing the velocity by the frequency and there you go no confusions you don't have to remember anything that's that's how i like to do this so this will be h times frequency would be the velocity velocity of light and we know velocity of light that's c so velocity of light is c divided by the wavelength the value of h we know that it's 6.6310 to the power minus 34 velocity of light is 3 times 10 power 8 wavelength is given so we can substitute let's do that so this will be 6.63 times 10 to the power minus 34 times 3 times 10 to the power 8 divided by lambda which is 430 nanometers so that will be 430 nano is 10 to the power minus 9. and so now all we have to do is plug all of that in in our calculator i'm going to quickly do that let me just calculate these numbers so here's my calculator 6.63 times 3 divided by 430 gives me 0.0462 so let me write that .0462 times 10 to the power how much do i get so i have a minus 34 then this 9 comes on the top and gets plus 9 plus 8 plus 9 is 17 so minus 34 plus 17 gives me minus 17 and if i shift to zeros i will get 18 19. so i get 4.62 times 10 to the power minus 19 and what's the unit joules this is the energy of the photon and if this energy is more than the work function photoelectric effect will happen otherwise it won't but now we see another problem this energy is in joules and the the minimum energy that's given here work function energy is given in electron volts and so i can't directly compare them so either i have to convert all of this in joules which i won't do or we can con convert this into electron world and that brings us now to the last question what exactly is an electron mold and how do you convert them between joules and again this could be confusing should i multiply should i divide by something well again here's how i think about it an electron volt one electron volt if i want to convert it into joules just substitute for that e there is an e over here so it's going to be 1 times e value we know is 1.6 times 10 to the power minus 19 coulomb and voltage volt now this coulomb volt is joules this itself is joules how you ask well remember volt or voltage or potential is work done per charge so joules is just c times v so this is how you convert so one electron volt is one point six times 10 power minus 19 joules but i want to convert joules to electron volt so what is one joule equal to will be one divided by so 1 joule will be 1 divided by 1.6 times 10 to the power minus 19 electron volt and so now i can convert this so in energy of the photon in electron volt will be 4.62 times 10 to the power minus 19 and to convert joule into electron volt i'll divide i'm going to write that over here 1.6 times 10 to the power minus 19 electron volt that makes sense so i can already joules to electron volt and this cancels out and if i again bring back my calculator there we go so now i'll do 4.62 divided by 1.6 and that gives me 2.887 i'll round it off to 2.89 let's do that so this will be 2.89 electron volt that's the energy of our photon now let's look at this when i insert this on cesium will there be photoelectric effect yes because the minimum energy is 2.14 less than 2.989 will happen what about here it will happen it will happen here it won't happen because the minimum energy is 3.2 the photon doesn't have enough energy won't happen here also it won't happen so these are the three metals that will show photoelectric effect