Comparing stopping potential, and max. speed: Solved example

when light of frequency two f naught falls on a metal the stopping potential is found to be v what would be the new stopping potential when the light of frequency for f naught is incident given that f naught is the threshold frequency since you're dealing with photoelectric effect the first thing that i like to do is write down the photoelectric equation because i know that any photoelectric effect problem numerical can be solved from the photoelectric equation so the photoelectric equation says the energy of the photon should equal the work function plus the maximum kinetic energy okay now let's go back to the equation and see what we are asked we are given the frequency of the light that is falling it's given to be two times the threshold frequency and for that light we are given the stopping voltage and we are asked what would happen to the stopping voltage when four times the threshold frequency light is shown on it so how do we do that since frequency and stopping voltages are given so i think we need to convert this equation into frequencies and stopping voltage how do i do that well energy of the photon we know is h times f thanks equation planck's constant times the frequency what is the threshold sorry what is the work function work function is the minimum energy so i can write that as h times the minimum frequency which is the threshold frequency plus how can i write kinetic energy in terms of stopping voltage well remember kinetic energy is equal to the stopping voltage but in electron volts so to be uh converted to joules it will be e times the stopping voltage and so we are given f in two cases one's two f naught so when f is two f naught we are given the stopping voltage to be v so the question is when f is four f naught what is the stopping voltage so i think we can build two equations and then we can compare them and then we can see what the new stopping voltage is going to be so why don't you pause the video and see if you can plug in and solve yourself first all right let's do this so when when the frequency f is two f naught in the first case we'll get this to be two h f naught that equals h times f naught plus e times the stopping voltage is just v let's just write that as v so what i can do now is i can simplify and find what and see what v is going to be so we will be well if i just rearrange 2h f naught minus h f naught will just give me h f naught divided by e now let's do the second case in the second case when f f is equal to four f naught let's see what happens to this equation you get four h f naught equals h f naught plus e times some new stopping voltage which i need to find out i don't know what the new stopping voltage is so let's call that as v dash and so if i now simplify v dash becomes let's see this minus this gives me 3 h f naught divided by e and now i can look at these two equations i can divide them or i can directly say hey h f naught e is v so v dash becomes 3 times this part is v this is v and that's our answer there we go so what we find is that when we double the stopping voltage sorry when we doubled the incident frequency the stopping voltage did not double it became triple so this is one of those tricky questions that you can find and you know sometimes if you don't we might try to do things very fast and you might say ah stopping sorry frequency doubles maybe stopping voltage also doubles it doesn't you can see that so it's better to always go back to the basics and then solve it that way let's try another one when photons of energy six electron volt is incident on a metal the speed of the most energetic electrons is found to be v and the question is if the photons of energy 11 electron voltage inside on the same metal what would be the speed of the most energetic electrons we are given the work function i think it's a very similar question compared to like like very similar to the previous one so why don't you pause now and see if you can try this yourself first all right let's do this so i start with the photoelectric equation it says the energy of the photon equals the work function plus the maximum kinetic energy now this time i look at the equation a question i say okay energy of the photons is given so i'm going to keep it that way work function is even i'm going to keep it that way and we are asked to calculate the more speed so let's convert this into speed how do i how do i convert kinetic energy into speed i can i can just say k is equal to half mv square so this would be half m v max squared and now i can plug in the values for the first case and second case and compare just like before so let's do that so in the first case the first case for six electron volt i get the speed to be v so when this is six six electron volt the work function is given to be two and that is the same for i mean in both cases it should be the same because it's the same metal plus half m v max squared or in fact they are calling that as v let me let's also call that as v so v squared so if i if i solve this i will get i can directly calculate what v square is i can simplify so i get 4 electron volt equals half m v squared in fact i'll just keep it this way because i know that i'll get something very similar and if i divide that half and m will get cancelled so i'll just keep it like this so that's my first equation in the second case i get uh energy of the photon is given to be 11 electron volt so 11 electron volt equals same work function to electron volt plus half m v dash square this is the new speed that i need to figure out and again if i equate this uh sorry if i solve this 11 minus 2 is 9 i get 9 electron volt equals half m v sorry v dash squared and now i can divide these two if i divide these two i get 9 by 4 equals on the right hand side half half and m cancels i get v dash squared divided by v squared and so i can now take square root on both sides so i get 3 by 2 equals v dash by v or i get v dash equals 3 by 2 is 1.5 1.5 times v and there we go that's the new speed