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### Course: Modern Physics (Essentials) - Class 12th > Unit 2

Lesson 6: Even electrons have wavelengths - How?# Comparing de Broglie wavelengths: Solved example

A proton and electron have the same kinetic energy, let's compare their de Broglie wavelength. Next, a proton, and an alpha particle are accelerated through the same potential difference, let's compare their de Broglie wavelength. Created by Mahesh Shenoy.

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- according to debroglie every particle behaves likes a wave the does that mean if we consider humans as a particle and if we shoot humans through double slit , then does human show interference pattern? 💀(1 vote)

## Video transcript

a proton and an electron have the same kinetic energy if the mass of the proton is 1800 times the mass of the electron find the ratio of their de broy wavelengths where do we begin well since we're dealing with debris wavelengths we could probably start by de roy's wavelength equation we've seen already in previous videos the debris wavelength that is the wavelength associated with any object having some momentum p can be written as lambda equals h the planck's constant divided by the momentum of the object and this is a pretty cool equation because it's saying that if you throw a ball which has some momentum it will have some wavelength a ball a moving cricket ball has a wavelength it behaves like a wave and that's the whole idea behind the wave particle duality that all things can behave like both waves and particles and where does this equation come from well we've derived this equation in a previous video and you can feel free to go back and check that out but i would recommend remembering this equation as a fundamental equation mainly because the derivation requires us to know about einstein's relationships of energy mass and momentum which is also beyond the scope of our syllabus so it's one of those few equations in physics that you can treat it as a fundamental equation and i would just recommend remembering it so this could be a starting point for whenever we're dealing with uh debris wavelengths all right so now we are what are we given we are asked to find the ratio of the debris wavelength so we have to find the ratio of wavelength of an electron to the wavelength of a proton and what are we given we are given that they have the same kinetic energy and we are given the ratio of their mass we have given that the mass of the proton is 1800 times the mass of the electron so what i'm thinking is we need to convert this equation in terms of kinetic energy and mass once we do that then we can compare them we can divide them and we can we can solve it so i want to get rid of momentum and bring mass and kinetic energy how do i do that we know what momentum is it is h divided by momentum is m times v so we brought mass into the picture so that's good uh but we want we don't want velocity we're not given any information about velocity we are given kinetic energy can i somehow get rid of velocity and bring kinetic energy into the picture i think we can and i want you to try to do this yourself first before we do it together all right let's see we know that kinetic energy of any object is half mv squared and so from this if i want to isolate v i can just rearrange and i get v square as 2k divided by m that is our v squared but then i can take square root on both sides and this is v so what i'll do is i'll just substitute that for v here and so v becomes square root of 2 k over m and if i simplify this what do i get i get m divided by root m is just root m so i get root of 2 k into root of m which is root of 2 km or root of 2 mk and there we have it we now have the equation for wavelength in terms of the mass and in terms of the kinetic energy and now we can build two equations and we can divide and and we can see what we get so again good idea to pause and try to do it yourself before we continue all right let's do it so let's divide wavelength of proton to wavelength of electron that will equal where length of proton would be i'll use the same equation it's going to be h divided by square root of 2 times mass of the proton times the kinetic energy of the proton divided by for electron i would get the same thing it'll be two times mass of the electron times the kinetic energy of the electron so h cancels the two cancels we know we're given that they have the same kinetic energy so that means the kinetic energy also cancels yay and so what we end up with now is this goes on top so we'll end up with both are under square root so i'll write a common square root over here we end up with mass of the electron divided by mass of the proton divided by mass of the proton and that happens to be let's see we know mass of the proton is 1800 times the mass of the electron so this would be i'll just keep mass of the electron as it is mass of the proton i can write that as 1800 times mass of the electron this cancels and i get this answer to be 1 over square root of 1800 and i can use my calculator i just take the square root of 1800 so 1800 square root that gives me about 42.4 something so i'll just write this as 1 by 42.4 so this will be 1 divided by 42.4 and so now i know the ratio i can say uh the the wavelength of electron has to be 42 times four times the wavelength of proton and there we go so we see that in this case the electron ends up having a higher wavelength compared to the proton and let's see if this makes sense if you look at the basic equation because they have the same kinetic energy whichever has smaller mass ends up having more wavelength we know electrons have tiny mass so we would expect the wavelengths to be higher and so this kind of makes sense and so it's like a sense check to make sure that i haven't made any mistakes over here in you know while solving it let's try another one a proton and alpha particle are accelerated through the same potential difference find the ratio of their debroy wavelength again we can start with the basic equation lambda is equal to h divided by p with the momentum and this time we are given a proton and an alpha particle so we we have to compare their wavelengths but we are given that they are accelerated through the same potential difference and i'm wondering how do i bring potential difference into the picture here i mean what's potential difference got to do with momentum like what is this well let's think about this they're accelerated through a same potential difference meaning there is an electric field probably and they're accelerated so let's try and draw let's try and make sense of it um let's assume that there are two plates over here this is positive play this is negative plate we are given that the the potential difference between the two plates is v so probably there is an electric field like this over here you know drawing like this will actually help us make sense of what's going on and now we are given that both of these particles let's consider any of these particles let's say there's a particle maybe this is a proton we're given that this is accelerated and that kind of makes sense if i take a proton and leave it in the electric field it will it will experience a force and it'll start accelerating it'll become faster and faster and faster and faster and faster and faster and by the time it comes over here it would have some kinetic energy and i know that if i if if i know the kinetic energy i can substitute like like what it did in the previous problem there is a connection between momentum and kinetic energy root of 2 mk i can do that so that means i need to find out i need to figure out how to calculate kinetic energy from potential difference if i can do that i'm done is that making sense so how do i do this how do i figure out what the kinetic energy is going to be well again why why don't you think about this this is going back to you know basic electrostatics so it's maybe try to give it a shot okay so i think of this as a ball falling down in in in gravitational field as its potential energy reduces its kinetic energy increases so this kinetic energy that it gained must be equal to the potential energy lost and so what is the potential energy lost well potential energy uh there is potential difference potential difference literally tells you how much is the potential energy lost per coulomb so this is one coulomb the potential energy lost is v if this was two coulombs the potential energy loss would be two v makes sense so this is this is a number this tells you how much is the potential is lost for one coulomb so if this was q coulombs the potential energy lost would be q times v so the potentially lost in this in this case must be q times v and that must be the kinetic energy gained and so now i can put this in this equation just like before and do i know the charge of the proton and alpha particle yes i don't need to know their charge i need to know their ratio and i think we do and we are given that they are excited to the same potential difference i think we have everything we need i think it will be a great time again to pause one last time and see if you can substitute everything and try okay let's do this again i don't remember any equation besides this so let me do it let me redo everything so momentum is mass times velocity and then i know that kinetic energy is half mv square just redoing everything we did before very quickly this time so v would be 2k divided by m so square root of that so when i substitute over here i get m divided by square root of 2 mk 2 m divided by k so that ends up giving me root of 2 mk just like what we got last time very similar and now for kinetic energy i will substitute q times v so this time i'll get h divided by square root of 2 m q v and now i can do the same thing as before i can divide the two wavelengths i can say wavelength of the proton divided by let's say wavelength of the alpha particle should equal h divided by square root of 2 times mass of the proton charge of the proton and the potential difference through which the proton was accelerated divided by h divided by square root of 2 times mass of the alpha particle charge of the alpha particle and the potential of the alpha particle potential through which the alpha particle was accelerated cancel stuff h cancels 2 cancels the potential difference through which they're excited was the same we can cancel that let's make some space and so this means we will get let's make some more space okay so this means we'll get this divided by lambda of alpha equals this comes on the top whole thing under root mass of alpha particle charge of the alpha particle divided by mass of the proton times charge of the proton okay now let's see if we know the ratio i know that alpha particle is basically the helium nucleus right so it has two protons and two neutrons so total four particles and this is one particle and mass of proton and mass of neutron are almost same so that means the alpha particle should be four times as heavy as the proton and so this should be let me just write that this should be mass of the alpha particle should be four times the mass of the proton what about the charge of the alpha particle i know it has two protons so it should have twice the charge of the proton so charge of the alpha particle will be two times the charge of the proton and there we go we can cancel stuff now and we get root 8 and we can calculate what that is but i'm just going to leave it as a rooted so that means we get the wavelength of the proton to be root 8 root a times the wavelength of the alpha particle and there we go that's our final answer