Through what potential difference must electrons be accelerated if they are to have (a) the same wavelength as an x ray of wavelength 0.220 nm and (b) the same energy as the x ray in part (a)?

Video transcript

Hey everyone. So this problem is working with wave particle duality, let's see what they're asking us. They want to know the electric potential difference required to accelerate a doubly ionized helium atom so that it has the same one. Part one is wavelength or part two is energy. And then they tell us that it is a 0.1 nanometer and it is X ray radiation is what we are comparing this doubly ionized helium atom to. So the first thing that we need to do is remember our electrical potential difference equation. And so that is given by the change in potential energy is equal to Q times delta V. Where delta V is that electric potential difference in this case, we know that all the electrical potential energy is converted to kinetic energy. So we can rewrite this as Q delta V equals one half B squared. So we're going to use this first equation here in both parts of the problem, but let's take this first part first. So it's asking us about the wavelength and that is a hint that we need to be thinking about. Debra Please equation where wavelength is given as H over P. H is Planck's constant and P is your momentum. We can also remember that momentum can be rewritten as M V, mass times the speed. So now we're looking at this term is very similar to this term. We are given wavelength in the problem. So we can kind of rearrange this equation, plug this in so that we can solve for potential for electrical potential difference. So the first thing we need to do is it's we have mass and velocity, not velocities squared. So we're gonna want to um square that mass term as well. Gonna multiply both sides by em. So it gives us A and B squared and then we can rearrange this in terms of um mass and velocity. So M V is equal to age over land up. Now. We can plug that back into where this term is and we have Mq delta B equals one half age squared over lambda squared. So from here we know we're still trying to solve for delta B, but we have everything that we need to plug into this equation. So let's look back at our um problem statement and make sure that we have everything that they're asking us for. So we have the mass that was given to us as 6.64 times 10 to the minus 27th kilograms Q. So Q. Here we need to recall is Q. For a doubly ionized helium atoms. That charge is going to be two times 1.6. So two because it's doubled um times 1.6 times 10 to the minus 19 columns. Okay. And then um we're solving for this term H that's recalled Plank's constant is 6.63 times 10 to the negative 30 for jewel seconds. And wavelength was given to us in the problem As 0.01 nm, Which we can rewrite as one times 10 to the -11 m. So from there it is a plug and chug to solve for delta V. So delta V equals plank's constant squared over. We have two times the wavelength squared times the mass, times the times Q. Alright. Which is 6 1.6 times 10 to the -19. Alright, plug that into our calculators and we get um Our electrical potential difference is 1.03V. And that is our answer for part one. So let's look at our potential solutions. Um So we can at this point kind of eliminate everything except for a But let's double check the answer for part two to make sure that we are on the right track. Alright, so in the same way that the wavelength was a hint for the first part of the problem. Looking at the energy is going to be our hint for the second part. So the first thing we can do is recall that the energy of a particle in terms of wavelength is given as E equals hc over lambda. We can rewrite that these are all constant. 6.36 point 63 planks constant times 10 to the minus joule seconds. Let's recall that the speed of light is three times 10 to the eighth meters per second. And then the wavelength was given to us in the problem which is one times 10 to the negative 11th meters. So the energy Of this, of a particle at this wavelength plug that in. We get 1.99 times 10 to the -14 jules. So we know that the this doubly ionized helium atom has the same energy here. And so we're gonna go back to that first equation where our electrical potential difference was given to us. Um That's Q delta B. And that energy is just the energy of that particles. That's E. So we can rewrite this in terms of delta v. That's just E over Q. Blood bad. And we just saw for E 1.99 times 10 to the minus jewels and Q two times 1.6 times 10 to the minus 19 poems. And that equals 6. times 10 to the fourth volts. And so we look back at this part to answer for a. That is this correct answer. So we know that A is the right choice for this problem. That's all I have for this video. We'll see you in the next one