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A particle in a box (V = 0 0<x<L; V=oohenvise) is in the ground state (n=I) 0). Find the probability that the partiele is located betwcen Land 2 L6).. Obtain ...

Question

A particle in a box (V = 0 0<x<L; V=oohenvise) is in the ground state (n=I) 0). Find the probability that the partiele is located betwcen Land 2 L6).. Obtain the expectation value <P Xp>.

A particle in a box (V = 0 0<x<L; V=oohenvise) is in the ground state (n=I) 0). Find the probability that the partiele is located betwcen Land 2 L 6).. Obtain the expectation value <P Xp>.



Answers

A particle is in the ground state of an infinite square well potential given by Equation $6-21 .$ Calculate the probability that the particle will be found in the region (a) $0<x<\frac{1}{2} L$ (b) $0<x<\frac{1}{3} L,$ and $(c) 0<x<\frac{3}{4} L$.

Okay, So in this problem, we must find the expectation value off the square off the position. First of all, let's remember what is the expectation value off square off the position. OK, so expectation value is the integral from zero to l. So is the integral off Sigh star. Which side star is the complex? Congregated off the wave function. Ah, the multiplies access square Sigh DX. Okay, so that's the integral must. So in this problem. And we must remember that the wavelengths over of ah particle the rate of function of a particle confined in a box sigh off X is described by two divided by L the square root of two divided by al the sign off, uh, and by X divided by l. So that's the way function of a particle gun fine in a box. And that's we must in in the left side. Here is the expectation value we must solve. But for let's move on. So these expectation of Adam is going to be being to grow from zero to l off two divided by l causes square. Okay, and here we have X square sign sign square off and by X divided by L D X. Okay, so two divided by our is a question. We can take it off of the integral and the function is just excess square. Sign off this argument in here on. I forgot to put something in here. Ah, we know that is the expectation value of a function of a particle in the third excited states. So since this is the third excited state and should be equal for okay, because the ground state is one, therefore, the third excited state is four. So that's precisely the integral we must solve. Okay, so how we managed to calculate this? First of all, let's remember the half uncle half angle trigger dramatically identity, which is that's put in here, the identity. So this is on observation. Remember that sign square off X is equal half off. One miners course sign off to X. Okay, So this is the identity where we should use therefore or integral here is going to be the integral from zero to l. We have here two divided by l. The multiplies excess square divided by two. Therefore, we can cross to here with this to what is left. This one uh, excess square there multiplies one minus. Go sign off to times the argument. So two times the argument is going to be equal eight by X, divided by l the X. Okay, so that's the integral. So we have two terms in here, and if we actually make a distribute TV distributive here, we're going to find that this integral is just one divided by L off the integral from zero to l off excess square miners excess square Go sign off. Eight by X L All this d x. Okay, So the first thing to grow is easily evaluated because is ah, power the power integral. So the 1st 1 let's put here, let's call this one and this this too. So the first thing to girl is just, uh, let me see. Being integral from zero to l off x square, the X this is going to be equal Ex cubic divided by three from zero to l, which gives us a final answer off l cubic. Divided by three. Okay, so that's the first answer. Now the second integral. That's the first thing to girl now, the second interest a little more complicated. So the second thing to grow. We must use integration by parts. So let's put here first of all or second integral. So we have here. Um, the integral from zero to L off Axis Square was signed eight by X L the X and we must use integration by parts. And we're going to call you equals access square. Do you equals two x the X And because of that, we have DV equals because signed off eight by X L the X and the integral off TV What gives us feet equals sign off eight by x l divided by eight by l. Okay, so that's or integration by parts. And the second integral is going to be second. Integral is going to be, uh you've u times v. So this is just X square times. That's what your L divided by 85 times this sign off eight by X l When we go from zero to l miners the integral off V do you? So we have zero to l v d you. So this is just too x l eight by sign off. Eight by X l d x. Okay. Therefore, if we look to the first German here off these integration by parts when we have this sign equals l, This is going to be zero, because the sign off eight by zero And when we have extra square equals, you know? Well, this is also zero, which means that this entire Barton here is equal zero. And what is left is this second term, which is another integral, which is another integral, which should solve by parts. Okay, therefore, the second integral again, We're going to have u equals X. Do you equals d X? Um, you're because x the U equals D X. Uh, DZ is going to be this sign off eight by acts L the X and V is going to be equal the negative co sign of eight by x l divided by eight by l. Okay. Actually, I forgot to mention that we Well, that is correct. That is correct. That is not correct. Therefore, or a second integral here is going to be simplified to just let's put here with you solving the integration to this is negative. Ah, integral from zero. Sorry. No need to grow now. So this is negative. You've is u times V. So this is No. Eight by the multiplies X negative story because sign off eight by X l All this in the interval from zero to l ah minus which is going to change, is going to be positive to divided by eight by L Square integral from zero to l off VD you So this is just negative. So we have here negative because sign off eight by expel d X and that's is something that we can solve. Therefore, let's poured into here These final into growing here is going to be zero. That's because the solution for the integral of course sign is a sign. So this is going to be grow negative two divided by eight by l cubic Sign off eight by X l When we go from zero to out And as we can see here when the interval is l we have sign off If I which is zero and when the interval zero we have signed of zero which is also zero that for this integral here is again zero And the final solution for the integral to second term is going to be just, uh, negative. Negative. So this is just l is square Sorry. L cubic. I forgot to put in here. Uh, square. So this is El Square. Al que big Sorry. Divide it by 32 by square. And if we some both of the terms in here because we have two terms, right, the first term is dissolution in the second term is second term is the solution in here. Therefore the expectation value off the square off the position is going to be equal. Al Cubic divided by three minus al cubic. Divided by attorney to buy square. And that's the final answer to this problem. Thanks, George.

Okay, So in this problem we have a particle in a three d box and the energy state. So let's see. The energy state is described by the quantum numbers and X equals three en y equals two and NZ Eco's one. So this is our state, and we must find what planes the probability function is. Zero. Okay, so how we managed to do this? First of all, we need to remember how is described the probability function. So the probability function is basically the way Function square equals zero because we want to find the probability function when she zero. Okay, so the probability function for this problem particular problem has the solution. L divided by two to the power off three. So this is the standard solution. The only thing we going to do is to substitute by the states. So the first solution for X is signed. Square off three because we are in the states three. Hi x l The solution for why is design square off to hi. Why? L And finally the solution for Z is the Sinus square off pie Z out. So this is the probability function for this problem. For this particle in a three D box. And since we want to describe where this probability what planes, this probability is zero. We just need to analyze these three functions here. All the three signs. Okay, so let's begin with the final one. Let's begin with the solution for Disease. So, looking to the last solution here, we know that this function will be zero. Where, when by Z l is going to be quote to zero to buy and so on. Okay. Therefore, we can took from this solution Two planes, the planes were Z equals zero NZ equals well, health and these two planes corresponds to the walls off the box. Okay, so this is the walls off the box. And these two are Blaine's. There are equal zero. But the problem alright. Told us that the probability function will be zero at the walls. So we must find out. Um, in what planes in addition to the walls is going to be zero. Okay, So, looking for C, we only took these two planes in here. Now let's look to the second solution. Let's look to the why. So the wide solution we have two pi. Why l equals zero by and so on. So from this we have that the solution is going to be why Eco's l divided by two the multiplies k were k Britain's to the to the natural. Okay, so what is the solutions in this particular case? Well, we could have que eco's zero or equals two But in these solutions we are in the walls off the box, So que eco's one is the only one that is different. Therefore, we can say that the first answer is the plane where why is going to be equal l divided by two. So this is the first plane. Now let's look to the final won the X So the axes just three pi x l equals again to sure okay, by therefore, from this we get that X is going to be equal Que el divided by tree again. Okay. Eco's zero and three represent the walls of the box. Therefore, the two different solutions to two additional solutions it's going to be acts the planes acts equals l divided by three and X equals to dirt's off health. So this is the two additional planes plus the the Y additional plans. So we have three additional plans for this particular case. Okay, These two. And this one. That's the final answer. Thanks for watching.

And this problem, we have to visualize the wave function. Ah, very specific energy level. This is a very important skill to have because now we can understand and visualize what the's energies would really look like. And this is based off of an equation called the stronger Equation. And if you've heard that word before or blended in class or you haven't, that's OK. We're just going to be talking about the visualization of it. So let's first talk about that. Were given n equals three. Remember? N is the principal quantum number and equals three means we have an energy level of three. And what that means is that we have to evenly spaced nodes. Remember, a node is the region where we would find zero probability of finding an electron. And so I squared Is the wave function of that energy level. So the wave function simply is graphing the energy or the wave of that specific energy orbital. So this is what our graph would look like. Remember that we have Sai as R y axis and our X axis. We have the positions of these energies and notice here how well over three and two L over three. Hit the what? Pardon me. The X axis. That means they're reach zero. Those are two nodes. There is no probability of finding an electron or a particle here. But notice here that this dash line is right in between the first zero point and the second zero point. So it's right halfway between that first peak in our wave. So what this tells us is that first are nodes are evenly spaced and the distance from the origin of our graph and this dash line is 1/6. So there is a one and 1/6 chance off finding a particle within the specific energy energy level using what we know about the nodes. So I hope that this problem helped you understand or maybe introduce you to the concept of the wave function and how we can use it toe, understand? Finding the probability of a particle in that specific energy orbital

So we're continuing on with quantum chemistry here. And so we're taking a look at the time independent Schrodinger equation that provides means for us to calculate the wave function from quantum mechanical particle. So here we have the right heart. We have the right hand half off the box, which is X one equals a over to on we have X two equals a And so the probability off the particle being that is no 0.5. So we continue on to the second part where we have x one equals a over three x two equals to 8/3. Now the probability of the particle for the middle side of the box 9.61


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