Hello. And in this question here will be looking at how to normalize the wave function and calculate the probability of X. Of the particles between some interval X. So the way function we've been given is equal to B times a squared minus x squared B times a squared minus x squared when X is less than a and greater than minus A. The wave function is equal to zero in all other cases. Yeah. So we want to find, we've been asked to find using normalization this real number B in terms of this real number A. Okay. Well, what what does the normalization condition mean? Well, this means that the probability of finding the way the finding the particle in a region from minus infinity to plus infinity is equal to one. Well, how do we write this? Well, the probability of finding the particle, Okay. In a region from minus infinity to plus infinity is given by this integral here. Where we're taking the complex conjugate of the wave function, multiplying it by the conjugate and integrating it from minus infinity to plus infinity. And this must equal one because we know the particle must be located somewhere between plus infinity and minus infinity. So if we were to compute this integral we will be able to find B. In terms of A. So to do this, I'm first going to break up the limits of integration in this integral. So the left hand side is equal to the integral of the wave of the. So one thing just to know before we start is that in this case so I squared is equal to the complex conjugate of sai time Cy but we're told in the question that both A and B are real. So sigh. The complex consecutive cy is equal to sigh. So this means we're just integrating the wave function to be squared. So this here is equal to the wave function to be squared from minus infinity to plus infinity. No rewriting over different limits of integration. So we're gonna say we start from minus infinity and go to minus a. The integral of size squared dx. And now we're going to integrate it from minus A. Two plus A. And we're integrating size squared dx. And finally we integrate from A to plus infinity of size squared dx. As you can see here we we've integrated this this way function to be squared in three separate places. And the reason why we've done this is because the wave function in the region from minus infinity to minus A is equal to zero. Similarly the wave function from plus to plus infinity is equal to zero. So an integrate the integral of zero is equal to zero. So the only bit of this integral on the left hand side that will contribute is the integral from minus a two plus A. So we just need to evaluate integral from minus a two plus A. Of size squared dx. Well this is equal to squaring outside. It's going to be b squared times a squared minus X squared all to be squared dx, expanding this square out and moving the constant be outside the integral. It's going to be equal to b squared times the integral from minus a two plus A of eight of the power of four plus X to the power of four minus two A squared x squared dx. Now when we're integrating something, all we do is we increase the power by one and divide by the new power. So this gives us b squared multiplied by A to the power of four times X Plus X. to the power of five, divided by five minus two. To the power minus two times a squared times X. To the power of three divided by three. And we evaluate this at a. From minus A to A. So filling in from A. To minus A. We get B squared is equal to eight of the power of five plus. Sorry. So B28 of power five because they're 118 of the power five will come from the plus a contribution and the second agent para five will come from the minus A. Then the next term is twice eight of the power of 5/5 and the final term is minus 4/3. 8 to the power of five. We can pull out a factor of eight to the power of five. To get eight of the power of five times B squared is equal to two plus two, divided by five minus four divided by three. And this is equal to eight of the power of five times B squared times 16/15. Now we said in the beginning that this integral here was equal to one. So because because it's normalized, so this means that eight of the power five times B squared is equal to 15/16. And this allows us to find B in terms of A and B is equal to The square root of 15/16. 8 to the power of five. And that there is the first part of the question. We now want to find the probability of finding a particle, the particle. So this is our X axis. And we're starting in a position X is equal to a over to. And we want to find the probability that the particle is within an interval of 0.1 Aid. That's something. So we want to, the probability of finding X within this interval is equal to the second. The so we're going to have X to be this value here and X one B. That value there. And we want to find out the probability that the particle is located between X one and X two. And this is just equal to the integral from X one to X two of size squared because the wave function is real. D. X. And we want to just evaluate this integral. Okay. And this will equal the probability. Yeah. Thus yeah, particle particle is within X one and X two. So what is X one and X two? Well if we look up here, The distance between X one and X two is equal to 0.1. A. So we're gonna add so X um X two will be equal to a over two plus 0.1 A divided by two. And this is apologies and this is equal to 101 over 200 a. And this here an X. Warm Is equal to a over 2 -0.1 a over two, which is equal to 99 over 200 a. Okay, So so this is equal to 0.495 a. And X two is equal to 0.5058. Okay, so we want to compute the integral of change colour Again, The integral of 0.05 a. So we're integrating from 0.4958 to 0.5058. The integral of the wave function to be squared which is b squared times a squared minus x squared all to be squared dx. But we evaluated this integral in the first part of the question and got that it was equal to this value here. But with the only difference is that we have now We have now changed our limits of integration from -8 a 2 0.495 a. At 2 0.5058. So this is equal to B squared times A. To the power of four. Just rewriting what we had before plus X. To the power of 5/5 minus 2/3 A squared excellent power of three Evaluated at 0.505 A and 0.4958. Okay, so just doing this evaluation, it's quite messy and it's not the most pleasant calculation to do. But I got an answer approximately equal to Um five. So it's sorry 0.56 to eight. to the power five B squared. So this is just by sobbing these values into we're just evaluating this year. And we also know from the first part of the question that A. to the power of five times b squared is equal to 15/16. So this means that we guess the integral is equal to 0.005-6 eight. So this means that the probability that the particle is located between the regions um The region 0.05 a. 0.495 a. is equal to. So this is a. So this answer here is intestinal. So we just need to multiply by 100 to convert to percentage. So it's equal 0.5-6. Okay. And finally the last part of the question is is we want to find the probability that the particle is located between A and a over two. Okay. Well We already explained how to calculate the particle the probability of the particle when it's located between X- one and X- two. So in this question, this part here we're going to be setting um X two is equal to a and X one is equal to a over two and we're going to integrate the wave function. Okay? Um to be squared with respect to X. Well, once again, this is just equal to a over to the integral from a over to a of B squared, multiplied by a squared minus X squared all to be squared dx expanding this out. And computing the integral like we've done early before the question we get B squared times eight to the power of four times X plus x to the power of 5/5 -2/3 times a squared over x cubes. And we're evaluating this at A and a over to okay, so stopping in these limits, We get that we have 53 over 480 A. to the power of five times b squared. And once again using the fat from the first part of the question that eight of the power five times B squared is equal to 15/16. Mhm. Way down there, this will give a final answer Of 53/15, or this is in decimal or approximately 10% chance of the particle being located between a over two and a. And that there will conclude the question.