5

Particle is represented by the following wave function:C(1 - 2l/L) 0 < Ix] < L/2 @() Ixl > L/2 where C is real positive constant_ Find the normalization co...

Question

Particle is represented by the following wave function:C(1 - 2l/L) 0 < Ix] < L/2 @() Ixl > L/2 where C is real positive constant_ Find the normalization constant C Compute the probability of finding the particle in an interval of width 0.010L at I = L/8, i between x 0.120L and 0.130L- (No integral is necessary for this calculation:) Compute the probability of finding the particle between L/6 and € = L/3. (d Calculate the average value of and the root-mean-square (rms) value of x_ i.

particle is represented by the following wave function: C(1 - 2l/L) 0 < Ix] < L/2 @() Ixl > L/2 where C is real positive constant_ Find the normalization constant C Compute the probability of finding the particle in an interval of width 0.010L at I = L/8, i between x 0.120L and 0.130L- (No integral is necessary for this calculation:) Compute the probability of finding the particle between L/6 and € = L/3. (d Calculate the average value of and the root-mean-square (rms) value of x_ i. @. Trms V (z2) .



Answers

A particle is represented by the following wave function: $$ \begin{aligned} \psi(x) &=0 & & x<-L / 2 \\ &=C(2 x / L+1) & &-L / 2<x<0 \\ &=C(-2 x / L+1) & & 0<x<+L / 2 \\ &=0 & & x>+L / 2 \end{aligned} $$ (a) Use the normalization condition to find $C$. ( $b$ ) Evaluate the probability to find the particle in an interval of width $0.010 L$ at $x=L / 4$ (that is, between $x=0.245 L$ and $x=0.255 L .$ (No integral is necessary for this calculation.) (c) Evaluate the probability to find the particle between $x=0$ and $x=+L / 4 .(d)$ Find the average value of $x$ and the rms value of $x: x_{\mathrm{rms}}=\sqrt{\left(x^{2}\right)_{\mathrm{ax}}}$

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.

So to normalize we take the wave function given and we integrate the square of that function within the limits negative eight to a. Which is the entire space. Um where the way functions contained said the integral called one. So apart from the integral um we get this term on the left here. 16/15 to the five B squared equals one. And solve for B. Okay we get the term on the right side. And for part B we take a small region Thin region of with .1 centered around half of it. So uh the integral people from has limits of over 2 -2 of the width .058. And the upper limit is over two plus .058. And we have the same rate function squared integrated. So um Uh in the wave function we replace the baby with the expression we found in part a. And perform the integral. And the numeric value we get is 0.0053. Then part C. We have a similar calculation here, but this time we change the limits. Lower limit is half of a. Upper limit is A. And the same function square. And the numeric value we get is 53 or five. Close, so that's the probability for part C.

So the way function is given as a times E to the minus absolute value of X, divided by a and we have a basic graph of the wave function right here. You see that on the right hand side, it looks like a decaying exponential function. And on the left hand side, you essentially have a mirror image of the function on the right. So this is actually an important observation to make about this wave function, because when you carry out integral the mirror image property about the vertical axis is important to take into consideration. So that's actually what we're gonna be doing next in getting a normalized wave function and determining the value of a So one thing to notice is I have plotted the function and made note where the values of a occur along the X axis and also the maximum value of the function. As you'll see when we get the normalized way function, you'll notice that the maximum value occurs at the origin at X equals zero, and the maximum value of the wave function is actually one divided by the square root of lower case A. So let's go ahead and get the normalized way function and we'll start with the usual. I will say that the integral from minus infinity positive infinity of the probability density is equal to one. And in doing that, this means that here we have a squared times, negative infinity to positive infinity of either the minus two times the absolute value of X designed by a Now we don't wanna have to integrate an absolute value inside of an exponential. But again, we can use the property off the symmetry of this function and this will allow us to rewrite our integral. So note that the area under this half of the curve is equal to the area under this half of the curve. So we can rewrite the integral in terms of a single exponential function, utilizing this symmetry property of the area and remember that a definite integral in this case for a one dimensional function is simply the area under the curve. So going forward this integral can then be written is two times the integral. So this two means that one of these areas multiplied by two will give us the entire area under the entire curve. So we write this as zero to infinity of each of the minus X over there. There's a two right here. Let me make this clear. You're the minus two X over a times dx again. This to just tells us that whereas this integral is the area under the curve on the right side, the two allows us to get the entire area because this area on the right equals the area on the left. So going forward now we have minus two a squared times, lower case a over to times the function and brackets because it's being evaluated from a definite integral. This will give us negative capital a squared times, lower case A and we here have either the minus two times x approaching infinity divided by a So we see that this is going to be zero because e to the minus X with X going to infinity gets incredibly large e to the minus. An incredibly large number must decay to zero. Here we have minus E to the negative to time zero over a and either the minus zero is one. So we have these two terms and going forward Now we have capital a square times Lower case A. We have zero minus one and this gives us capital a squared times. Lower case A. That's a multiplication. And remember, this must equal one because we're normalising. So we see then that capital a squared equals one over lower case A and therefore capital a must equal one over the square root of lower case A So one eagles, a square root lower case A. And one other thing we're going to dio is calculated probability in the interval from minus a Do you a too positive A. And this is going to equal capital a squared minus infinity or rather minus a. Remember, we're just looking at a finite interval now, minus a toe positive A. And we have our the square of our way function and this is written is so again we can use our trick though this is symmetric about the vertical axis so we can use her trick and substitute in our value for a squared capital a square and we now have a two. Because this way we can get rid of the absolute value in the exponential. We have a because we squared capital a squared and we now have negative lower case. A positive lower case. A either the minus two x over a d. X And this is equal to negative to camp. A place where times Oh, I'm sorry. This is not supposed to take in a capital a square. We already had that instead, What we should have is negative to a and then another a over two from integrating the exponential either minus two x over a evaluated from minus a too positive a appear. And what will happen next? Oh, the other thing is, I've made another mistake. I meant to make this a zero in the lower limit of integration. Remember that in working with the symmetry of the function that has the absolute value in it, you replace the lower limit of integration with zero. And so you no longer have the minus. A is your lower limit, you have zero. That's very important, because otherwise you've done nothing with the integral. So that becomes a zero. And I'm sorry if that interrupted, you're taking notes or anything like that. But now that we have that corrected, we can properly evaluate the function that was integrated. And so we see that we have minus 2/8 times A over to that just gives us a minus one. These will cancel each other. And now we have either the minus two X over a evaluated A and zero. And this is minus one times ease of the negative, too. A over a minus either. The minus two times zero. Now things get a lot easier for us. So we still have the minus one on the outside. But notice that this A in this able cancel. So that simply leaves us with eat of the minus two. And, of course, E to the minus. Two times zero is just e to the minus zero. This becomes one. This is simply one. So now we have minus one. And with this negative one outside, multiplying that these two terms inside parentheses will swap. And so we now have one minus either the minus two. This is the probability we want and you can write this as a decimal is 0.865 So this is our final result. The probability is equal 2.86

In order to find the probability off the party. Coal, Uh, we first should make sure that the way function given fulfills the normalization condition. So first of all, let us integrate the square off the absolute value of this way function. So the way function is or should I say the the square absolute value of three function is be over by times X squared plus B square. That's the X. So you can see that this is the integral for the ark 10. So this gives us be over by times when we were being comes from this sector. That's one over V you, Acton, or you can read it like this. 10 to the minus one with X over B. This is from minus infinity to infinity. Well, if you remember, the are 10 graph. It goes from something like this. Well, little that are drawn and brooks inmates by over to infinity minus pi over two in mine isn't anything. So this gives us one over by the bees. Cancel. Then we get by over two minus minus fiber tools. We get floods by over two. This is indeed one. Okay, so no, to know the probability that the particle is between minus BNB. We did the same except to integrate between B where a minus B and B So the square absolute value of the way function. This will simply give the same. Be over by one Overbey the first time It's a little bit city. No. So x over being reminds me to be And your, um the are 10 off one on minus one. He's respectively. So let's say one is here minus one. Is here these value? Yes, my over four. So this is minus by over four. So get by over four plus by over four BCE cancel so he kids 1/2 which means probability is 50%.


Similar Solved Questions

5 answers
Match the moleculeimage0
Match the molecule image 0...
5 answers
0.5 pts various taxa of Searen 1 8 for instance: 1 eukaryotes Hl UW 1 ] tablilzing: are because eukaryotes inherited eukaryotic cells from earlier ancestors: ancient to be [ animals; Help eukaryotic cells unite all eukaryotic cells in animals are Mopuim the examine the tree U Bookmarks Question H because because Yes; Yes, Yes, 2 No because istory
0.5 pts various taxa of Searen 1 8 for instance: 1 eukaryotes Hl UW 1 ] tablilzing: are because eukaryotes inherited eukaryotic cells from earlier ancestors: ancient to be [ animals; Help eukaryotic cells unite all eukaryotic cells in animals are Mopuim the examine the tree U Bookmarks Question H be...
5 answers
What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?...
5 answers
Use the direction field given beloi to sketch the graphs the golutions that gatisfy the glven Inltial conditions.
Use the direction field given beloi to sketch the graphs the golutions that gatisfy the glven Inltial conditions....
3 answers
1_ (20 pts) Find the rank and the nullity of the given matrices. In each case, give a basis for the image and the kernel.2 0A =2 -21B = 3 8 |2 (5 pts) Find the coordinates of the vector (1,-1) with respect to the basis 3 = {(1,1) , (1,3)}.
1_ (20 pts) Find the rank and the nullity of the given matrices. In each case, give a basis for the image and the kernel. 2 0 A = 2 -2 1 B = 3 8 | 2 (5 pts) Find the coordinates of the vector (1,-1) with respect to the basis 3 = {(1,1) , (1,3)}....
4 answers
Thu raly n utech Choeen Muy nrand VirtyIkL tvU & nrrdolehlkoo Vch Genebesteuormalon alhoamn"ulutuhsulMntennsDicllet hd tucotu(n teet rruk
Thu raly n utech Choeen Muy nrand Virty IkL tvU & nrrdol ehlkoo Vch Gene besteuormalon alhoamn "ulutuhsul Mntenns Dicllet hd tucotu (n teet rruk...
5 answers
Locus A has four different alleles with restriction sites for BamH as shown in the diagram (below, left) . Sizes between each restriction site is Ipo given Ind Ind in kilobasepairs (kbp)_ 0.5 2.5 3.5 Allele A1Allele A20.52.5Allele A33,50.51Allele A4152.5ProbeYou design a probe for the locus of interest (locus A) and probe a genomic Southern blot: On the Southern blot is genomic DNA from individuals X, and Z that has been digested with BamH1+15
Locus A has four different alleles with restriction sites for BamH as shown in the diagram (below, left) . Sizes between each restriction site is Ipo given Ind Ind in kilobasepairs (kbp)_ 0.5 2.5 3.5 Allele A1 Allele A2 0.5 2.5 Allele A3 3,5 0.51 Allele A4 15 2.5 Probe You design a probe for the loc...
5 answers
A link $A B$ is moving in a vertical plane. At a certain instant when the link is inclined $60^{circ}$ to the horizontal, the point $A$ is moving horizontally at $3 mathrm{~m} / mathrm{s}$, while $B$ is moving in the vertical direction. What is the velocity of $B$ ?(a) $frac{1}{sqrt{3}} mathrm{~m} / mathrm{s}$(b) $2 sqrt{3} mathrm{~m} / mathrm{s}$(c) $sqrt{3} mathrm{~m} / mathrm{s}$(d) $frac{sqrt{3}}{2} mathrm{~m} / mathrm{s}$
A link $A B$ is moving in a vertical plane. At a certain instant when the link is inclined $60^{circ}$ to the horizontal, the point $A$ is moving horizontally at $3 mathrm{~m} / mathrm{s}$, while $B$ is moving in the vertical direction. What is the velocity of $B$ ? (a) $frac{1}{sqrt{3}} mathrm{~m} ...
5 answers
Find the particular solution of y"' + Y = 55 x25 X5 cos(x) +5 sin(x)
Find the particular solution of y"' + Y = 5 5 x2 5 X 5 cos(x) +5 sin(x)...
2 answers
Question 2. [2+6pt] Let X have a logistic distribution with PDFfx(x) <x < O (1+e-=)2 (2 Show Y = is a uniform random variable over (0,1) 1+eVerify it is indeed PDF_
Question 2. [2+6pt] Let X have a logistic distribution with PDF fx(x) <x < O (1+e-=)2 (2 Show Y = is a uniform random variable over (0,1) 1+e Verify it is indeed PDF_...
5 answers
+ -6x+2 Find Iim X+0 2x3 _ 41/3=1/2none of the other answers1/2-1/3
+ -6x+2 Find Iim X+0 2x3 _ 4 1/3 =1/2 none of the other answers 1/2 -1/3...
4 answers
Sample 20 students who had recently taken elementary Hewlett Packard, Casio, Sharp):tistics yielded the following information on brandcalculator ownedTexas Instruments;c HT T 5 T T 5 $ 5 H 5Estimate the true proportion of all such students who own Texas Instruments calculatorOf the students who owned TI calculator had graphing calculators_ Estimate the proportion of students who do not own TI graphing calculatorNeed Help?Rend tTelkte ar
sample 20 students who had recently taken elementary Hewlett Packard, Casio, Sharp): tistics yielded the following information on brand calculator owned Texas Instruments; c H T T 5 T T 5 $ 5 H 5 Estimate the true proportion of all such students who own Texas Instruments calculator Of the students w...
5 answers
Question 4Not yet answeredMarked out of 1Flag questionThe x-intercept of the linear function f(x) = 13x + 5 is55.20 1.15 -1.92 -0.38
Question 4 Not yet answered Marked out of 1 Flag question The x-intercept of the linear function f(x) = 13x + 5 is 55.20 1.15 -1.92 -0.38...
5 answers
Simplify and express in & + bi form:(2 ~ 5i)2
Simplify and express in & + bi form: (2 ~ 5i)2...
5 answers
Question 3stereolsomien possible for 2-bromo 4 Adlchlorohexane The toLM nurber
Question 3 stereolsomien possible for 2-bromo 4 Adlchlorohexane The toLM nurber...
5 answers
[1.32/2 Points]DETAILSPREVIOUS ANSWERSTANAPCALCBR1O 4.2.008_ 3/10 Submissions UsedMY NOTESASK YOUR TEACHERYau are given the graph of functionDetermine the intervals where the graph fis concave upward and whereconcave downwaro (Enter Your answers using interval notation:}concave upwardconcave downward3,0 )Find the inflection point of f. (If an answer does not exist; enter DNE;)(x, Y) =(0,1 )Need Help?RendUnichTalkto Jutor
[1.32/2 Points] DETAILS PREVIOUS ANSWERS TANAPCALCBR1O 4.2.008_ 3/10 Submissions Used MY NOTES ASK YOUR TEACHER Yau are given the graph of function Determine the intervals where the graph fis concave upward and where concave downwaro (Enter Your answers using interval notation:} concave upward conca...
5 answers
Point) Book Problem 11Use the Midpoint Rule withto approximate the function Vx' + 19 on tho interval from [-25.15]_Approximation
point) Book Problem 11 Use the Midpoint Rule with to approximate the function Vx' + 19 on tho interval from [-25.15]_ Approximation...

-- 0.020381--