Consider the electron wave function

ψ(x)={√c 1−x^2 |x|≤1 cm
0 ...

Question

Consider the electron wave function
 
ψ(x)={√c 1−x^2 |x|≤1 cm
 0 |x|≥1 cm

where x is in cm.

d. If 10^4 electrons are detected, how many will be in the interval 0.00 cm≤x≤0.50 cm?

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with quantum mechanics. When this pro problem particles in a one dimensional space in the region of minus two decimeters is less than or equal to Y is less than or equal to two decimeters. Obey the wave function of si of Y is equal to a multiplied by the square root of the quantity of three minus 0.5 decimeters to the negative two Y squared. If there are 100 62 particles in the range of minus two decimeters is less than or equal to Y is less than or equal to two decimeters. How many particles are in the range of minus one decimeters is less than or equal to Y is less than or equal to one decimeter. We're given four possible choices as our answers. For choice A we have 300 particles for choice B we have 100 40 particles for choice C we have 98 particles and for choice D we have 60 particles. Now, in order to figure out how many particles are in this range from minus one decimeter to one decimeter, we first need to find the probability of finding one particle in that same range. So recall the your probability formula for uh finding a particle in a region. And that is P is equal to the integral over the region. In this case, we're going from minus one decimeter to one decimeter of the modules of size squared dy. Now we were given a function for size so we can go ahead and plug that in. So we'll have P is equal to the integral going from minus one decimeter 21 decimeter. And here we need to take the modulus of si squared. But I here is a real function. So I don't need to worry about taking the complex conjugate. So I can just square the function. So I'll get a squared multiplying the quantity of three minus 0.5 decimeters to the negative two multiplied by Y squared. That whole quantity is multiplied by Dy. Now, the function that I'm integrating over is an even function. So I can simplify the integration a bit by multiplying our entire integral by two. But changing the limits of integration to go from 0 to 1 decimeter. So when I do that I'll have P is equal to two, I'm going to go ahead and pull up the constant A squared outside the interval. And that's gonna be multiplying the integral of zero going from 0 to 1 decimeter of the quantity of three minus 0.5 decimeters to the negative two. Why squared, the whole quantity is multiplied by dy. So I can go ahead and do the integration, we'll have P equal to two A squared multiplying the quantity. When I integrate three, I get three Y, they have the minus the 0.5 decimeters to the negative two. And when I integrate Y squared, I get Y cubed divided by three. And that's gonna be evaluated at the limits of zero and one decimeter. Now, at this point before I plug in the limits of integration, I noticed that I still have an unknown in this um equation. I don't know what the A is now A is the normalization constant of our wave function. So we're gonna use our normalization condition in order to find that quantity. So recall your normalization condition that is the integral going from minus infinity to infinity of the modulus of S squared dy has to be equal to one. So basically this is saying the probability of finding the particle anywhere is going to be one. So we're gonna go ahead and plug in our value our value for I but heresy is only defined between the minus two decimeters and the two decimeters is zero. Otherwise that means our limits of integration are going to change. So we're gonna be integrating from minus two decimeters to two decimeters. And we plug in um for the modules of sci square, we'll get the same thing that we had before it's gonna be the A squared multiplying the quantity of three minus 0.5 decimeters to the negative two Y squared. That quantity is multiplied by DY and this all has to equal one. So we do the same sort of math tricks as we did before. Um I can change the limit set of integration to go from 0 to 2 decimeters by multiplying everything by two. And so I'll get two A squared multiplying the integral going from 0 to 2 decimeters of the quantity of three minus the 0.5 decimeters to the negative two Y squared. That quantity is multiplied by dy and this has to be equal to one. We do the same integration as before and we'll get the exact same thing. Then we'll have two A squared multiplying the quantity of three Y minus the 0.5 decimeters to the negative two multiplied by Y cubed divided by three. This is gonna be evaluated at the limits of zero and two decimeters. And this all has to be equal to one. So I can go ahead and plug it in my limbs of integration. So I'll have two A squared multiply in the quantity of three multiplied by two decimeters minus 0.5 decimeters to the negative two multiplying the quantity of two decimeters cubed divided by three. And when I plug in the lower limit I get zero. And so this whole thing has to be equal to one So at this point I can solve four A and so to solve for A, I'm just going to divide over everything multiplying by A and then take the square root. So get A is equal to the square root of one divided by the quantity of two multiplying the quantity I'll have three multiplied by two decimeters. So it's gonna be six decimeters. Then I'll have minus I'll have eight decimeters cubed from the two decimeters cubed. But that's gonna be multiplying the 0.5 decimeters to the negative two. So leave me with four decimeters that's still gonna be divided by three. And we can actually simplify this will have a is equal to the square root of three divided by 28. And that will have units of decimeters to the negative one half. So now I have a value for A, I can plug that back into my probability formula and complete that calculation to here I will have P is equal to two. And here we'll multiply that by A squared that's gonna get multiplied by three divided by 28 they'll have units of decimeters to the negative one. And this is going to multiply the quantity of three multiplied by one decimeter minus the 0.5 decimeters to the negative two multiplying one decimeter cubed divided by three. And here when I plug in the lower limit, I get zero again. So at this point, I just actually have numbers on the right hand side. So I can calculate that probability P and it's going to be equal to 0.6071. And notice that the units of decimeters actually cancel out. So now I have the probability I can use it to calculate the number of particles because the number of particles in that region we'll call that in is going to be the number of particles I have in the larger region which is the 162. And then I need to multiply that by the probability of finding a particle in the region that we want, which is between the minus one decimeter to the one decimeter. And so that's the probability with, we just calculated which is the zero point six 071. And so that means that my N is going to be equal to 98. And so when I look at my answer choices, this corresponds to answer choice C so just a quick little recap of what we've done here, we use our formula for finding the probability of a particle within a region. But in order to use that, we also had to find the normalization constant for our um wave function just using our normalization condition. But once we found the probability of finding a particle in the region that we want, we just multiplied the total number of particles that we had by that probability to get the number of particles that we'd find in our expected region. So I hope that this has been useful and I'll see you in the next video.

Consider the electron wave function ψ(x)={√c 1−x^2 |x|≤1 cm 0 |x|≥1 cm where x is in cm. d. If 10^4 electrons are detected, how many will be in the interval 0.00 cm≤x≤0.50 cm?

Key Concepts

Wave Function

Probability Density

Normalization

Watch next