CALC A particle of mass m has the wave function ψ(x) = Ax exp ...

Question

CALC A particle of mass m has the wave function ψ(x) = Ax exp (−x²/a²) when it is in an allowed energy level with E = 0. At what value or values of x is the particle most likely to be found?

Accepted Answer

Hello, let's go through this practice problem. A quantum oscillator with a mass of M obeys the wave function si of Y equals by squared multiplied by the exponential of negative Y squared divided by B squared. For all permissible energy levels where E equals zero, determine the most likely positions to find the particle option A Y equals plus or minus BB plus or minus B divided by two C plus or minus one divided by B or D plus or minus two divided by B. So the key thing to know with this problem is to understand that the most likely position to find the particle is where the probability density has reached its maximum. So we're going to need to take the function for probability density and then maximize it, which means we're gonna have to bring some calculus into this. So remember the probability density is the square of the wave function. And in order to maximize a function, we need to find where the derivative of that function. So the derivative of the probability density is equal to zero, that's where the maximum is gonna be. So our method for this problem is we're going to take the wave function that was given to us, we're going to square it to find the probability density function. Then we're gonna take the derivative of it and find where that derivative is equal to zero. So first off the square of the probability of the wave function is gonna be equal to B squared multiplied by. And then we're squaring the Y squared. So that becomes Y to the power of four. And then we're squaring the exponential function which just means we're gonna be adding them together because when we multiply exponents together, they just add. So negative Y squared divided by B squared plus negative Y squared divided by B squared is just going to be negative two Y surviving the negative two in front of it. So negative two Y squared divided by B squared. So now we're taking the derivative with respect to why of B squared Y to the power of four multiplied by E to the power of negative two Y squared divided by B squared. So now let's take this derivative, not the easiest derivative in the world, but it's not a terribly hard one either. First, let's pull out the B squared since that is a constant. So take out the B squared and put it outside of a parenthesis that'll contain everything else. So what's left is A Y raised to the power of four multiplied by an exponential function that also contains A Y. So since we have two different parts of this that contain the variable of differentiation. We're gonna want to use the product rule and recall the product rule of differentiation states that we will not differentiate one part of it, differentiate the other part and then add to it the reverse. So I'm gonna leave the Y race the power of or the same. But then we will multiply by the derivative of the exponential function. And recall that to take the derivative of the exponential, you leave it the same. So the exponential of negative two Y squared divided by B squared, but then apply the chain rule and multiply by the derivative of what's inside the exponential function. So it's inside is negative two Y squared divided by B squared. So we will apply the power rule to this part by multiplying the exponent of two by the coefficient. So we multiply by negative or Y divided by B squared. Now let's add the opposite. Now we'll differentiate the Y to the power of four term. So again, we apply the power rule here. So that is we, we subtract the exponent by one and bring it down. So that is four Y cubed, then we just multiply by the exponential function and don't change anything about that. So negative Y squared divided by B squared and we want to set all of this equal to zero. Now, since the B squared term cannot be zero, we can ignore that while trying to maximize this function. Remember, we're looking for the values of Y that'll make this equal to zero. So it's OK to ignore some variables if they aren't going to help us in that, in that regard. So now just looking at the parentheses, we have Y to the power of four multiplied by negative four Y divided by B squared, which can be more simply written as putting the Ys together. So it's Y to the power of five negative divided by B squared. And don't forget the four will multiply by the exponential function. So exponential of negative two Y squared divided by B squared plus four Y cubed multiplied by the exponential function negative two Y squared divided by B squared. Now notice that both of the terms that we're left with contain the exponential function. So they can cancel out especially since that term can't be equal to zero either. So now we have negative four Y to the power of five divided by B squared plus four Y cubed equals zero. So a few more things we can do to simplify this. There's now a four in every term that can be canceled out, we have a, a term containing A Y to the power of five. And the other term contains Y to the power of three. So we can cancel those out. I'm gonna add the, I'm gonna add the negative Y to the power of five divided by B squared to both sides of the equation just to get rid of the negative sign. And so we can see more clearly what's happening. So Y cubed equals Y to the power of five divided by B squared. Then I'm gonna multiply B squared by both sides of the equation. So B squared, Y cubed equals Y to the power of five. And then the Y cubed can cancel out with three of the Ys on the right hand side. So we're left with B squared equals Y squared, which clearly means that Y can be either positive or negative B because whether B is positive or negative it can be squared to give us the same value for why? So that is our answer to the problem which corresponds to option A and that is our answer. I hope this video helped you out. If it did, please check out our other tutoring videos, which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

CALC A particle of mass m has the wave function ψ(x) = Ax exp (−x²/a²) when it is in an allowed energy level with E = 0. At what value or values of x is the particle most likely to be found?

Key Concepts

Wave Function

Probability Density

Maximizing the Probability Density

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