$10...$

Question

10.0

-g marble is gently placed on a horizontal tabletop that is

1.75

m wide.

(a) What is the maximum uncertainty in the horizontal position of the marble?

(b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble?

(c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately $14$ billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

Accepted Answer

Hello everyone. Let's take a look at this practice problem. A rook of mass 15 g is placed smoothly on the open E file of a chessboard. The chessboard is a square with a side of 50 centimeters. Consider the E file line as being in the vertical position. Part one, calculate the maximum vertical uncertainty of the rook's position. Part two, calculate the minimum uncertainty in the vertical velocity of the rook. Part three. What would be the maximum duration for which the rook could stay on the chessboard option? A part 1.25 m, part 21.41 and multiplied by 10 to the power of negative 32 m per second and part 31.39 years. Option B part 1.5 m, part 27.03 multiplied by 10 to the power of negative 33 m per second. Part 32.26 years. Option C part 1.25 m, part 21.41 multiplied by 10 to the power of negative 32 m per second. And part 31.39 multiplied by 10 to the power of 24 years. Option D part 1.5 m, part 27.03 multiplied by 10 to the power of negative 33 m per second and part 32.26 multiplied by 10 to the power of 24 years. This is kind of an interesting problem because we're asked about finding the uncertainties of the position of something and connecting that to the uncertainty of its velocity. So from that, it makes sense that we'd want to use Heisenberg's uncertainty principle. And Heisenberg's uncertainty principle is a tenant of quantum mechanics. But what's kind of funny about this problem is that while quantum mechanics generally applies towards very tiny things in the universe, this is a problem where we're asked to apply that to a piece on a chessboard. And we wouldn't really expect the rules of quantum mechanics to really apply to macroscopic objects. So this is kind of a funny problem where the results will basically show us why quantum mechanics starts to sort of break down when we're dealing with larger objects. So let's start with part A where we're asked to simply find the maximum vertical uncertainty of the rook's position. This part is fairly simple because we know that the file line has the same length as one side length of the chessboard and the rook can be anywhere along that file. So logically the maximum uncertainty is just going to be equal to the side length of the chessboard. So to kind of put this in a picture, you kind of crudely drawing the chessboard, even if the rook was measured to be at one side of the board, its maximum uncertainty could be as large as the side length of the board itself, which is 50 centimeters. So the correct answer to part A is 50 centimeters or 0.5 m. So that is our answer to part A part B asks about the minimum uncertainty in the vertical velocity of the rook. So part A had us find the maximum, the maximum uncertainty in the position as 0.5 m. And now part B of the problem is asking us to find the minimum uncertainty in the velocity. Since we're trying to connect velocity uncertainty and position uncertainty, we can compare the two using the Heisenberg uncertainty principle. And recall that the Heisenberg uncertainty principle states that the uncertainty in position of an object multiplied by the uncertainty in momentum or alternatively, the mass of the object multiplied by its uncertainty and velocity is greater than or equal to the plank constant divided by four pi. So to actually solve for the uncertainty and velocity, there are a few ways we have to rewrite this equation. First of all, of course, we're going to have to algebraically rewrite it to solve for delta V which we can do by dividing both sides of the equality by delta Y and M. But the other thing to know is that since we're specifically looking for the minimum uncertainty in the velocity, we're, we're looking for the case where the two terms are equal rather than having the greater than or equal sign. So delta V is equal to the plant constant H divided by four pi M delta Y. So to solve for the uncertainty and the velocity, we simply have to plug in our values, the uncertainty and the velocity is equal to the plank constant 6.626 multiplied by 10 to the power of negative 34 jewel seconds divided by four pi multiplied by the mass of the particle which is 15 g or point 0 1 5 kg multiplied by the uncertainty in the vertical position which we found in part A to be 0.5 m. And if we plug this into a calculator, then we find an uncertainty and velocity of about 7.03 multiplied by 10 to the power of negative 33 m per second. And so this is our uncertainty or the velocity. Finally part C asks us to find the minimum or maximum duration for which the rook could stay on the chessboard. Now, this pro part of the problem especially shows us why quantum mechanics doesn't really usually apply to macroscopic objects. Because you might be thinking well, technically the rook could stay on the chess board for infinity because it could have a measured velocity of 0 m per second. But since one of the core principles of quantum mechanics and, and the Heisenberg uncertainty equation is that we can't know the exact value of a posi of a particle's position and its velocity at the same time, that's going to create some issues here because the uncertainty is more important than any expected exact value. So one way to look at this is that the measured speed is going to be equal to whatever its exact it it's supposed exact speed is plus or minus the uncertainty like the one we found in part B. So even if we assume that the rook is at rest and has an exact speed of zero, it's a minimum speed according to quantum mechanics still is going to be the negative of the uncertainty value. So that means that the minimum measured speed of the rook is going to be the uncertainty that we found in part B 7.03 multiplied by 10 to the power of negative 33 m per second. So if we want to find the maximum amount of time that the rook can stay on the board, we're going to have to invoke our good old speed equation where the speed of a particle is equal to the distance it travels divided by the unit time, which consequently indicates that time is equal to distance divided by speed. Now, as it relates to this problem, we're looking for the maximum amount of time that the Rook can stay on the board. And according to the maximum case, the maximum case, the Rook has to slide across the entire length of the board before falling off. So the maximum time is going to be equal to the maximum distance that it might need to travel to fall off or 0.5 m divided by its minimum possible speed, which is 7.03 multiplied by 10 to the power of negative 33 m per second. And if we put this into our calculator, we find a maximum time of 7.11 multiplied by 10 to the power of 31 seconds. Though, our multiple choice options are all written with units of years. So we're going to have to do a unit conversion on this to get this in years. So recall that one year is made up of about 3.154 multiplied by 10 to the power of seven seconds. So if we put this into a calculator, then we find a time of 2.26 multiplied by 10 to the power of 24 years. So this is our answer to part C. And if we look at our multiple choice options, we can see that the only option that agrees with our answers for all three parts is option D. So option D is our answer to this problem. And that is all for this video. I hope this video helped you out and please consider checking out some of our other tutoring videos that'll give you more experience with these types of problems, that's all for now and I hope you all have a lovely day. Bye bye.

Key Concepts

Heisenberg Uncertainty Principle

Quantum Mechanics

Measurement and Observables

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