## Problem 1B.25

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Chloe Shamtob 3I
Posts: 42
Joined: Wed Sep 30, 2020 9:55 pm

### Problem 1B.25

What is the minimum uncertainty in the speed of an electron confined within a lead atom of diameter 350. pm? Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom.

How do you start this problem?

Isabella Chou 1E
Posts: 45
Joined: Wed Sep 30, 2020 9:52 pm
Been upvoted: 1 time

### Re: Problem 1B.25

For this problem, you would use Heisenberg's uncertainty equation: Δp x Δx ≥ h/(4π). We also know that Δp = m x Δv. Since the problem is asking for minimum uncertainty in the speed of an electron, we are trying to solve for Δv. The problem gives you Δx = 350. pm (but you should convert this to meters), so you can solve for Δp first, and then find Δv using the mass (in kg) of an electron for m. I hope this helps!

Thomas Vu 3A
Posts: 46
Joined: Wed Sep 30, 2020 10:08 pm
Been upvoted: 1 time

### Re: Problem 1B.25

This might be kind of a side question, but what exactly does the "minimum uncertainty" mean? Is it supposed to represent a range or like a "margin of error" of some sorts?

Alan Huang 2C
Posts: 34
Joined: Wed Sep 30, 2020 9:37 pm

### Re: Problem 1B.25

Thomas Vu 3A wrote:This might be kind of a side question, but what exactly does the "minimum uncertainty" mean? Is it supposed to represent a range or like a "margin of error" of some sorts?

Yes, uncertainty would mean the range of possible values within which the true value of the measurement lies, so minimum uncertainty would mean the smallest range.

clairehathaway 3L
Posts: 45
Joined: Wed Sep 30, 2020 9:38 pm

### Re: Problem 1B.25

Isabella Chou 1E wrote:For this problem, you would use Heisenberg's uncertainty equation: Δp x Δx ≥ h/(4π). We also know that Δp = m x Δv. Since the problem is asking for minimum uncertainty in the speed of an electron, we are trying to solve for Δv. The problem gives you Δx = 350. pm (but you should convert this to meters), so you can solve for Δp first, and then find Δv using the mass (in kg) of an electron for m. I hope this helps!

I'm a little confused on the different variables in the equation, what does Δx represent? and how did you know that it was equal to 350. pm?

Isabella Chou 1E
Posts: 45
Joined: Wed Sep 30, 2020 9:52 pm
Been upvoted: 1 time

### Re: Problem 1B.25

clairehathaway 3L wrote:
Isabella Chou 1E wrote:For this problem, you would use Heisenberg's uncertainty equation: Δp x Δx ≥ h/(4π). We also know that Δp = m x Δv. Since the problem is asking for minimum uncertainty in the speed of an electron, we are trying to solve for Δv. The problem gives you Δx = 350. pm (but you should convert this to meters), so you can solve for Δp first, and then find Δv using the mass (in kg) of an electron for m. I hope this helps!

I'm a little confused on the different variables in the equation, what does Δx represent? and how did you know that it was equal to 350. pm?

Δx represents the indeterminacy in position of the electron. We know that it is equal to 350. pm because the electron is confined within the diameter of the atom, which is 350. pm, and the problem tells us to model the atom as a one-dimensional box where the length is equal to the diameter.

Raashi Chaudhari 2D
Posts: 35
Joined: Wed Sep 30, 2020 9:31 pm

### Re: Problem 1B.25

Since we cannot ask the same question twice I was also having difficulty with this one.
On the solutions manual, the solution is using a different equation than the one listed on our formula sheet. However, even if I use a different formula I should be getting the same answer, however, that is not the case.
I am doing this:
$\Delta V\geq \frac{h}{4\pi \Delta x\Delta m}$
My h is (6.626 x 10^-34 J/s), my Delta X is (350 x 10^-12 m), and my Delta M is (9.109 x 10^-31 kg)

When I plug all those values in I get 165387 m/s which is not the one listed on the solutions manual.