## Uncertainty Principle Problem

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

Doreen Liu 4D
Posts: 54
Joined: Wed Sep 11, 2019 12:17 am

### Uncertainty Principle Problem

What is the minimum uncertainty in the speed of an electron confined within an atom diameter of 3.50 pm?
Is the delta x in the equation the diameter? If so, do I need to change it to meters before I can use the diameter?

Mansi_1D
Posts: 50
Joined: Fri Aug 02, 2019 12:15 am

### Re: Uncertainty Principle Problem

Yes, delta x is 3.50 pm in the equation and you would need to convert it to meters because that's the SI unit for position.

Alexis Webb 2B
Posts: 124
Joined: Thu Jul 11, 2019 12:15 am

### Re: Uncertainty Principle Problem

Yes, the change in position (delta x) is equal to the diameter, and if you convert to meters, it will cancel out with the meters in the units for J (kg•m^2•s^-1).

Megan Jung 3A
Posts: 50
Joined: Thu Jul 11, 2019 12:17 am

### Re: Uncertainty Principle Problem

For this problem, delta x, or uncertainty in position, would be 3.50 pm however you would have to convert it to meters to use the Heisenberg Indeterminacy equation. (10^12pm=1m).