Heisenberg Uncertainty, Need Some Clarification. TIME SESITIVE, HELP

AI Thread Summary
The discussion focuses on applying the Heisenberg uncertainty principle to calculate the uncertainty in position (Δx) for a ball with a mass of 100 g and a velocity uncertainty (Δv) of 0.645 m/s. The initial calculation yielded an incorrect result of 1.23 * 10^-32 m, prompting the user to seek clarification on their approach. It was noted that dividing by the ball's diameter was unnecessary, and removing that step led to a new result of 8.174 * 10^-34 m. However, the accuracy of this value as the approximate uncertainty was questioned, indicating that it may represent the minimum possible uncertainty rather than the expected answer. The conversation emphasizes the need for careful application of the uncertainty principle in calculations.
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Heisenberg Uncertainty, Need Some Clarification. TIME SESITIVE, HELP!

Homework Statement


Use the Heisenberg uncertainty principle to calculate Deltax for a ball (mass = 100 g, diameter = 6.65 cm) with Deltav = 0.645 m/s.


Homework Equations



PX = h/(4*3.14)

The Attempt at a Solution



So, I took the equation and converted the values with my
numbers:((h/4π)/(.1kg * .645 m/s))/(.0665m) and yielded an
answer. The answer I got with these numbers is 1.23 *
10^-32 m. This answer is wrong and I am not sure why. I only have two hours left to answer this question. I would really appreciate it if someone could guide me on what my mistake is?
 
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Hello asl3589,

Why did you divide by the diameter of the ball? :confused:

[Edit: Also, you seem to be using the more formal \sigma _x \sigma _p \geq \frac{\hbar}{2} where \hbar = \frac{h}{2 \pi}. But keep in mind that relationship is not an equality. If you want an approximate value with a \approx sign and using \Delta x and \Delta p, there is a slightly different version of the relation.]
 
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Thanks for looking at it. I just assumed that the diameter factors into somehow. Is that unneccesary. If I take that step out the answer is just 8.174 * 10^-34 m. Is that right?
 


asl3589 said:
Thanks for looking at it. I just assumed that the diameter factors into somehow. Is that unneccesary. If I take that step out the answer is just 8.174 * 10^-34 m. Is that right?

That would give you the minimum possible uncertainty in position. But that's not necessarily the approximate uncertainty. The minimum possible uncertainty in position might be the answer your instructor is looking for, but I'm uncertain about that.
 
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