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Chemistry 8 Online
OpenStudy (anonymous):

Heisenburgs Uncertainty Principle. I keep getting the wrong answer and it is driving me crazy. Using Heisenbergs uncertanty priciple, calculate the uncertanty in the position of a) A 1.50-mg mosquito moving at a speed of 1.40 m/s if the speed is known to within 0.01 m/s; and b) which I will try to figure out on my own if I fully understand this one. The following post will show what I have done.

OpenStudy (anonymous):

\[h=6.626*10^{−34}\] \[Δx=h/(4π∗Δp)\] \[m=1.5mg=1.5∗10^{-6}kg\] \[Δv=1.4m/s±0.01m/s=1.40∗10^{-2}\] I would assume, here that I have everything and it should just be plug and chug, however I am not getting the correct answer and do not know why.

OpenStudy (vincent-lyon.fr):

Simply \(\Delta v=0.01 \;m/s \)

OpenStudy (anonymous):

Even using that I still have the wrong answer...

OpenStudy (anonymous):

To one sig fig, it is correct... let me ask a few questions...

OpenStudy (anonymous):

So, the speed that is given at 1.4 m/s is not used or even needed, just the uncertanty in the speed itself. And this will find the uncertanty in position correct?

OpenStudy (anonymous):

and that the mass should always be in kg? and velocity should always be in m/s, right?

OpenStudy (vincent-lyon.fr):

Check with your textbook. There are different definitions of "\(\Delta\)something" and they can lead to slightly different expressions of Heisenberg's inequalities by a factor 2 or a factor \(\pi\).

OpenStudy (vincent-lyon.fr):

Correct for kg and m/s.

OpenStudy (vincent-lyon.fr):

"So, the speed that is given at 1.4 m/s is not used or even needed, just the uncertanty in the speed itself. And this will find the uncertanty in position correct?" - Correct!

OpenStudy (anonymous):

This is a Chem book, not very well defined in this area. Going though my Physics book would be overkill with the chapters I would have to go back through. (Which I plan on doing prior to the next Chem class)

OpenStudy (anonymous):

That is strange that is is only the uncertanty that we need, and not the actual speed itself. This helps clear things up a bit. Thanks!!

OpenStudy (vincent-lyon.fr):

You're welcome :))

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