uncertainty question


by Pepsi24chevy
Tags: uncertainty
Pepsi24chevy
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#1
Dec6-05, 03:01 PM
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In an atom, an electron is confined to a space of roughly 10^-10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty in its momentum?


delta(x)delta(p)>h (this is how it was derived in the previous parts to the problem)
Ok so i know px = mvx and m = 9.11X10^-31 kg and that delta px = 10^-10(px). I guess where i am stuck is, what do i do with the velocity term?
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Galileo
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Dec6-05, 03:28 PM
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You are not asked about the velocity. You are asked about the uncertainty in momentum. So you got [itex]\Delta x \Delta p \geq \hbar[/itex] (or without the bar, depening on what version your book has).

So you know [itex]\Delta x[/itex], that was given and you are asked the minimum value [itex]\Delta p[/itex] can take without violating the inequality. You should be able to tell that immediately from the equation. Or least least without reference to other relations.
Pepsi24chevy
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Dec6-05, 04:04 PM
P: 65
Quote Quote by Galileo
You are not asked about the velocity. You are asked about the uncertainty in momentum. So you got [itex]\Delta x \Delta p \geq \hbar[/itex] (or without the bar, depening on what version your book has).
So you know [itex]\Delta x[/itex], that was given and you are asked the minimum value [itex]\Delta p[/itex] can take without violating the inequality. You should be able to tell that immediately from the equation. Or least least without reference to other relations.
am i missing some or would it be 1.05e-34J/10e-10?

Galileo
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Dec6-05, 05:18 PM
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uncertainty question


That's right. Simply [itex]\Delta p_{min}=\hbar/\Delta x[/itex]. What would you be missing?
Pepsi24chevy
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Dec6-05, 06:32 PM
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well the answer should be in kg*m/s thats why i feel i am missing something, but its probably just a conversion or something?
Galileo
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Dec6-05, 06:37 PM
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The answer is in kg m/s, since that is the unit of momentum. The units of h is Js. Since J=Nm=kgm^2/s^2 you can work it out.


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