
#1
Dec605, 03:01 PM

P: 65

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? 



#2
Dec605, 03:28 PM

Sci Advisor
HW Helper
P: 2,004

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. 



#3
Dec605, 04:04 PM

P: 65





#4
Dec605, 05:18 PM

Sci Advisor
HW Helper
P: 2,004

uncertainty question
That's right. Simply [itex]\Delta p_{min}=\hbar/\Delta x[/itex]. What would you be missing?




#5
Dec605, 06:32 PM

P: 65

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?



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