# uncertainty question

by Pepsi24chevy
Tags: uncertainty
 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?
 HW Helper Sci Advisor P: 2,004 You are not asked about the velocity. You are asked about the uncertainty in momentum. So you got $\Delta x \Delta p \geq \hbar$ (or without the bar, depening on what version your book has). So you know $\Delta x$, that was given and you are asked the minimum value $\Delta p$ 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.
P: 65
 Quote by Galileo You are not asked about the velocity. You are asked about the uncertainty in momentum. So you got $\Delta x \Delta p \geq \hbar$ (or without the bar, depening on what version your book has). So you know $\Delta x$, that was given and you are asked the minimum value $\Delta p$ 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?

HW Helper
That's right. Simply $\Delta p_{min}=\hbar/\Delta x$. What would you be missing?