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?
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.
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?
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.