Minimum Uncertainty of Electron's Momentum

[Pepsi24chevy]

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?

 

[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.

 

[Pepsi24chevy]

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?

 

[Galileo]

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

 

[Pepsi24chevy]

well the answer should be in kg*m/s that's why i feel i am missing something, but its probably just a conversion or something?

 

[Galileo]

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.