Homework #19

[206190205]

What is the uncertainty in an electron's velocity if the position is known within 13 A.

What is the minimum uncertainty in a helium atom's velocity if the position is known within 1.4 A.
I was able to find the min for the first one, but I keep getting the second answer as 5.66 x 10^-45. I don't know what I'm doing wrong. Could I be plugging it into the calculator wrong?

[Hao Tam Tran 3H]

Hi!

The Heisenberg uncertainty principle is ∆p_x * ∆x ≥ h/(4pi), but we can substitute ∆p_x with m*∆v (equation for momentum). Thus, when we solve for ∆v, the equation becomes ∆v ≥ h/(4pi*m*∆x). To find the mass of a helium atom, we would take its molar mass (4.003 g/mol), convert it to kg (0.004003 kg/mol) and divide it by avogadro's number (6.023 x 10^23 atoms/mol) to get 6.646*10^-27 kg/atom. The given value for ∆x is 1.4Å or 1.4*10^-10 m. Now, all we need to do is plug everything in:

∆v ≥ (6.626*10^-34 J•s)/(4pi * (6.646*10^-27 kg/atom) * (1.4*10^-10 m)), which should give you a minimum uncertainty in velocity of approximately 56.6 m/s. You most likely forgot to convert the mass to kg and/or forgot that Å means 10^-10.