How Does Electron Uncertainty Compare to Helium Atom Size?

[littlebearrrr]

Homework Statement

(NOTE: I solved the first two parts of this problem, but I included it anyway so you have background knowledge for the third question)

The average kinetic energy of an electron in a ground-state helium atom is 2.4x10^3 kJ/mol.

A. What is the corresponding electron velocity?

B. An experiment measures this velocity with an uncertainty of 10%. Calculate the minimum uncertainty in the position of the electron for this experiment.

C. The effective radius of a helium atom is 130 pm. Is the uncertainty in the position you calculated in B a significant fraction of this radius?

Homework Equations

A. KE=(1/2)mv^2

B. Δx*Δv≥ h/(4pi)m

C. N/A

The Attempt at a Solution

A. Just multiplied 2.4e3 kJ/mol by (1 mole/6.022e23 electrons) and used the above equation to solve for v. Got 3.0e6 m/s.

B. Took 10% of my velocity, plugged that into the above inequality. Obtained Δx=2.0e-10 m (can someone check if this is right?)

C. If the above answer in B is right, do I just take that answer and divide it by 1.30e-10 m to get my answer for C?

 

[DrClaude]

B. Took 10% of my velocity, plugged that into the above inequality. Obtained Δx=2.0e-10 m (can someone check if this is right?)

That looks correct.

C. If the above answer in B is right, do I just take that answer and divide it by 1.30e-10 m to get my answer for C?

I don't think that C calls for a numerical result, rather an appreciation of the result of B compared to the effective radius.

 

[littlebearrrr]

Hey DrClaude, thanks for your help/checking my answers again :)