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Kinetic energy of electron

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    (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?

    2. Relevant equations

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

    B. Δx*Δv≥ h/(4pi)m

    C. N/A

    3. 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?
  2. jcsd
  3. Oct 28, 2013 #2


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    Staff: Mentor

    That looks correct.

    I don't think that C calls for a numerical result, rather an appreciation of the result of B compared to the effective radius.
  4. Oct 28, 2013 #3
    Hey DrClaude, thanks for your help/checking my answers again :)
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