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Calculating Spring Constant using Minimum Energy State of hydrogen atom

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A hydrogen chloride molecule may be modeled as a hydrogen atom (mass: 1.67 x10^-27 kg ) on a spring; the other end of the spring is attached to a rigid wall (the massive chlorine atom).

    If the minimum photon energy that will promote this molecule to its first excited state is 0.358 eV, find the "spring constant."

    I'm not sure which equation to use. is it E=1/2kA^2? If so, what would the value of A be?

    Thank you for your help!
  2. jcsd
  3. Nov 23, 2009 #2
    Shouldn't you be using the energy eigenvalues of the quantum harmonic oscillator, and not the classical potential energy of a spring?
  4. Nov 23, 2009 #3
    I see...so what does that mean? i'm kind of confused
  5. Nov 23, 2009 #4
    I assume by this comment that you haven't covered the http://galileo.phys.virginia.edu/classes/252/SHO/SHO.html" [Broken]. I'd recommend taking a glimpse over that link, it's a pretty simplistic essay on the quantum harmonic oscillator. But basically the energy eigenvalues of the QHO are given by


    where [itex]\omega^2=k/m[/itex]. So in the ground state, your energy equation would be


    So you can use this equation to solve for your spring constant [itex]k[/itex]. (Just in the off chance you don't know what it is, [itex]\hbar=1.054\times10^{-34}\,\mathrm{m^2kg/s}[/itex] and is the reduced Planck constant)

    Edit: that was silly of me, forgot the power in [itex]\hbar[/itex]!
    Last edited by a moderator: May 4, 2017
  6. Nov 23, 2009 #5
    hmmm I used that formula and got 7.7 x 10^40..which is incorrect..am I using the wrong units? Thanks for your help btw
  7. Nov 23, 2009 #6
    You need to do two things:
    (1) convert eV into J: [itex]0.358 \,\mathrm{eV}=5.73\times10^{-20}\mathrm{m^2kg/s^2}[/itex].
    (2) let [itex]n=1[/itex] so that you can have the first excited state. I didn't catch this one earlier, but your energy eigenvalue should be


    leading to

    k=\frac{4mE_1^2}{9\hbar^2}=\frac{4\cdot1.67 \times10^{-27} \mathrm{kg}\cdot(5.73\times10^{-20}\mathrm{m^2kg/s^2})^2}{9(1.05\times10^{-34}\mathrm{m^2kg/s^2})^2}\approx220
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