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10.28: Hydrogen Atom Calculation Assuming the Electron is a Particle in a Sphere of Radius R

  • Page ID
    136978
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    Trial wave function:

    \[ \Phi (r, R) := \frac{1}{ \sqrt{2 \pi R}} \frac{ \sin ( \frac{ \pi r}{R})}{r} \nonumber \]

    Integral:

    \[ \int_{0}^{ \infty} \blacksquare 4 \pi r^2 dr \nonumber \]

    Kinetic energy operator:

    \[ T = \frac{1}{2r} \frac{d^2}{dr^2} (r \blacksquare ) \nonumber \]

    Potential energy operator:

    \[ V = \frac{1}{r} \nonumber \]

    Demonstrate the wave function is normalized.

    Set up the variational energy integral.

    \[ E(R) := \int_{0}^{R} \Phi (r, R) [ \frac{-1}{2r} \frac{d^2}{dr^2} (r \Phi (r, R))] 4 \pi r^2 dr + \int_{0}^{R} \Phi (r, R) \frac{-1}{r} \Phi (r, R) 4 \pi r^2 dr] \nonumber \]

    Minimize the energy with respect to the variational parameter R.

    R := 1 R := Minimize(E, R) R = 4.049 E(R) = -0.301

    The exact ground state energy for the hydrogen atom is -.5 Eh. Calculate the percent error.

    \[ \frac{-.5 - E(R)}{-.5} = 39.793 \nonumber \]

    Compare optimized trial wave function with the exact solution by plotting the radial distribution functions.

    \[ S(r) := \frac{1}{ \sqrt{ \pi}} exp(-r)~~~r := 0,.02..4.2 \nonumber \]

    Screen Shot 2019-02-14 at 1.13.18 PM.png


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