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GR0877 Problem 26

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) The probability that the electron will be found between r and r+dr is P = |\psi|^2 dV = |\psi|^2 4 \pi r^2 dr = p(r)dr. The most probable value is given by the maximum of the probability density p(r) (take the second derivative if you want to convince yourself that it is indeed the maximum): dP/dr = 4 \pi r^2d|\psi|^2/dr + 8\pi r|\psi|^2 = 0. Plugging in: \displaystyle -\frac{4 \pi r^2}{\pi a_0^3} \frac{2}{a_0} e^{-2r/r_0} + \frac{8 \pi r}{\pi a_0^3} e^{-2r/r_0} = 0. From this, one has r = a_0, which is nothing else but the Bohr radius.

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  1. Steve
    09.11.2011 at 02:58

    For those (like me) who don’t wanna do that calculation on the test, the most probably radius for the ground state of H is a_o, the Bohr radius. Makes perfect sense. The expectation value is 1.5a_o, though, due to the tiny contributions from the finite probabilities of finding the electron at a larger radius. Don’t ask me how to calculate that; I just memorized it!

  2. Sam Fischer
    17.10.2012 at 06:06

    Where does the 4*pi*r^2*dr come from as dV?

    • 17.10.2012 at 16:34

      It’s a volume of a spherical shell between r and r+dr.

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