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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. 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. 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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