Solution to 2001 Problem 93

The probability that the electron is between r and r + dr is

\begin{align*}\left|\frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \right|^2 4 \pi r^2 dr\end{align*}
So, the probability is proportional to f(r) = r^2 e^{-2 r/(a_0)}. We can find the maximum value of the probability by differentiating f with respect to r and setting the derivative equal to 0.

\begin{align*}\frac{d f}{dr} = 2 r e^{-2 r/a_0} - \frac{2 r^2}{a_0} e^{-2 r/a_0} = 0 \Rightarrow r = a_0\end{align*}

We can check that this is a local maximum by finding the second derivative

\begin{align*}\frac{d^2 f}{dr^2} = 2 e^{-2 r/a_0} - \frac{4 r}{a_0} e^{-2 r/a_0} - \frac{4 r}{a_0 }e^{-2 r/a_0}  + \frac{4 r^...
If we plug in r = a_0, we find that

\begin{align*}\frac{d^2 f}{dr^2} = 2 e^{-2} - 8 e^{-2} + 4e^{-2} = -2 e^{-2} < 0 \end{align*}
Therefore, r = a_0 is indeed a local maximum of the radial probability density function. Therefore, the correct answer is a_0.

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