Solution to 2008 Problem 26

The probability that the electron is in a shell of radius r and width dr about the nucleus is

\begin{align*}4 \pi \left|\psi_{100}\left(r\right)\right|^2 r^2 dr = \frac{4}{a_0^3} e^{-2 r/ a_0} r^2 dr\end{align*}
To find the maximum value of this probability, we find the zeroes of the derivative of
\begin{align*}f(r) = \frac{4}{a_0^3} e^{-2 r/ a_0} r^2\end{align*}
\begin{align*}f'(r) = \frac{8}{a_0^3} e^{-2 r/ a_0} r + \frac{-8}{a_0^4} e^{-2 r/ a_0} r^2\end{align*}
\begin{align*}f'(r) = 0 \Longrightarrow \frac{8}{a_0^3} e^{-2 r/ a_0} r = \frac{8}{a_0^4} e^{-2 r/ a_0} r^2 \Longrightarrow 1...
To check that this is a maximum, we compute that second derivative of f,
\begin{align*}f''(r) = \frac{8}{a_0^3} e^{-2 r/ a_0}  -\frac{32}{a_0^4}e^{-2 r/ a_0} r + \frac{16}{a_0^5} e^{-2 r/ a_0} r^2\L...
The negative second derivative implies that r = a_0 is a maximum. Therefore, answer (D) is correct.

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