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\begin{document}


{\bf Question}

The wave function for the ground state of the hydrogen atom is
given by $$\psi(r,\theta,\phi)=\frac{1}{\sqrt{\pi
a_0^{3}}}e^\frac{-r}{a_0}.$$ The probability of finding the
electron at a distance $r$ from the nucleus is given by

$$p(r)=\int_{\phi=0}^{\phi=2\pi} \! \int_{\theta=0}^{\theta=\pi}
(\psi(r,\theta,\phi))^2r^2\sin(\theta)\,d\theta\,d\phi.$$
Calculate $p(r)$ and show that it has a maximum when $r=a_0$.


{\bf Answer}

\begin{eqnarray*} p(r) &amp; = &amp; \frac{1}{\pi a_0^3}
\int_{\phi=0}^{\phi=2\pi} \! \int_{\theta=0}^{\theta=\pi} r^2\sin
\theta e^{\frac{-2r}{a_0}} \,d\theta \,d\phi \\ &amp; = &amp;
\frac{r^2e^{\frac{-2r}{a_0}}}{\pi a_0^3} \int_0^{2\pi} \,d\phi
\int_0^{\pi}\sin \theta \,d\theta \\ &amp; = &amp;
\frac{r^2e^{\frac{-2r}{a_0}}}{\pi a_0^3} [\phi]_0^{2\pi}[-\cos
\theta]_0^{\pi}\\ &amp; = &amp; \frac{r^2e^{\frac{-2r}{a_0}}}{\pi
a_0^3}(2\pi)(1+1)\\ &amp; = &amp; \frac{4r^2e^{\frac{-2r}{a_0}}}{a_0^3}
\end{eqnarray*}

To find stationary points, calculate $\frac{\partial p}{\partial
r}$. Using chain rule:

$$\frac{\partial p}{\partial
r}=\frac{8re^{\frac{-2r}{a_0}}}{a_0^3}-\frac{8r^2e^{\frac{-2r}{a_0}}}{a_0^4}
= \frac{8r(a_0-r)e^{-\frac{2r}{a_0}}}{a_0^4}$$

So $\frac{\partial p}{\partial r}=0$ when $r=0$ or $r=a_0\
\left(e^{\frac{-2r}{a_0}} \ne 0\right)$

\begin{description}
\item[\underline{$r=0$}]

As r increases through $r=0$,\ $\frac{\partial p}{\partial r}$
changes from negative to positive.  So $p(r)$ has a minimum at
$r=0$.

PICTURE

\item[\underline{$r=a_0$}]

As r increases through $r=a_0$,\ $\frac{\partial p}{\partial r}$
changes from positive to negative.  So $p(r)$ has a maximum at
$r=a_0$.

PICTURE

\end{description}

\end{document}
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