Question 39.54: The wave function for the hydrogen-atom quantum state repres......
The wave function for the hydrogen-atom quantum state represented by the dot plot shown in Fig. 39-21, which has n = 2 and \ell = m_\ell = 0, is
\psi_{200}(r)=\frac{1}{4 \sqrt{2 \pi}} a^{-3 / 2}\left(2-\frac{r}{a}\right) e^{-r / 2 a}, ,
in which a is the Bohr radius and the subscript on \psi (r) gives the values of the quantum numbers n, \ell , m_\ell . (a) Plot \psi_{200}^2(r) and show that your plot is consistent with the dot plot of Fig. 39-21. (b) Show analytically that \psi_{200}^2(r) has a maximum at r = 4a. (c) Find the radial probability density P_{200}(r) for this state. (d) Show that
\int_0^{\infty} P_{200}(r) d r=1
and thus that the expression above for the wave function \psi_{200}(r) has been properly normalized.
(a) The plot shown below for \left|\psi_{200}(r)\right|^2 is to be compared with the dot plot of Fig. 39-21. We note that the horizontal axis of our graph is labeled “r,” but it is actually r/a (that is, it is in units of the parameter a). Now, in the plot below there is a high central peak between r = 0 and r = 2a, corresponding to the densely dotted region around the center of the dot plot of Fig. 39-21. Outside this peak is a region of near-zero values centered at r = 2a, where \psi_{200}=0 . This is represented in the dot plot by the empty ring surrounding the central peak. Further outside is a broader, flatter, low peak that reaches its maximum value at r = 4a. This corresponds to the outer ring with near-uniform dot density, which is lower than that of the central peak.
(b) The extrema of \psi^2(r) for 0 < r < ∞ may be found by squaring the given function, differentiating with respect to r, and setting the result equal to zero:
-\frac{1}{32} \frac{(r-2 a)(r-4 a)}{a^6 \pi} e^{-r / a}=0
which has roots at r = 2a and r = 4a. We can verify directly from the plot above that r =
4a is indeed a local maximum of \psi_{200}^2(r) . As discussed in part (a), the other root (r = 2a) is a local minimum.
(c) Using Eq. 39-43 and Eq. 39-41, the radial probability is
d V=\left(4 \pi r^2\right) d r (39-41)
P(r) d r=\psi^2(r) d V . (39-43)
P_{200}(r)=4 \pi r^2 \psi_{200}^2(r)=\frac{r^2}{8 a^3}\left(2-\frac{r}{a} \right)^2 e^{-r / a} .
(d) Let x = r/a. Then
\begin{aligned} \int_0^{\infty} P_{200}(r) d r & =\int_0^{\infty} \frac{r^2}{8 a^3}\left(2-\frac{r}{a}\right)^2 e^{-r / a} d r=\frac{1}{8} \int_0^{\infty} x^2(2-x)^2 e^{-x} d x=\int_0^{\infty}\left(x^4-4 x^3+4 x^2\right) e^{-x} d x \\ & =\frac{1}{8}[4 !-4(3 !)+4(2 !)]=1 \end{aligned}
where we have used the integral formula \int_0^{\infty} x^n e^{-x} d x=n ! .
