Answer
P(r) has its maximum value at r=4a.
Work Step by Step
At the r where P(r) has its maximum value, $\frac{dP}{dr} =0$.
Using the wavefunction given in the problem, $P(r)=\frac{1}{24a^5}r^4e^{-r/a}$.
Find the derivative and set it equal to zero.
$\frac{dP}{dr} = \frac{1}{24a^5}(4r^3- \frac{r^4}{a})e^{-r/a}$
This is zero when $4r^3- \frac{r^4}{a}=0$, or r = 4a.
P(r) has its maximum value at the distance of the electron from the nucleus in the Bohr model, where $r_n=n^2a$.
Our result also agrees with Figure 41.8 in the textbook.
