Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The radial wave function for the 2p state is given by:
$$
R_{2p}(r) = \frac{ A_{2p} }{2a_B}\;r\; e^{-r/2a_B}
$$
From the graph below of $ R_{2p}(r) $, we can see that it has a single maximum, indicating that the radial probability density reaches a peak at some point.
$$\color{blue}{\bf [b]}$$
To find the maximum, we first take the derivative of $ R_{2p}(r) $ with respect to $ r $ and set it equal to zero:
$$
\frac{dR_{2p}(r)}{dr} =\frac{d }{dr}\left[ \frac{ A_{2p} }{2a_B}\;r\; e^{-r/2a_B}\right]=0$$
So,
$$
\frac{ A_{2p}}{2a_B} \left[ e^{-r/2a_B} - \frac{r}{2a_B} e^{-r/2a_B} \right]=0
$$
Factoring out common terms:
$$
e^{-r/2a_B} \left[1- \frac{r}{2a_B} \right] =0$$
Since $ e^{-r/2a_B} \neq 0 $, so
$$
1 - \frac{r}{2a_B} = 0
$$
Thus,
$$
\boxed{r = 2a_B}
$$
Thus, the radial wave function $ R_{2p}(r) $ reaches its maximum at $ r = 2a_B $.
Therefore,
$$
(R_{2p}(r=2a_B ))_{\rm max} = \frac{ A_{2p} }{2a_B}\;[2a_B ]\; e^{-[2a_B ]/2a_B}=\boxed{{\bf 0.368}\;A_{2p}}
$$
$$\color{blue}{\bf [c]}$$
The probability density is proportional to the square of the radial wave function:
$$
P_{2p}(r) = 4\pi r^2 |R_{2p}(r)|^2
$$
For the absolute square of the radial wave function $ |R_{2p}(r)|^2 $, it represents the probability per unit volume of finding the electron at a given distance from the origin. The figure shows that as we move along the radial direction and consider differential volume elements $ \delta V = \delta r^3 $, we can measure the probability of finding the electron at different distances $ r $.
The probability density $ P_{2p}(r) $ gives the probability per unit length of finding the electron in a thin shell of thickness $ \delta r $ and radius $ r $. This probability can be visualized as we inspect the shell's volume to find where the electron is most likely to be.
The equation for the radial probability density, considering the surface area of a spherical shell, is:
$$
P_{2p}(r) = 4 \pi r^2 |R_{2p}(r)|^2
$$
The factor of $ r^2 $ accounts for the increased volume available in a spherical shell of radius $ r $. This is why $ P_{2p}(r) $ peaks at $ r = 4a_B $, which is further from the nucleus than the peak of $ R_{2p}(r) $, which occurs at $ r = 2a_B $.
The peak of $ P_{2p}(r) $ occurs at $ r = 4a_B $ because the surface area of the spherical shell increases as $ r $ increases, and this compensates for the decay in the wave function at larger distances. Hence, the electron is most likely to be found at a distance of $ 4a_B $ from the nucleus.
