Answer
See the detailed answer below.
Work Step by Step
We know that the radial probability density $ P(r) $ is related to the square of the radial wave function $ R(r) $ and the factor $ r^2 $:
$$
P(r) =4\pi r^2 |R_{1s}(r)|^2\tag 1
$$
For the hydrogen atom in the $ 1s $ state, the radial wave function is:
$$
R_{1s}(r) = \frac{1}{\sqrt{\pi a_B^3}} e^{-r/a_B}
$$
Plug into (1);
$$
P(r) =4\pi r^2 \left[ \frac{1}{\sqrt{\pi a_B^3}} e^{-r/a_B}\right]^2
$$
$$
P(r) =4 \color{red}{\bf\not}\pi r^2 \cdot \frac{e^{-2r/a_B} }{ \color{red}{\bf\not}\pi a_B^3}
$$
Simplifying:
$$
P(r) = \frac{4e^{-2r/a_B}}{a_B^3}\; r^2\tag 2
$$
To find the value of $ r $ where the radial probability density peaks, we need to take the derivative of $ P(r) $ with respect to $ r $, set it equal to zero, and solve for $ r $.
$$
\frac{d}{dr} P(r) = \frac{d}{dr} \left( \frac{4e^{-2r/a_B}}{a_B^3}\; r^2\right)=0
$$
We can factor out the constant $ \dfrac{4}{a_B^3} $, so the derivative becomes:
$$
\frac{4}{a_B^3} \frac{d}{dr} \left( r^2 e^{-2r/a_B} \right)=0
$$
$$
\frac{d}{dr} \left( r^2 e^{-2r/a_B} \right)=0
$$
So
$$
2r e^{-2r/a_B} + r^2 \left( -\frac{2}{a_B} \right) e^{-2r/a_B}=0
$$
Factor out the common term $ e^{-2r/a_B} $.
$$
2r + \left( -\frac{2}{a_B} \right) r^2 =0
$$
Factor out the common term $2r$;
$$
2r\left[ 1 -\frac{r}{a_B} \right] =0
$$
This gives two solutions:
$ r = 0 $ which corresponds to the minimum probability at the origin, or
$ r = a_B $ at which the value of the radial probability density is maximum.
