Answer
${\bf 2\sqrt{5}}\; a_B$
Work Step by Step
To find the distance between the two peaks in the radial probability density of the $ 2s $ state of hydrogen, we need to start with the radial probability distribution function for the $ 2s $ state.
So, we need to analyze the radial probability density $ P_r(r) $ for this state.
The radial probability density $ P_r(r) $ for the 2s state is given by
$$
P_r(r) = 4\pi r^2 |R_{2s}(r)|^2\tag 1
$$
where $R_{2s}(r)$ is given by
$$
R_{2s}(r) = $$
Plug into (1);
$$
P_r(r) = 4\pi r^2 \left[\frac{1}{\sqrt{8\pi a_B^3}} \left(1 - \frac{r}{2a_B}\right) e^{-r/2a_B}\right]^2
$$
$$
P_r(r) = \dfrac{ 1}{ 2 a_B^3} \left(1 - \frac{r}{2a_B}\right)^2r^2 e^{-r/a_B}
$$
To find the most probable distance, we need to locate the maximum of the radial probability density function. This requires taking the derivative of $ P_r(r) $ with respect to $ r $ and setting it equal to zero.
$$
\frac{dP_r}{dr} =\frac{d}{dr} \left[\dfrac{ 1}{ 2 a_B^3} \left(1 - \frac{r}{2a_B}\right)^2r^2 e^{-r/a_B} \right]=0
$$
where $\dfrac{ 1}{ 2 a_B^3}$ is constant.
$$ \frac{d}{dr} \left[ \left(1 - \frac{ r}{ a_B}+\frac{r^2}{4a_B^2}\right) r^2 e^{-r/a_B} \right]=0 $$
$$ \frac{d}{dr} \left[ \left(r^2 - \frac{ r^3}{ a_B }+\frac{r^4}{4a_B^2}\right) e^{-r/a_B} \right]=0 $$
$$= \left(2r - \frac{ 3r^2}{ a_B }+\frac{r^3}{ a_B^2}\right) e^{-r/a_B} +\\
\left(r^2 - \frac{ r^3}{ a_B }+\frac{r^4}{4a_B^2}\right) \left[\dfrac{-1}{a_B}\; e^{-r/a_B} \right]=0 $$
Since both terms include the exponential factor $ e^{-r/a_B}$, and this function is never zero, we can eliminate the exponential term:
$$
\left( 2r - \frac{3r^2}{a_B} + \frac{r^3}{ a_B^2} \right) - \frac{1}{a_B} \left( r^2 - \frac{r^3}{a_B} + \frac{r^4}{4a_B^2} \right) = 0
$$
$$ 2r - \frac{3r^2}{a_B} + \frac{r^3}{ a_B^2} - \frac{r^2}{a_B} + \frac{r^3}{a_B^2} - \frac{r^4}{4a_B^3} = 0
$$
$$ 2r - \frac{4r^2}{a_B} + \frac{2r^3}{ a_B^2} - \frac{r^4}{4a_B^3} = 0
$$
$$ r \left[ 2 - \frac{4r }{a_B} + \frac{2r^2}{ a_B^2} - \frac{r^3}{4a_B^3} \right] = 0
$$
we can see that $r=0$ which is dismissed here and not reasonable.
$$2 - \frac{4r }{a_B} + \frac{2r^2}{ a_B^2} - \frac{r^3}{4a_B^3} =0$$
Solving the cubic equation numerically using any software calculator.
$ r_1= (3-\sqrt{5})a_B=0.764 \,a_B $ (maximum)
$ r_2= 2.0\,a_B $ (minimum)
$ r_3 = (3+\sqrt{5})a_B= 5.236 \,a_B $ (maximum)
We also can see the data in Figure 41.8, which matches the numerical solution.
The first and third roots correspond to the two maxima in the radial probability density. The second root represents a minimum between the two peaks.
So, the separation between the two peaks (the two maxima) is simply the difference between the two maxima:
$$
\Delta r = r_3 - r_1 = (3+\sqrt{5})a_B - (3-\sqrt{5})a_B$$
$$\Delta r =2\sqrt{5}\;a_B= \color{red}{\bf 4.472}\; a_B
$$
