Find the distance, in terms of $a_B$, between the two peaks in the radial probability density of the $2s$state of hydrogen.

Hint: This problem requires a numerical solution.

The radial probability density of an electron in a hydrogen atom for state $nl$as,

role="math" $P_r\left(r\right)=4{\mathrm{\pi r}}^2{\left|{\mathrm{R}}_{\mathrm{nl}}\left(\mathrm{r}\right)\right|}^2$

Here, $r$is the distance of the electron from nucleus, $R_{nl}$is the radial wave function.

For $2s$state,

$P_r\left(r\right)=4{\mathrm{\pi r}}^2{\left|{\mathrm{R}}_{2s}\left(\mathrm{r}\right)\right|}^2$

Substitute role="math" localid="1648806903608" $\frac{1}{\sqrt{8{\mathrm{\pi a}}_{\mathrm{B}}^3}}\left(1-\frac{r}{2a_B}\right)e^{\frac{-r}{2a_B}}$for $R_{2s}\left(r\right)$

$P_r\left(r\right)=4{\mathrm{\pi r}}^2{\left[\frac{1}{\sqrt{8{\mathrm{\pi a}}_{\mathrm{B}}^3}}\left(1-\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}}\right]}^2$

$$\begin{gathered} =\frac{4{\mathrm{\pi r}}^2}{8{\mathrm{\pi a}}_{\mathrm{B}}^3}{\left(1-\frac{r}{2a_B}\right)}^2{e}^{\frac{-2r}{2a_B}} \\ =2\frac{r^2}{a_B^3}{\left(1-\frac{r}{2a_B}\right)}^2{e}^{\frac{-2r}{2a_B}} \end{gathered}$$

Differentiate the radial probability density$P_r\left(r\right)$with respect to $r$.

$$\begin{gathered} \frac{d}{dr}\left[p_r\left(r\right)\right]=\frac{d}{dr}\left[2\frac{r^2}{a_B^3}{\left(1-\frac{r}{2a_B}\right)}^2{e}^{\frac{-2r}{a_B}}\right] \\ =\frac{2}{a_B^3}\left[2r{\left(1-\frac{r}{2a_B}\right)}^2{e}^{\frac{-2r}{a_B}}+r^2\left(2\right)\left(1-\frac{r}{2a_B}\right)\left(\frac{-1}{2a_B}\right)e^{\frac{-2r}{a_B}}+r^2{\left(1-\frac{r}{2a_B}\right)}^2\left(\frac{-2}{2a_B}\right)e^{\frac{-2r}{a_B}}\right] \\ =\frac{\left(2\right)\left(2\right)}{a_B^3}\left(r\right)\left(1-\frac{r}{2a_B}\right)\left(e^{\frac{-2r}{2a_B}}\right)\left[1-\frac{r}{2a_B}-\frac{r}{2a_B}+r\left(1-\frac{r}{2a_B}\right)\left(\frac{-1}{2a_B}\right)\right] \\ =\frac{4}{a_B^3}\left(r\right)\left(1-\frac{r}{2a_B}\right)\left(e^{\frac{-2r}{2a_B}}\right)\left[1-\frac{r}{2a_B}-\frac{r}{2a_B}-\frac{r}{2a_B}+\frac{r^2}{4{a^2}_B}\right] \\ =\frac{4}{a_B^2}\left(r\right)\left(1-\frac{r}{2a_B}\right)\left(e^{\frac{-2r}{2a_B}}\right)\left[1-\frac{3^2}{2a_B}+\frac{r^2}{4{a^2}_B}\right] \end{gathered}$$

The radial probability density $P_r\left(r\right)$is maximum or minimum when $\frac{d}{dr}\left[p_r\left(r\right)\right]=0$. Now,

$\frac{d}{dr}\left[p_r\left(r\right)\right]=0$

$\frac{4}{a_B^2}r\left(1-\frac{r}{2a_B}\right)e^{\frac{-2r}{2a_B}}\left(1-\frac{3r}{2a_B}+\frac{r^2}{4{a^2}_B}\right)=0$

From the equation $r=0$or $1-\frac{r}{2a_B}=0$or $e^{\frac{-2r}{2a_B}}=0$or $1-\frac{3r}{2a_B}+\frac{r^2}{4a_B^2}=0$. But when,

$$\begin{gathered} 1-\frac{r}{2a_B}=0 \\ r=2a_B \end{gathered}$$

So, $r=2a_B$corresponds to minimum.

The radial probability density $P_r\left(r\right)$is a maximum when

$$\begin{gathered} 1-\frac{3r}{2a_B}+\frac{r^2}{4a_B^2}=0 \\ {4}_B^2-6a_{B}r+r^2=0 \\ {r}^2-6a_{B}r+4a_B^2=0 \end{gathered}$$

This equation is quadratic equation in terms of $r$. The roots of $r$are

$$\begin{gathered} r=\frac{-(-6a_B)\pm \sqrt{{(-6a_B)}^2-4\left(1\right)\left(4a_B^2\right)}}{2\left(1\right)} \\ =\frac{6a_B\pm \sqrt{36{a_B}^2-16a_B^2}}{2} \\ =\frac{6a_B\pm \sqrt{20{a_B}^2}}{2} \\ =\frac{6a_B\pm 2\sqrt{5}{a}_B}{2} \\ =(3\pm \sqrt{5})a_B \end{gathered}$$

We have the conditions $r=\left(3+\sqrt{5}\right)a_B$or $r=\left(3-\sqrt{5}\right)a_B$when $P_r\left(r\right)$having maximum peaks. The distance between the two peaks in the radial probability density of $2s$state of hydrogen is

$$\begin{gathered} \left(3a_B+\sqrt{5}{a}_B\right)-\left(3a_3-\sqrt{5}{a}_B\right) \\ =2\sqrt{5}{a}_B \end{gathered}$$

Therefore, the distance between two peaks is$2\sqrt{5}{a}_B$