Question

Prove that the most probable distance from the proton of an electron in the$2s$state of hydrogen is $5.236a_B$.

Short answer

The most probable distance from the proton of an electron in the$2s$ state of hydrogen is $5.236a_B$.

Step-by-step solution

Given information

We need to prove that the most probable distance from the proton of an electron in the $2s$ state of hydrogen is $5.236a_B$.

Explanation

We need to prove that$r=5.236a_B$where $r$is most probable distance.

For that we will find radial probability density for $2s$state:

role="math" localid="1650381765979" $$\begin{gathered} Pr\left(r\right)=4{\mathrm{\pi r}}^2{\left|{\mathrm{R}}_{2\mathrm{s}}\left(\mathrm{r}\right)\right|}^2 \\ \mathrm{Pr}\left(\mathrm{r}\right)=4\mathrm{\pi }{{\mathrm{A}}_{2\mathrm{s}}}^2{\mathrm{r}}^2{\left(1-\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}\right)}^2\mathrm{e}\raisebox{1ex}{$-\mathrm{r}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{a}}_{\mathrm{B}}$}\right. \end{gathered}$$.

We can treat as constant $4{{\mathrm{\pi A}}_{2\mathrm{s}}}^2{\mathrm{r}}^2=\mathrm{C}$and rearranging the equation on right hand side ,

$Pr\left(r\right)=C\left(r^2-\frac{r^3}{a_B}+\frac{r^4}{4{a^2}_B}\right)e\raisebox{1ex}{$-r$}\!\left/ \!\raisebox{-1ex}{$a_B$}\right.$.

Now we will differentiate it with respect to $r$$$\begin{gathered} \frac{d{P}_r}{dr}=C\left[\left(2r-\frac{3r^2}{a_B}+\frac{r^3}{{a_B}^2}\right)e^{\frac{-r}{a_B}}-\frac{1}{a_B}\left(r^2-\frac{r^3}{a_B}+\frac{r^4}{4{a_B}^2}\right)e^{\frac{-r}{a_B}}\right] \\ \frac{d{P}_r}{dr}=C\left[2-\frac{4r}{a_B}+\frac{2r^2}{a_B}-\frac{r^3}{{4a_B}^2}\right]r{e}^{\frac{-r}{a_B}} \end{gathered}$$

Rearranging the terms , we get

$$\begin{gathered} 0=\left[2-4\left(\frac{r}{a_B}\right)+2{\left(\frac{2}{a_B}\right)}^2-\frac{1}{4}{\left(\frac{r}{a_B}\right)}^3\right] \\ 0=2-4\left(5.236\right)+2{\left(5.236\right)}^2-\frac{1}{4} \end{gathered}$$

$r=5.236a_B$