91Ó°ÊÓ

The radial wave function of electron in the $1s$state of hydrogen atom is given as follows:

$R_{1s}\left(r\right)=\frac{1}{\sqrt{{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}{e}^{\frac{r}{a_B}}$

The radial probability density for a $1s$is given as follows:

localid="1648803801176" $P_r\left(r\right)=4{\mathrm{Ï€°ù}}^2{\left|{\mathrm{R}}_{1\mathrm{s}}\left(\mathrm{r}\right)\right|}^2$

Substitute $\frac{1}{\sqrt{{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}{e}^{\frac{r}{e^{a_B}}}$for .

$$\begin{gathered} P_r\left(r\right)=4{\mathrm{Ï€°ù}}^2{\left|\frac{1}{\sqrt{{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}\mathrm{e}\frac{\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}\right|}^2 \\ =4{\mathrm{Ï€°ù}}^2\frac{1}{{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}} \\ =\frac{4}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{r}}^2\mathrm{e}\frac{2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}} \end{gathered}$$

The average value of $r$is given as follows:

localid="1648803378410" $$\begin{gathered} r_{avg}={∫}_0^{∞}r{P}_r\left(r\right)dr \\ ={∫}_0^{∞}r\frac{4}{a_B^3}{r}^2{e}^{\frac{2r}{a_B}}dr \\ =\frac{4}{a_B^3}{∫}_0^{∞}{r}^3{e}^{\frac{2r}{a_B}}dr \end{gathered}$$

Use standard integration formula ${∫}_0^{∞}{x}^n{e}^{-ax}dx=\frac{n!}{a^{n+1}}$

$$\begin{gathered} r_{avg}=\frac{4}{a_B^3}\frac{a_0^4}{16} \\ =\frac{24}{a_b^3}\frac{a_0^4}{16} \\ =1.5a_B \end{gathered}$$

Therefore, the average value of $r$for $1s$state is localid="1649751144354" $r_{avg}=1.5a_B$

The radial wave function of electron in the 2p state of hydrogen atom is,

$R_{2p}\left(r\right)=\frac{1}{\sqrt{24{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}\left(\frac{r}{2a_B}\right)e^{\frac{r}{2a_B}}$

The radial probability density for a $2p$is,

$P_r\left(r\right)=4{\mathrm{Ï€°ù}}^2{\left|{\mathrm{R}}_{2p}\left(\mathrm{r}\right)\right|}^2$

Substitute $\frac{1}{\sqrt{24{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}\left(\frac{r}{2a_B}\right)e^{\frac{r}{2a_B}}$for $R_{2p}\left(r\right)$,

$$\begin{gathered} P_r\left(r\right)=4{\mathrm{Ï€°ù}}^2{\left|\frac{1}{\sqrt{24{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}}\left(\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}\right){\mathrm{e}}^{\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}}\right|}^2 \\ =4{\mathrm{Ï€°ù}}^2\frac{1}{24{\mathrm{Ï€²¹}}_{\mathrm{B}}^3}{\left(\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}\right)}^2{\mathrm{e}}^{\frac{\mathrm{r}}{2{\mathrm{a}}_{\mathrm{B}}}} \\ =\frac{1}{24{\mathrm{Ï€²¹}}_{\mathrm{B}}^5}{\mathrm{r}}^4{\mathrm{e}}^{\frac{\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}} \end{gathered}$$

The average value of $r$is,

$$\begin{gathered} r_{avg}={∫}_0^{∞}r{P}_r\left(r\right)dr \\ ={∫}_0^{∞}r\frac{1}{24a_B^5}{r}^4{e}^{\frac{r}{a_B}}dr \\ =\frac{1}{24a_B^5}{∫}_0^{∞}{r}^5{e}^{\frac{r}{a_B}}dr \end{gathered}$$

Use standard integration formula ${∫}_0^{∞}{x}^n{e}^{-ax}dx=\frac{n!}{a^{n+1}}$.

$$\begin{gathered} r_{avg}=\frac{1}{24a_B^5}\frac{5!}{{\left({\displaystyle \frac{1}{a_B}}\right)}^{5+1}} \\ =5a_B \end{gathered}$$

Therefore, the average value of$r$for$1s$state is$r_{avg}=5a_B$uncaught exception: Http Error #503

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