Suppose you have a machine that gives you pieces of candy when you push a button. Eighty percent of the time, pushing the button gets you two pieces of candy. Twenty percent of the time, pushing the button yields $10$pieces. The average number of pieces per push is $N_{avg}=2\times 0.80+10\times 0.20=3.6$. That is, $10$pushes should get you, on average, $36$pieces. Mathematically, the average value when the probabilities differ is $N_{avg}=\sum (N_i\times Probabilityofi)$. We can do the same thing in quantum mechanics, with the difference that the sum becomes an integral. If you measured the distance of the electron from the proton in many hydrogen atoms,

you would get many values, as indicated by the radial probability density. But the average value of $r$would be

$r_{avg0}={\int }_0^{\infty }r{P}_r\left(r\right)dt$

Calculate the average value of $r$in terms of $a_B$for the electron in the $1s$and the $2p$states of hydrogen.

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

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

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

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

Substitute $\frac{1}{\sqrt{{\mathrm{\pi a}}_{\mathrm{B}}^3}}{e}^{\frac{r}{e^{a_B}}}$for .

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

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

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

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

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

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,

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

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

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

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

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

The average value of $r$is,

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

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

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

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

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