Question

What is the probability of finding a $1s$hydrogen electron at distance $r>a_B$from the proton?

Short answer

The probability of finding a $1s$hydrogen electron at a $r>a_B$is $67.7\%$

Step-by-step solution

The radial wave function:

Let the radial wave function of the electron in $1s$state of hydrogen is,

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

Here, $A$is the normalization constant and $a_B$is the Bohr radius.

Radial probability density of electron in the $1s$state is,

localid="1648724818627" $P\left(r\right)=4{\mathrm{\pi r}}^2{\left|{\mathrm{R}}_{\mathrm{is}}\left(\mathrm{r}\right)\right|}^2$

Probability of finding is hydrogen electron at a distance $r>a_B$from the proton is given by,

$(r>a_B)={\int }_{a_B}^{\infty }{p}_r\left(r\right)dr$

Here, $P\left(r\right)=4{\mathrm{\pi r}}^2{\left|{\mathrm{R}}_{\mathrm{is}}\left(\mathrm{r}\right)\right|}^2$

$$\begin{gathered} =4{\mathrm{\pi r}}^2{\left|\frac{1}{\sqrt{{\mathrm{\pi a}}_{\mathrm{B}}^3}}{\mathrm{e}}^{\frac{\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}}\right|}^2 \\ =\frac{4{\mathrm{\pi r}}^2}{{\mathrm{\pi a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}} \\ =\frac{4{\mathrm{r}}^2}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}} \\ (\mathrm{r}>{\mathrm{a}}_{\mathrm{B}})={\int }_{{\mathrm{a}}_{\mathrm{B}}}^{\infty }\frac{4{\mathrm{r}}^2}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}}\mathrm{dr} \\ ={\int }_0^{\infty }\frac{4{\mathrm{r}}^2}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}}\mathrm{dr}-{\int }_0^{{\mathrm{a}}_{\mathrm{B}}}\frac{4{\mathrm{r}}^2}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}}\mathrm{dr} \end{gathered}$$

Further calculation:

But we have$={\int }_0^{\infty }{x}^n{e}^{-ax}\mathrm{dr}=\frac{n!}{{\alpha }^{n+1}}$

role="math" localid="1648725359957" $$\begin{gathered} ={\int }_{{\mathrm{a}}_{\mathrm{B}}}^{\infty }\frac{4{\mathrm{r}}^2}{{\mathrm{a}}_{\mathrm{B}}^3}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_{\mathrm{B}}}}\mathrm{dr}=\frac{4}{a_B^3}\frac{2!}{{\left({\displaystyle \frac{2}{a_B}}\right)}^3} \\ =\frac{8a_B^3}{8a_B^3} \\ =1 \end{gathered}$$

Now$$\begin{gathered} ={\int }_0^{a_B}\frac{4r^2}{a_B^3}{e}^{\frac{-2r}{a_B}}dr=\frac{4}{a_B^3}\left[\left(r^2\frac{e^{{\displaystyle \frac{-2r}{a_B}}}}{{\displaystyle \frac{-2}{a_B}}}\right)-{\int }_0^{a_B}2r\frac{e^{{\displaystyle \frac{-2r}{a_B}}}}{{\displaystyle \frac{-2}{a_B}}}dr\right] \\ =\frac{4}{a_B^3}\left[\left(\frac{a_B^3{e}^{-2}-0}{\left(\frac{-2}{a_B}\right)}\right)+a_B\left\{\left(\frac{r{e}^{\frac{-2r}{a_B}}}{\frac{-2}{a_B}}\right)-{\int }_0^{a_B}\frac{e^{\frac{-2r}{a_B}}}{\frac{-2}{a_B}}\right\}\right] \\ =\frac{4}{a_B^3}\left[\left(\frac{-a_B}{2}\right)\left(a_B^2{e}^{-2}\right)+a_B\left\{\left(\frac{a_B{e}^{\frac{-2a_B}{a_B}}}{\frac{-2}{a_B}}-0\right)+\frac{a_B}{2}{\left(\frac{e^{\frac{-2r}{a_B}}}{\frac{-2}{a_B}}\right)}^{a_B}\right\}\right] \\ =\frac{4}{a_B^3}\left[\left(\frac{-{a_B}^3}{2}\right)\left(e^{-2}\right)+a_B\left\{\left(-\frac{a_B^2}{2}{e}^{-2}\right)-\frac{a_B}{4}\left(e^{-2}-e^0\right)\right\}\right] \\ =\frac{4}{a_B^3}\left[\left(\frac{{a_B}^3{e}^{-2}}{2}\right)+a_B\left\{\frac{-a_B^2{e}^{-2}}{2}-\frac{a_B^2{e}^{-2}}{4}+\frac{a_B^2}{4}\right\}\right] \\ =\frac{4}{a_B^3}\left[-\frac{{a_B}^3{e}^{-2}}{2}-\frac{a_B^3{e}^{-2}}{2}-\frac{a_B^3{e}^{-2}}{4}+\frac{a_B^3}{4}\right] \\ =\frac{4}{a_B^3}\left[-\frac{5}{4}{a}_B^3{e}^{-2}+\frac{a_B^3}{4}\right] \\ =(-5e^{-2}+1) \\ =(1-5e^{-2}) \end{gathered}$$

Find probability:

The probability of finding a $1s$hydrogen electron at a distance $r>a_B$is,

$$\begin{gathered} (r>a_B)=1-(1-5e^{-2}) \\ =5e^{-2} \\ {a}_B=\left(5\right)(0.135335) \\ =0.67667 \\ =67.7\% \end{gathered}$$

Therefore, the probability is $67.7\%$