Chapter 42, Problem 78AP

(a)

To determine

An expression for the probability that an electron in ground state will be found outside a sphere of radius $r$.

Answer to Problem 78AP

The expression for the probability that an electron in ground state will be found outside a sphere of radius $r$ is $\underset{\_}{P=\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)e^{\frac{-2r}{a_0}}}$.

Explanation of Solution

Write the expression for the radial probability density function for the Hydrogen atom in ground state.

    $P_{1s}\left(r^{\prime }\right)=\left(\frac{4{r^{\prime }}^2}{a_0^3}\right)e^{\frac{-2r^{\prime }}{a_0}}$                                                                                             (I)

Here, $P_{1s}\left(r^{\prime }\right)$ is the radial probability density function for the Hydrogen atom in ground state, $a_0$ is the Bohr radius, and $r^{\prime }$ is the radial distance between nucleus and electron.

Write the expression to find the probability to for an electron in grounds state to be found outside a sphere of radius $r$.

    $P={\displaystyle \int P_{1s}\left(r^{\prime }\right)d{r}^{\prime }}$                                                                                                      (II)

The probability outside a sphere is to be found. So integration must be done from radius of sphere to infinity.

    $P={\displaystyle \underset{r}{\overset{\infty }{\int }}{P}_{1s}\left(r^{\prime }\right)d{r}^{\prime }}$                                                                                                       (III)

Use expression (I) in (III) to find $P$.

    $\begin{array}{c}P={\displaystyle \underset{r}{\overset{\infty }{\int }}\left(\frac{4{r^{\prime }}^2}{a_0^3}\right)e^{\frac{-2r^{\prime }}{a_0}}d{r}^{\prime }}\\ =\frac{4}{a_0^3}{\displaystyle \underset{r}{\overset{\infty }{\int }}{r^{\prime }}^2{e}^{\frac{-2r^{\prime }}{a_0}}d{r}^{\prime }}\end{array}$                                                                                             (IV)

Integrate expression (IV) by integration by parts.

    $\begin{array}{c}P=\frac{4}{a_0^3}{\displaystyle \underset{r}{\overset{\infty }{\int }}{r^{\prime }}^2{e}^{\frac{-2r^{\prime }}{a_0}}d{r}^{\prime }}\\ =\frac{4}{a_0^3}\left[-{r^{\prime }}^2{e}^{-\frac{2r}{a_0}}\left(\frac{a_0}{2}\right)-{\displaystyle \underset{r}{\overset{\infty }{\int }}2r^{\prime }\times e^{\frac{-2r^{\prime }}{a}}\left(-\frac{a}{2}\right)d{r}^{\prime }}\right]\\ =\frac{4}{a_0^3}\left[-{r^{\prime }}^2{e}^{-\frac{2r}{a_0}}\left(\frac{a_0}{2}\right)+{\displaystyle \underset{r}{\overset{\infty }{\int }}{r}^{\prime }{a}_0{e}^{-\frac{2r}{a_0}}d{r}^{\prime }}\right]\\ =\frac{4}{a_0^3}\left[-{r^{\prime }}^2{e}^{-\frac{2r}{a_0}}\left(\frac{a_0}{2}\right)+a_0\left[r^{\prime }{e}^{\frac{-2r^{\prime }}{a_0}}\left(-\frac{a_0}{2}\right)+\frac{a}{2}{e}^{\frac{-2r^{\prime }}{a_0}}\left(-\frac{a_0}{2}\right)\right]\right]\end{array}$                                    (V)

Simplify expression (V) to find $P$.

    $\begin{array}{c}P=\frac{4}{a_0^3}\left[-{r^{\prime }}^2{a}_0{e}^{-\frac{2r^{\prime }}{a_0}}-\frac{a_0^2{r}^{\prime }{e}^{-\frac{2r^{\prime }}{a_0}}}{2}-\frac{a_0^3}{4}{e}^{-\frac{2r^{\prime }}{a_0}}\right]\\ =\left[-\frac{4{r^{\prime }}^2}{a_0^2}\frac{e^{-\frac{2r^{\prime }}{a_0}}}{2}-\frac{2}{a_0}{r}^{\prime }{e}^{-\frac{2r^{\prime }}{a_0}}-e^{-\frac{2r^{\prime }}{a_0}}\right]\\ =e^{-\frac{2r^{\prime }}{a_0}}\left[-\left(\frac{2{r^{\prime }}^2}{a_0^2}+\frac{2r^{\prime }}{a_0}+1\right)\right]\end{array}$                                                                 (VI)

Conclusion:

Apply limits from $r\text{to}\infty$ in equation (VI).

    $\begin{array}{c}P=e^{-\frac{2r^{\prime }}{a_0}}{\left[-\left(\frac{2{r^{\prime }}^2}{a_0^2}+\frac{2r^{\prime }}{a_0}+1\right)\right]}_r^{\infty }\\ =e^{-\frac{2r}{a_0}}\left[-\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)\right]\end{array}$

Therefore, the expression for the probability that an electron in ground state will be found outside a sphere of radius $r$ is $\underset{\_}{P=\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)e^{\frac{-2r}{a_0}}}$.

(b)

To determine

Graph the relation between probability and $\frac{r}{a_0}$.

Answer to Problem 78AP

The graph between probability and $\frac{r}{a_0}$ is shown in figure 1 below.

Explanation of Solution

The expression for probability as a function of $r$ is derived in part (a) of the question.

    $P=\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)e^{\frac{-2r}{a_0}}$

The expression is exponential in nature. The graph will be exponentially decreasing as the value of $\frac{r}{a_0}$ increases.

Assume values between $0\text{and 5}$ for $\frac{r}{a_0}$ and plot the corresponding probability. $X$ axis shows $\frac{r}{a_0}$ and $Y$ shows the variation of probability.

Figure 1 below shows the plot between probability and $\frac{r}{a_0}$.

Conclusion:

Therefore, The graph between probability and $\frac{r}{a_0}$ is shown in figure 1.

(c)

To determine

The value of $r$ for which the probability to find electron inside sphere is equal to probability to find electron outside the sphere.

Answer to Problem 78AP

The value of $r$ for which the probability to find electron inside sphere is equal to probability to find electron outside the sphere is $\underset{\_}{1.34a_0}$.

Explanation of Solution

The probability to detect an electron inside or outside a sphere of radius $r$ is equal to $\frac{1}{2}$.

Use expression for probability derived in part (a).

    $P=\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)e^{\frac{-2r}{a_0}}$

Substitute $\frac{1}{2}$ for $P$ in above equation.

    $\left(\frac{2r^2}{a_0^2}+\frac{2r}{a_0}+1\right)e^{\frac{-2r}{a_0}}=\frac{1}{2}$                                                                              (VII)

Put $z=\frac{2r}{a_0}$ in expression (VII) and rewrite.

    $z^2+2z+2=e^z$                                                                                              (VIII)

Conclusion:

Equation (VIII) is a transcendental equation. Solving it the value of $r$ is $1.34a_0$.

Therefore, the value of $r$ for which the probability to find electron inside sphere is equal to probability to find electron outside the sphere is $\underset{\_}{1.34a_0}$.