Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between $r$ and $r+\Delta r$ .

Answer to Problem 41P

The probability of finding the electron in the region between $r$ and $r+\Delta r$ is $9.20\times {10}^{-4}$ .

Explanation of Solution

Given:

The value of $\Delta r$ is $0.02a_o$ .

The value of $r$ is $a_o$ .

The value of $n$ is $2$ .

The value of $l$ is $0$ .

The value of $m_l$ is $0$ .

Formula used:

The expression for probabilityis given by,

  $P=P\left(r\right)\Delta r$

The expression for $P\left(r\right)$ is given by,

  $P\left(r\right)=4\pi r^2{\left|\psi \right|}^2$

The expression for $\psi$ at $n=2$ is given by,

  $\psi =C\left(2-\frac{Zr}{a_o}\right)e^{-Zr/2a_o}$

The expression for $C$ at $n=2$ is given by,

  $C=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}$

Calculation:

The probability is calculated as,

  $\begin{array}{c}P=P\left(r\right)\Delta r\\ =4\pi r^2{\left|\psi \right|}^2\Delta r\\ =4\pi r^2{\left|C\left(2-\frac{Zr}{a_o}\right)e^{-Zr/2a_o}\right|}^2\times 0.02a_o\\ =0.08\pi a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left|\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}\right|}^2\end{array}$

Solving further as,

  $\begin{array}{c}P=0.08\pi a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left|\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}\right|}^2\\ =0.0025a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left(\frac{Z}{a_o}\right)}^3\end{array}$

For $r=a_o$ and $Z=1$ the above expression is,

  $\begin{array}{c}P=0.0025a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left(\frac{Z}{a_o}\right)}^3\\ =0.0025a_o{\left(a_o\right)}^2{\left(2-\frac{a_o}{a_o}\right)}^2{e}^{-a_o/a_o}{\left(\frac{1}{a_o}\right)}^3\\ =9.2\times {10}^{-4}\end{array}$

Conclusion:

Therefore, the probability of finding the electron in the region between $r$ and $r+\Delta r$ is $9.20\times {10}^{-4}$ .

(b)

To determine

The probability of finding the electron in the region between $r$ and $r+\Delta r$ .

Answer to Problem 41P

The probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0$ .

Explanation of Solution

Given:

The value of $\Delta r$ is $0.02a_o$ .

The value of $r$ is $2a_o$ .

The value of $n$ is $2$ .

The value of $l$ is $0$ .

The value of $m_l$ is $0$ .

Formula used:

The expression for probability is given by,

  $P=P\left(r\right)\Delta r$

The expression for $P\left(r\right)$ is given by,

  $P\left(r\right)=4\pi r^2{\left|\psi \right|}^2$

The expression for $\psi$ at $n=2$ is given by,

  $\psi =C\left(2-\frac{Zr}{a_o}\right)e^{-Zr/2a_o}$

The expression for $C$ at $n=2$ is given by,

  $C=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}$

Calculation:

The probability is calculated as,

  $\begin{array}{c}P=P\left(r\right)\Delta r\\ =4\pi r^2{\left|\psi \right|}^2\Delta r\\ =4\pi r^2{\left|C\left(2-\frac{Zr}{a_o}\right)e^{-Zr/2a_o}\right|}^2\times 0.02a_o\\ =0.08\pi a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left|\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}\right|}^2\end{array}$

Solving further as,

  $\begin{array}{c}P=0.08\pi a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left|\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_o}\right)}^{3/2}\right|}^2\\ =0.0025a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left(\frac{Z}{a_o}\right)}^3\end{array}$

For $r=2a_o$ and $Z=1$ the above expression is,

  $\begin{array}{c}P=0.0025a_o{r}^2{\left(2-\frac{Zr}{a_o}\right)}^2{e}^{-Zr/a_o}{\left(\frac{Z}{a_o}\right)}^3\\ =0.0025a_o{\left(2a_o\right)}^2{\left(2-\frac{2a_o}{a_o}\right)}^2{e}^{-2a_o/a_o}{\left(\frac{1}{a_o}\right)}^3\\ =0\end{array}$

Conclusion:

Therefore, the probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0$ .