Chapter 42, Problem 81AP

(a)

To determine

The radial probability density for the $2s$ state of Hydrogen atom.

Answer to Problem 81AP

The radial probability density for the $2s$ state of Hydrogen atom is $\underset{\_}{P\left(r\right)=\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}}$.

Explanation of Solution

Write the expression for the wave function of electron in the $2s$ state of Hydrogen.

    ${\psi }_{2s}\left(r\right)=\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[2-\frac{r}{a_0}\right]e^{-\frac{r}{2a_0}}$                                                              (I)

Here, ${\psi }_{2s}\left(r\right)$ is the wave function of electron in the $2s$ state of Hydrogen, $r$ is the radial distance between the nucleus and electron of atom, and $a_0$ is the Bohr radius of the atom.

Write the expression for the probability density for $2s$ state of Hydrogen atom.

    $\begin{array}{c}P\left(r\right)=4\pi r^2{\left|{\psi }_{2s}\right|}^2\\ =4\pi r^2{\psi }_{2s}{\psi }_{2s}^{\ast }\end{array}$                                                                                           (II)

Here, $P\left(r\right)$ is the probability density for $2s$ state of Hydrogen atom, and ${\psi }_{2s}^{\ast }$ is the complex conjugate of the wave function.

Write the complex conjugate of ${\psi }_{2s}$.

    ${\psi }_{2s}^{\ast }=\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[2-\frac{r}{a_0}\right]e^{-\frac{r}{2a_0}}$                                                                           (III)

Conclusion:

Use expressions (III), and (I) in expression (II) to find $P\left(r\right)$.

    $\begin{array}{c}P\left(r\right)=4\pi r^2\left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[2-\frac{r}{a_0}\right]e^{-\frac{r}{2a_0}}\right)\left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[2-\frac{r}{a_0}\right]e^{-\frac{r}{2a_0}}\right)\\ =\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}\end{array}$

Therefore, the radial probability density for the $2s$ state of Hydrogen atom is $\underset{\_}{P\left(r\right)=\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}}$.

(b)

To determine

The derivative of the radial probability density with respect to $r$.

Answer to Problem 81AP

The derivative of the radial probability density with respect to $r$ is $\underset{\_}{\frac{dP\left(r\right)}{dr}=\frac{r}{8a_0^5}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[r^2-6r{a}_0+4a_0^2\right]}$.

Explanation of Solution

Write the expression for $P\left(r\right)$ as derived from subpart (a).

    $P\left(r\right)=\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}$                                                                               (IV)

Take the derivative of expression (IV) with respect to $r$.

    $\begin{array}{c}\frac{dP\left(r\right)}{dr}=\frac{d\left(\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}\right)}{dr}\\ =\frac{1}{8a_0^3}\left[2r{\left(2-\frac{r}{a_0}\right)}^2-2r^2\left(\frac{1}{a_0}\right)\left(2-\frac{r}{a_0}\right)-r^2{\left(2-\frac{r}{a_0}\right)}^2\left(\frac{1}{a_0}\right)\right]e^{-\frac{r}{a_0}}\\ =\frac{1}{8a_0^3}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[4r-\frac{2r^2}{a_0}-\frac{2r^2}{a_0}-\left(\frac{2r^2}{a_0}-\frac{r^3}{a_0^2}\right)\right]\end{array}$               (V)

Simplify expression (V).

    $\begin{array}{c}\frac{dP\left(r\right)}{dr}=\frac{r}{8a_0^5}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[4a_0^2-6r{a}_0+r^2\right]\\ =\frac{r}{8a_0^5}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[r^2-6r{a}_0+4a_0^2\right]\end{array}$

Conclusion:

Therefore, the derivative of the radial probability density with respect to $r$ is $\underset{\_}{\frac{dP\left(r\right)}{dr}=\frac{r}{8a_0^5}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[r^2-6r{a}_0+4a_0^2\right]}$.

(c)

To determine

Three values of $r$ that represents minima in the function.

Answer to Problem 81AP

The three values possible for $r$ for getting minimum value of function is $\underset{\_}{r=0,r=2a_0,\text{and }r=\infty }$.

Explanation of Solution

Minimum points of the function can be found by equating the derivative to zero.

    $\begin{array}{c}\frac{dP\left(r\right)}{dr}=\frac{r}{8a_0^5}\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}\left[r^2-6r{a}_0+4a_0^2\right]\\ =0\end{array}$                                                     (VI)

The derivative equals to zero when any of the product terms equals to zero. Thus $r=0$ is a possible solution.

First term of expression $\frac{r}{8a_0^5}$ cannot be equal to zero.

Equate the second term to zero.

    $\left(2-\frac{r}{a_0}\right)=0$                                                                                                     (VII)

Solve expression (VII) to find $r$.

    $\begin{array}{c}2=\frac{r}{a_0}\\ r=2a_0\end{array}$

Equate third of expression (VI) to zero.

    $e^{-\frac{r}{a_0}}=0$                                                                                                         (VIII)

Expression (VII) is true if and only if $r=\infty$.

Conclusion:

Therefore, the three values possible for $r$ for getting minimum value of function is $\underset{\_}{r=0,r=2a_0,\text{and }r=\infty }$.

(d)

To determine

The two values of $r$ that is the maximum point of the function.

Answer to Problem 81AP

The two values of $r$ that is the maximum point of the function are $\underset{\_}{r=\left(3+\sqrt{5}\right)a_0}$ and $\underset{\_}{r=\left(3-\sqrt{5}\right)a_0}$.

Explanation of Solution

The maximum points of the function can be found by equating the quadratic equation of the derivative to zero.

    $r^2-6r{a}_0+4a_0^2=0$                                                                                           (IX)

Write the general expression for a quadratic expression.

    $a{r}^2+br+c=0$                                                                                              (X)

Write the expression to find the solution of expression (X).

    $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$                                                                                               (XI)

Conclusion:

Compare expressions (IX) and (X).

    $\begin{array}{c}a=1\\ b=-6r\\ c=a_0^2\end{array}$                                                                                                                 (XII)

Substitute values of $a,b,\text{and }c$ in equation (XI) to find $r$.

    $\begin{array}{c}r=\frac{-\left(-6a_0\right)\pm \sqrt{{\left(-6a_0\right)}^2-4\left(1\right)\left(4a_0^2\right)}}{2}\\ =\frac{6a_0\pm \sqrt{20a_0^2}}{2}\\ =\left(3\pm \sqrt{5}\right)a_0\end{array}$

Therefore, the two values of $r$ that is the maximum point of the function are $\underset{\_}{r=\left(3+\sqrt{5}\right)a_0}$ and $\underset{\_}{r=\left(3-\sqrt{5}\right)a_0}$.

(e)

To determine

The value of $r$ that gives maximum probability.

Answer to Problem 81AP

The value of $r$ that gives the maximum probability is $\underset{\_}{r=\left(3+\sqrt{5}\right)a_0}$.

Explanation of Solution

Write the expression for probability as found in subpart (a).

    $P\left(r\right)=\frac{r^2}{8a_0^3}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{r}{a_0}}$                                                                             (XIII)

Conclusion:

Substitute $r=\left(3-\sqrt{5}\right)a_0$ in expression (XIII) to find $P\left(r\right)$.

    $\begin{array}{c}P\left(r\right)=\frac{{\left(\left(3-\sqrt{5}\right)a_0\right)}^2}{8a_0^3}{\left(2-\frac{\left(3-\sqrt{5}\right)a_0}{a_0}\right)}^2{e}^{-\frac{\left(3-\sqrt{5}\right)a_0}{a_0}}\\ =\frac{\left(0.764a_0^2\right)}{8a_0^3}{\left(2-\frac{0.764a_0}{a_0}\right)}^2{e}^{-0.764}\\ =\frac{0.0519}{a_0}\end{array}$

Substitute $r=\left(3+\sqrt{5}\right)a_0$ in expression (XIII) to find $P\left(r\right)$.

    $\begin{array}{c}P\left(r\right)=\frac{{\left(\left(3+\sqrt{5}\right)a_0\right)}^2}{8a_0^3}{\left(2-\frac{\left(3+\sqrt{5}\right)a_0}{a_0}\right)}^2{e}^{-\frac{\left(3+\sqrt{5}\right)a_0}{a_0}}\\ =\frac{\left(5.236a_0^2\right)}{8a_0^3}{\left(2-\frac{5.236a_0}{a_0}\right)}^2{e}^{-5.236}\\ =\frac{0.191}{a_0}\end{array}$

Thus the value of probability is greater when $r=\left(3+\sqrt{5}\right)a_0$.

Therefore, the value of $r$ that gives the maximum probability is $\underset{\_}{r=\left(3+\sqrt{5}\right)a_0}$.