Chapter 5, Problem 5.8P

(a)

To determine

The expression for orbital radii and energies.

Answer to Problem 5.8P

The expression for orbital radii is $\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}$ and the expression for orbital energy is $-\frac{{\mu }^2{z}^2{e}^4}{32{\pi }^2{\epsilon }_0^2{n}^2{h}^2}$.

Explanation of Solution

Write the expression for force between two charges.

    $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$        (1)

Here, $q_1$ and $q_2$ are the electric charges, $r$ is the distance between the charges are $z$ is the number of proton in the atom.

Write the expression for electric potential energy between two charges.

    $U=-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$        (2)

Write the expression for kinetic energy of the electron.

    $K=\frac{1}{2}\mu v^2$        (3)

Here, $\mu$ is the mass of the electron and $v$ is the velocity of the electron.

Write the expression for centripetal force between electron and the nucleus.

    $F_c=\frac{\mu v^2}{r}$        (4)

Write the expression for Bohr’s quantization of angular momentum.

    $\mu vr=nh$        (5)

Here, $n$ is the number of orbit.

Write the expression for total orbital energy.

    $E=U+K$        (6)

Conclusion:

Substitute $e$ for $q_1$, $ze$ for $q_2$ in the equation (1).

    $F=\frac{1}{4\pi {\epsilon }_0}\frac{z{e}^2}{r^2}$        (7)

The centripetal force is equal to force between two charges.

Substitute $\frac{\mu v^2}{r}$ for $F$ in the equation (7)

    $\begin{array}{c}\frac{\mu v^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{z{e}^2}{r^2}\\ \mu v^2=\frac{1}{4\pi {\epsilon }_0}\frac{z{e}^2}{r}\\ \frac{1}{2}\mu v^2=\frac{1}{8\pi {\epsilon }_0}\frac{z{e}^2}{r}\end{array}$        (8)

Substitute $\frac{nh}{\mu r}$ for $v$ in the equation (8).

    $\begin{array}{c}\frac{1}{2}\mu {\left(\frac{nh}{\mu r}\right)}^2=\frac{1}{8\pi {\epsilon }_0}\frac{z{e}^2}{r}\\ \frac{1}{2}\frac{n^2{h}^2}{\mu r^2}=\frac{1}{8\pi {\epsilon }_0}\frac{z{e}^2}{r}\\ r=\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}\\ =0.0529\frac{n^2}{Z}\end{array}$

Substitute $\frac{1}{2}\mu v^2$ for $K$ and $-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$ for $U$ in the Equation (6).

    $E=\frac{1}{2}\mu v^2-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$        (9)

Substitute $\frac{nh}{\mu r}$ for $v$, $\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}$ for $r$ in the equation (9).

    $\begin{array}{c}E=\frac{1}{2}\mu {\left(\frac{nh}{\mu r}\right)}^2-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{\left(\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}\right)}\\ =\frac{1}{2}\frac{n^2{h}^2}{\mu {\left(\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}\right)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{\left(\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}\right)}\\ =-\frac{{\mu }^2{z}^2{e}^4}{32{\pi }^2{\epsilon }_0^2{n}^2{h}^2}\end{array}$

Thus, the expression for orbital radii is $\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu z{e}^2}$ and the expression for orbital energy is $-\frac{{\mu }^2{z}^2{e}^4}{32{\pi }^2{\epsilon }_0^2{n}^2{h}^2}$.

(b)

To determine

The radius and energy of ground state orbit and ionization energy of the singly ionized helium.

Answer to Problem 5.8P

The radius of ground state orbit is $0.0265\text{nm}$ and the ionization energy of the singly ionized helium is $54.4\text{eV}$.

Explanation of Solution

Write the expression for radius of ground state orbit.

    $r=0.0529\frac{n^2}{Z}$        (10)

Write the expression for energy of ground state orbit.

    $E=-\left(13.6\text{eV}\right)\frac{Z^2}{n^2}$        (11)

Conclusion:

Substitute $2$ for $Z$ and $1$ for $n$ in the equation (10).

    $\begin{array}{c}r=0.0529\text{nm}\frac{{\left(1\right)}^2}{\left(2\right)}\\ =0.0265\text{nm}\end{array}$

Substitute $2$ for $Z$ and $1$ for $n$ in the equation (11).

    $\begin{array}{c}E=-\left(13.6\text{eV}\right)\frac{{\left(2\right)}^2}{{\left(1\right)}^2}\\ =-54.4\text{eV}\end{array}$

Thus, the radius of ground state orbit is $0.0265\text{nm}$ and the ionization energy of the singly ionized helium is $54.4\text{eV}$.

(c)

To determine

The radius and energy of ground state orbit or ionization energy of the doubly ionized lithium.

Answer to Problem 5.8P

The radius of ground state orbit is $0.0176\text{nm}$ and the ionization energy of the doubly ionized lithium is $122.4\text{eV}$.

Explanation of Solution

The equation $\left(10\right)$ and the equation $\left(11\right)$ is expression for ground state radius and the ionization energy.

Conclusion:

Substitute $3$ for $Z$ and $1$ for $n$ in the equation (10).

    $\begin{array}{c}r=0.0529\text{nm}\frac{{\left(1\right)}^2}{\left(3\right)}\\ =0.0176\text{nm}\end{array}$

Substitute $3$ for $Z$ and $1$ for $n$ in the equation (11).

    $\begin{array}{c}E=-\left(13.6\text{eV}\right)\frac{{\left(3\right)}^2}{{\left(1\right)}^2}\\ =-122.4\text{eV}\end{array}$

Thus, the radius of ground state orbit is $0.0176\text{nm}$ and the ionization energy of the doubly ionized lithium is $122.4\text{eV}$.