Chapter 4, Problem 54P

(a)

To determine

Calculate $f_{\text{orb}}$ in the ground state, by using $f_{\text{classical}}$ in Equation (4.34)

Answer to Problem 54P

The orbital frequency is $6.54\times {10}^{15}\text{Hz}$.

Explanation of Solution

Write the expresion for $f_{\text{classical}}$ to calculate  $f_{\text{orb}}$ in the ground state.

    $f_{\text{orb}}=\frac{m{e}^4}{4{\epsilon }_0^2{h}^3}$        (I)

Here, $f_{\text{orb}}$ is the orbital frequency of electron around the nucleus, $m$ is the mass of the electron, $e$ is the charge in the electron, $h$ is the Planck’s constant and ${\epsilon }_0$  the permittivity of free space.

Conclusion:

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m$ , $1.60\times {10}^{-19}\text{C}$ for $e$ , $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ and $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (I).

    $\begin{array}{c}{f}_{\text{orb}}=\frac{\left(9.11\times {10}^{-31}\text{kg}\right){\left(1.60\times {10}^{-19}\text{C}\right)}^4}{4{\left(8.85\times {10}^{-12}\text{F/m}\right)}^2{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)}^3}\\ =6.54\times {10}^{15}\text{Hz}\end{array}$

Thus, the orbital frequency is $6.54\times {10}^{15}\text{Hz}$.

(b)

To determine

Calculate $f_{\text{orb}}$ in the ground state. By using equation (4.33a), but first calculate $v$ and $r$ .

Answer to Problem 54P

The frequency $f_{\text{orb}}$ in the ground state is $6.59\times {10}^{15}\text{Hz}$.

Explanation of Solution

If an electron rotates around the nucleus then the expression for speed is given below:

    $v=\frac{e}{\sqrt{4\pi {\epsilon }_{0}mr}}$

Here, $v$ is the speed of an electron, $e$ is the charge on electron, $r$ is the radius and $m$ is the mass of electron.

Multiply by $c$ in both numerator and denominator.

    $v=\frac{ec}{\sqrt{4\pi {\epsilon }_0}\sqrt{m{c}^{2}r}}$        (II)

Value of $m{c}^2$ in electron Volt is:

    $\begin{array}{c}m{c}^2=\frac{9.1\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2}{1.6\times {10}^{-19}}\text{eV}\\ =\text{511875}\text{eV}\end{array}$

Write the expression for $a_0$.

    $a_0=\frac{4\pi {\epsilon }_0{\sout{h}}^2}{m{e}^2}$        (III)

Here, $a_0$ is the Bohr’s radius,  and $\left(\sout{h}\right)$ is the $h$ - bar constant .

Write the expression for frequency.

    $f=\frac{v}{2\pi r}$        (IV)

Here, $v$ is the velocity and $r$ is the radius of the orbit

Conclusion:

Substitute $\text{511875}\text{eV}$ for $m{c}^2$, $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ , $1.6\times {10}^{-19}$ for $e$  and $5.29\times {10}^{-11}\text{m}$ for $r$ in equation (II).

    $\begin{array}{c}v=\frac{\sqrt{\left(9\times {10}^9\times 1.6\times {10}^{-19}\text{eV}\text{.m}\right)}c}{\sqrt{\left(\text{511875}\text{eV}\right)\left(5.29\times {10}^{-11}\text{m}\right)}}\\ =\frac{\sqrt{1.44\text{eV}\cdot \text{nm}}\left(3\times {10}^8\right)}{\sqrt{\left(\text{511875}\text{eV}\right)\left(0.0529\text{nm}\right)}}\\ =2.19\times {10}^6\text{m/s}\end{array}$

Substitute $9.10938356\times {10}^{-31}\text{kg}$ for $m$, $1.055\times {10}^{-34}\text{kg}\cdot {\text{m/s}}^2$ for $\left(\sout{h}\right)$, $1.6021733\times {10}^{-19}\text{C}$ for $e$ and $8.8541878\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ in equation (III).   

$\begin{array}{c}{a}_0=\frac{4\pi \left(8.8541878\times {10}^{-12}\text{F/m}\right){\left(1.055\times {10}^{-34}\text{kg}\cdot {\text{m/s}}^2\right)}^2}{\left(9.10938356\times {10}^{-31}\text{kg}\right){\left(1.6021733\times {10}^{-19}\text{C}\right)}^2}\\ =5.2918\times {10}^{-11}\text{m}\end{array}$

Substitute $2.19\times {10}^6\text{m/s}$ for $v$ and  $5.2918\times {10}^{-11}\text{m}$ for $r$ in equation (IV).

    $\begin{array}{c}f=\frac{2.19\times {10}^6\text{m/s}}{2\pi \left(5.2918\times {10}^{-11}\text{m}\right)}\\ =6.59\times {10}^{15}\text{Hz}\end{array}$

Thus, the frequency $f_{\text{orb}}$ in the ground state is $6.59\times {10}^{15}\text{Hz}$.

(c)

To determine

Show that the mean value $K$ is equal to the absolute value of the electron-nucleus system total energy and that this is $13.6\text{ eV}$ . Use this value of $K$  to determine $f_{\text{orb}}$  from the relation for $K$.

Answer to Problem 54P

The mean value $K$ is equal to the absolute value of the electron-nucleus system total energy and that this is $13.6\text{ eV}$ proved below and value of $f_{\text{orb}}$ from the relation for $K$ is $6.58\times {10}^{15}\text{Hz}$.

Explanation of Solution

Write the expression for $E$.

    $E=-\frac{e^2}{8\pi {\epsilon }_0{a}_0}$

Here, $E$ is the Energy.

Substitute $5.29\times {10}^{-11}\text{m}$ for $a_0$, $1.60\times {10}^{-19}\text{C}$ for $e$ and $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ in above expression.

$\begin{array}{c}{E}_0=\frac{{\left(1.60\times {10}^{-19}\text{C}\right)}^2}{8\pi \left(8.85\times {10}^{-12}\text{F/m}\right)\left(5.29\times {10}^{-11}\text{m}\right)}\\ =2.179\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{\left(1.60\times {10}^{-19}\text{J}\right)}\right)\\ =\text{13}\text{.60 eV}\end{array}$

Write the expression for kinetic energy.

    $\begin{array}{c}K=\frac{e^2}{8\pi {\epsilon }_0{a}_0}\\ =\left|E\right|\end{array}$

Here, $K$ is the kinetic energy.

From above equations it can be concluded that mean value $K$ is equal to the absolute value of the electron-nucleus system total energy and that this is $13.6\text{ eV}$ .

Write the expression for kinetic energy.

    $K=\frac{nh{f}_{\text{orb}}}{2}$

Rearrange above equation for $f_{\text{orb}}$ and for $n=1$ .

    $f_{\text{orb}}=\frac{2K}{h}$        (V)

Conclusion:

Substitute $13.6\text{ eV}$ for $K$ and $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (V).

    $\begin{array}{c}{f}_{\text{orb}}=\frac{2\left(13.6\text{ eV}\right)}{6.63\times {10}^{-34}\text{J}\cdot \text{s}\left(\frac{1\text{eV}}{1.60\times {10}^{-19}\text{J}}\right)}\\ =6.58\times {10}^{15}\text{Hz}\end{array}$

Thus, the mean value $K$ is equal to the absolute value of the electron-nucleus system total energy and that this is $13.6\text{ eV}$ proved below and value of $f_{\text{orb}}$ from the relation for $K$ is $6.58\times {10}^{15}\text{Hz}$.