Chapter 4, Problem 36P

(a)

To determine

The frequency of revolution and the orbit radius of the electron in the Bohr model of hydrogen for $n=100,n=1000,n=10000$.

Answer to Problem 36P

The frequency of revolution in the Bohr model of hydrogen for $n=100$ is $\underset{\_}{6.62\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{6.62\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{6.62\times {10}^3\text{Hz}}$ and the orbit radius of the electron in the Bohr model of hydrogen for $n=100$ is $\underset{\_}{0.529\times {10}^{-6}\text{m}}$, $n=1000$ is $\underset{\_}{0.529\times {10}^{-4}\text{m}}$, $n=10000$ is $\underset{\_}{0.529\times {10}^{-2}\text{m}}$.

Explanation of Solution

Write the expression for the frequency of the revolution of the electron in the Bohr model of the hydrogen atom.

    $f_{\text{revolution}}=\frac{v}{2\pi r_n}$        (I)

Here, $f_{\text{revolution}}$ is the frequency of the revolution, $v$ is the velocity, $r$ is the radius of the orbit.

Write the expression for the $v$.

    $v=\frac{n\hslash }{m_e{r}_n}$        (II)

Here, $n$ is the energy level, $\hslash$ is the reduced Planck constant, $m_e$ is the mass of the electron.

Write the expression for $r_n$.

    $r_n=n^2{a}_0$        (III)

Here, $a_0$ is the Bohr radius.

Use equation (III) and (II) in (I) to solve for $f_{\text{revolution}}$.

    $f_{\text{revolution}}=\frac{\hslash }{2\pi m_e}{\left(a_0\right)}^2{n}^3$        (IV)

Conclusion:

Substitute $100$ for $n$, $1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $\hslash$, $0.0529\text{nm}$ for $a_0$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (IV) to find the frequency of the revolution for $n=100$.

    $\begin{array}{c}{f}_{\text{revolution}}=\frac{\left(1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}\right)}{2\pi \left(9.1\times {10}^{-31}\text{kg}\right)}{\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^2{\left(100\right)}^3\\ =6.62\times {10}^9\text{Hz}\end{array}$

Substitute $1000$ for $n$, $1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $\hslash$, $0.0529\text{nm}$ for $a_0$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (IV) to find the frequency of the revolution for $n=1000$.

    $\begin{array}{c}{f}_{\text{revolution}}=\frac{\left(1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}\right)}{2\pi \left(9.1\times {10}^{-31}\text{kg}\right)}{\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^2{\left(1000\right)}^3\\ =6.62\times {10}^6\text{Hz}\end{array}$

Substitute $10000$ for $n$, $1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $\hslash$, $0.0529\text{nm}$ for $a_0$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (IV) to find the frequency of the revolution for $n=10000$.

    $\begin{array}{c}{f}_{\text{revolution}}=\frac{\left(1.05\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}\right)}{2\pi \left(9.1\times {10}^{-31}\text{kg}\right)}{\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^2{\left(10000\right)}^3\\ =6.62\times {10}^3\text{Hz}\end{array}$

Substitute $100$ for $n$, $0.0529\text{nm}$ for $a_0$, in equation (III) to find the orbital radius of electron in the Bohr model hydrogen atom.

    $\begin{array}{c}{r}_{100}=\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right){\left(100\right)}^2\\ =0.529\times {10}^{-6}\text{m}\end{array}$

Substitute $1000$ for $n$, $0.0529\text{nm}$ for $a_0$, in equation (III) to find the orbital radius of electron in the Bohr model hydrogen atom.

    $\begin{array}{c}{r}_{1000}=\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right){\left(1000\right)}^2\\ =0.529\times {10}^{-4}\text{m}\end{array}$

Substitute $10000$ for $n$, $0.0529\text{nm}$ for $a_0$, in equation (III) to find the orbital radius of electron in the Bohr model hydrogen atom.

    $\begin{array}{c}{r}_{100}=\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right){\left(10000\right)}^2\\ =0.529\times {10}^{-2}\text{m}\end{array}$

Therefore, the frequency of revolution in the Bohr model of hydrogen for $n=100$ is $\underset{\_}{6.62\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{6.62\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{6.62\times {10}^3\text{Hz}}$ and the orbit radius of the electron in the Bohr model of hydrogen for $n=100$ is $\underset{\_}{0.529\times {10}^{-6}\text{m}}$, $n=1000$ is $\underset{\_}{0.529\times {10}^{-4}\text{m}}$, $n=10000$ is $\underset{\_}{0.529\times {10}^{-2}\text{m}}$.

(b)

To determine

The photon frequency for transitions from the level $n$ to $n-1$ states for the same values of $n$ as in part (a) and compare with the revolution frequencies found in part (a).

Answer to Problem 36P

The photon frequency for transitions from the level $n$ to $n-1$ states for the values of $n=100$ is $\underset{\_}{6.5798\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{6.5798\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{6.5798\times {10}^3\text{Hz}}$ and the difference in frequency for $n=100$ is $\underset{\_}{-0.05\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{-0.04\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{-0.05\times {10}^3\text{Hz}}$.

Explanation of Solution

Write the expression for the photon frequency for transitions from the $n$ to $n-1$ states.

    $f_{\text{photon}}=\frac{\Delta E}{h}$        (V)

Here, $f_{\text{photon}}$ is the photon frequency, $\Delta E$ is the energy difference between the states, $h$ is the Planck’s constant.

Write the expression for the $\Delta E$.

    $\Delta E=13.6Z^2\left[\frac{1}{{\left(n-1\right)}^2}-\frac{1}{n^2}\right]$        (VI)

Here, $Z$ is the atomic number of the atom.

Use equation (VI) in (V) to solve for $f_{\text{photon}}$.

    $f_{\text{photon}}=\frac{13.6Z^2}{h}\left[\frac{1}{{\left(n-1\right)}^2}-\frac{1}{n^2}\right]$        (VII)

Write the expression for the difference in frequency.

    $\Delta f=f_{\text{revolution}}-f_{\text{photon}}$        (VIII)

Conclusion:

Substitute $1$ for $Z$, $100$ for $n$, $6.636\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (VII) to find $f_{\text{photon}}$ for $n=100$.

    $\begin{array}{c}{f}_{\text{photon}}=\left(\frac{13.6{\left(1\right)}^2\text{eV}}{6.636\times {10}^{-34}\text{J}\cdot \text{s}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}}\right)\left[\frac{1}{{\left(100-1\right)}^2}-\frac{1}{{100}^2}\right]\\ =6.5798\times {10}^9\text{Hz}\end{array}$

Substitute $1$ for $Z$, $1000$ for $n$, $6.636\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (VII) to find $f_{\text{photon}}$ for $n=1000$.

    $\begin{array}{c}{f}_{\text{photon}}=\left(\frac{13.6{\left(1\right)}^2\text{eV}}{6.636\times {10}^{-34}\text{J}\cdot \text{s}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}}\right)\left[\frac{1}{{\left(1000-1\right)}^2}-\frac{1}{{1000}^2}\right]\\ =6.5798\times {10}^6\text{Hz}\end{array}$

Substitute $1$ for $Z$, $10000$ for $n$, $6.636\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (VII) to find $f_{\text{photon}}$ for $n=10000$.

    $\begin{array}{c}{f}_{\text{photon}}=\left(\frac{13.6{\left(1\right)}^2\text{eV}}{6.636\times {10}^{-34}\text{J}\cdot \text{s}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}}\right)\left[\frac{1}{{\left(10000-1\right)}^2}-\frac{1}{{10000}^2}\right]\\ =6.5798\times {10}^3\text{Hz}\end{array}$

Substitute $6.62\times {10}^9\text{Hz}$ for $f_{\text{revolution}}$, $6.5798\times {10}^9\text{Hz}$ for $f_{\text{photon}}$ in equation (VIII) to find $\Delta f$ for $n=100$.

    $\begin{array}{c}\Delta f=6.62\times {10}^9\text{Hz}-6.5798\times {10}^9\text{Hz}\\ =-0.05\times {10}^9\text{Hz}\end{array}$

Substitute $6.62\times {10}^6\text{Hz}$ for $f_{\text{revolution}}$, $6.5798\times {10}^6\text{Hz}$ for $f_{\text{photon}}$ in equation (VIII) to find $\Delta f$ for $n=1000$.

    $\begin{array}{c}\Delta f=6.62\times {10}^6\text{Hz}-6.5798\times {10}^6\text{Hz}\\ =-0.04\times {10}^6\text{Hz}\end{array}$

Substitute $6.62\times {10}^3\text{Hz}$ for $f_{\text{revolution}}$, $6.5798\times {10}^3\text{Hz}$ for $f_{\text{photon}}$ in equation (VIII) to find $\Delta f$ for $n=100$.

    $\begin{array}{c}\Delta f=6.62\times {10}^3\text{Hz}-6.5798\times {10}^3\text{Hz}\\ =-0.05\times {10}^3\text{Hz}\end{array}$

Therefore, the photon frequency for transitions from the level $n$ to $n-1$ states for the values of $n=100$ is $\underset{\_}{6.5798\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{6.5798\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{6.5798\times {10}^3\text{Hz}}$ and the difference in frequency for $n=100$ is $\underset{\_}{-0.05\times {10}^9\text{Hz}}$, $n=1000$ is $\underset{\_}{-0.04\times {10}^6\text{Hz}}$, $n=10000$ is $\underset{\_}{-0.05\times {10}^3\text{Hz}}$.

(c)

To determine

Verify the correspondence principle in the transitions.

Answer to Problem 36P

Verified that $f_{\text{photon}}$ tends to $f_{\text{revolution}}$ in the limit of large quantum numbers.

Explanation of Solution

The correspondence principle states that “predictions of quantum theory must correspond to the predictions of classical physics in the region of sizes where classical theory is known to hold”.

If the quantum number becomes large because of increased size or mass, we may state the correspondence principle symbolically as

    $\underset{n\to \infty }{\lim }\left[\text{quantum physics}\right]=\left[\text{classical physics}\right]$        (IX)

As the quantum physics tends to classical physics in the limit of large quantum numbers implies the $f_{\text{photon}}$ tends to $f_{\text{revolution}}$ in the limit of large quantum numbers.

The value of equation (IV) and (VI) tends to almost equal to same value as the value of $n$ becomes large. These results show that $f_{\text{photon}}$ tends to $f_{\text{revolution}}$ in the limit of large quantum numbers $\left(n=10000\right)$ and macroscopic sizes $\left(r~\frac{1}{2}\text{ cm}\right)$.

Conclusion:

Therefore, it is verified that $f_{\text{photon}}$ tends to $f_{\text{revolution}}$ in the limit of large quantum numbers.