Chapter 4, Problem 31P

(a)

To determine

To show that the frequency of the electron’s orbital motion is $f_e=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\left(\frac{1}{n^3}\right)$ .

Answer to Problem 31P

It is showed that the frequency of the electron’s orbital motion is $f_e=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\left(\frac{1}{n^3}\right)$ .

Explanation of Solution

Write the expression for the conservation of angular momentum of the electron.

  $m_{e}vr=n\hslash$

Here, $m_e$ is the mass of the electron, $v$ is the orbital speed of the electron, $r$ is the radius of the orbit, $n$ is quantum number of the orbit and $\hslash$ is the reduced Planck’s constant.

Rewrite the above equation for $v$ .

  $v=\frac{n\hslash }{m_{e}r}$        (I)

Write the equation for the frequency of the electron.

  $f_e=\frac{v}{2\pi r}$

Here, $f_e$ is the frequency of the electron’s orbital motion.

Put equation (I) in the above equation.

  $\begin{array}{c}{f}_e=\frac{\frac{n\hslash }{m_{e}r}}{2\pi r}\\ =\frac{n\hslash }{2\pi m_e{r}^2}\end{array}$        (II)

Write the equation for the radius of an orbit.

  $r=a_0{n}^2$

Here, $a_0$ is the Bohr radius.

Put the above equation in equation (II).

  $\begin{array}{c}{f}_e=\frac{n\hslash }{2\pi m_e{\left(a_0{n}^2\right)}^2}\\ =\frac{\hslash }{2\pi m_e{a}_0^2{n}^3}\end{array}$        (III)

Write the equation for the Bohr radius.

  $a_0=\frac{{\hslash }^2}{m_{e}Zk{e}^2}$        (IV)

Here, $Z$ is the atomic number, $k$ is the Coulomb constant and $e$ is the magnitude of the charge of the electron.

Put the above equation in equation (III).

  $\begin{array}{c}{f}_e=\frac{\hslash }{2\pi m_e{\left(\frac{{\hslash }^2}{m_{e}Zk{e}^2}\right)}^2{n}^3}\\ =\frac{\hslash m_e^2{Z}^2{k}^2{e}^4}{2\pi m_e{\hslash }^4{n}^3}\\ =\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\left(\frac{1}{n^3}\right)\end{array}$

Conclusion:

Therefore, it is showed that the frequency of the electron’s orbital motion is $f_e=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\left(\frac{1}{n^3}\right)$

(b)

To determine

To show that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $f_e=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{n_i+n_f}{2n_i^2{n}_f^2}\left(n_i-n_f\right)$ and $\frac{1}{n_i^3}<\frac{n_i+n_f}{2n_i^2{n}_f^2}<\frac{1}{n_f^3}$ for electronic transitions between adjacent orbits.

Answer to Problem 31P

It is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $f_{\text{photon}}=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{n_i+n_f}{2n_i^2{n}_f^2}\left(n_i-n_f\right)$ and $\frac{1}{n_i^3}<\frac{n_i+n_f}{2n_i^2{n}_f^2}<\frac{1}{n_f^3}$ for electronic transitions between adjacent orbits.

Explanation of Solution

Write the equation for the energy of the photon emitted.

  $\Delta E=h{f}_{\text{photon}}$

Here, $\Delta E$ is the energy of the photon emitted, $h$ is the Planck’s constant and $f_{\text{photon}}$ is the frequency of the photon.

Write the equation for the energy of the photon emitted when an electron jumps from an outer to inner orbit.

  $\Delta E=h\left(\frac{k{Z}^2{e}^2}{2a_{0}h}\right)\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Here, $n_i$ is the quantum number of the initial level and $n_f$ is the quantum number of the final level.

Compare the above two equations to write the expression for $f_{\text{photon}}$ .

  $\begin{array}{c}{f}_{\text{photon}}=\left(\frac{k{Z}^2{e}^2}{2a_{0}h}\right)\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =\left(\frac{k{Z}^2{e}^2}{2a_{0}h}\right)\left(\frac{n_i^2-n_f^2}{n_i^2{n}_f^2}\right)\\ =\left(\frac{k{Z}^2{e}^2}{2a_{0}h}\right)\frac{\left(n_i+n_f\right)\left(n_i-n_f\right)}{n_i^2{n}_f^2}\end{array}$

Put the equation for Bohr radius in the above equation.

  $\begin{array}{c}{f}_{\text{photon}}=\left(\frac{k{Z}^2{e}^2}{2\left(\frac{{\hslash }^2}{m_{e}k{e}^2}\right)h}\right)\frac{\left(n_i+n_f\right)\left(n_i-n_f\right)}{n_i^2{n}_f^2}\\ =\frac{m_e{k}^2{Z}^2{e}^4}{2{\hslash }^{2}h}\frac{\left(n_i+n_f\right)\left(n_i-n_f\right)}{n_i^2{n}_f^2}\\ =\frac{m_e{k}^2{Z}^2{e}^4}{2{\hslash }^2\left(\frac{h}{2\pi }\right)\left(2\pi \right)}\frac{\left(n_i+n_f\right)\left(n_i-n_f\right)}{n_i^2{n}_f^2}\\ =\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{\left(n_i+n_f\right)\left(n_i-n_f\right)}{2n_i^2{n}_f^2}\end{array}$        (V)

Write the relationship between $n_i$ and $n_f$ for transitions between adjacent orbits.

  $n_i-n_f=1$

Put the above equation in equation (V).

  $\begin{array}{c}{f}_{\text{photon}}=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{\left(n_i+n_f\right)\left(1\right)}{2n_i^2{n}_f^2}\\ =\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{\left(n_i+n_f\right)}{2n_i^2{n}_f^2}\end{array}$        (VI)

Conclusion:

For $n_i=2,n_f=1,n_i=3,n_f=2,\text{ etc}\text{.,}$ using equation (VI) gives that $\frac{1}{n_i^3}<\frac{n_i+n_f}{2n_i^2{n}_f^2}<\frac{1}{n_f^3}$ .

Therefore, it is showed that frequency of the photon emitted when an electron jumps from an outer to inner orbit can be written as $f_{\text{photon}}=\frac{m_e{k}^2{Z}^2{e}^4}{2\pi {\hslash }^3}\frac{n_i+n_f}{2n_i^2{n}_f^2}\left(n_i-n_f\right)$ and $\frac{1}{n_i^3}<\frac{n_i+n_f}{2n_i^2{n}_f^2}<\frac{1}{n_f^3}$ for electronic transitions between adjacent orbits.

(c)

To determine

The conclusion regarding the frequency of the emitted radiation compared with the frequencies of orbital revolution in the initial and final states and the result as $n_i\to \infty$ .

Answer to Problem 31P

Frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $n_i\to \infty$ .

Explanation of Solution

The expression for the orbital frequency of the electron is found in part (a). The equation for the frequency of the emitted photon is found in part (b). Comparison of the two equations shows that the frequency of the emitted radiation is in between the initial orbital frequency and the final orbital frequency.

As $n_i\to \infty$ , the initial orbital frequency and the final orbital frequency squeeze closer together. This will make the frequency of the emitted radiation equal to the orbital frequency. This result is in agreement with Bohr’s correspondence principle.

Conclusion:

Thus, the frequency of the emitted radiation is in between the initial and the final orbital frequency and the frequency of the emitted radiation becomes the orbital frequency as $n_i\to \infty$ .