Chapter 42, Problem 91CP

(a)

To determine

Show that the frequency of the light emitted when electron moves from $n$ state to $n-1$ state is $f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\frac{2n-1}{n^2{\left(n-1\right)}^2}$.

Answer to Problem 91CP

It is shown that the frequency of the light emitted when electron moves from $n$ state to $n-1$ state is $\underset{\_}{f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\frac{2n-1}{n^2{\left(n-1\right)}^2}}$.

Explanation of Solution

Write the expression for the allowed energies of an atom.

    $E_n=\frac{k_e{e}^2}{2a_0}\left(\frac{1}{n^2}\right)$                                                                                             (I)

Here, $E_n$ is the possible energies, $k_e$ is the Coulomb’s constant, $e$ is the charge of electron, $a_0$ is the Bohr radius, and $n$ is the principle quantum number corresponding to each energy level.

Write the expression for Bohr radius.

    $a_0=\frac{h^2}{4{\pi }^2{m}_e{k}_e{e}^2}$                                                                                               (II)

Here, $m_e$ is the mass of electron, and $h$ is the Planck’s constant.

Use expression (II) in (I).

    $\begin{array}{c}{E}_n=\frac{k_e{e}^2}{2}\frac{4{\pi }^2{m}_e{k}_e{e}^2}{h^2}\left(\frac{1}{n^2}\right)\\ =\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{1}{n^2}\right)\end{array}$                                                                                (III)

Here the transition of electron takes place from $n$ to $n-1$ state.

Write the expression for the energy released during a transition from $n$ to $n-1$ state.

    $\Delta E=hf$                                                                                                (IV)

Here, $\Delta E$ is the energy released during a transition from $n$ to $n-1$ state, and $f$ is the frequency of the photon.

Use expression (III) to find the energy released during a transition from $n$ to $n-1$ state.

    $\Delta E=\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{1}{{\left(n-1\right)}^2}-\frac{1}{n^2}\right)$                                                                        (V)

Simplify expression (V).

    $\begin{array}{c}\Delta E=\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{1}{{\left(n-1\right)}^2}-\frac{1}{n^2}\right)\\ =\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{n^2-\left(n^2-2n+1\right)}{n^2{\left(n-1\right)}^2}\right)\\ =\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{2n-1}{n^2{\left(n-1\right)}^2}\right)\end{array}$                                                                       (VI)

Equate expression (VI) and (IV) and solve for $f$.

    $\begin{array}{c}hf=\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^2}\left(\frac{2n-1}{n^2{\left(n-1\right)}^2}\right)\\ f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\left(\frac{2n-1}{n^2{\left(n-1\right)}^2}\right)\end{array}$

Conclusion:

Therefore, it is shown that the frequency of the light emitted when electron moves from $n$ state to $n-1$ state is $\underset{\_}{f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\frac{2n-1}{n^2{\left(n-1\right)}^2}}$.

(b)

To determine

Show that when $n\to \infty$, the expression for frequency varies as $\frac{1}{n^3}$ and also that it reduces to the classical limit.

Answer to Problem 91CP

It is shown that when $n\to \infty$, the expression for frequency varies as $\frac{1}{n^3}$ and also that it reduces to the classical limit.

Explanation of Solution

Write the expression for the frequency of light emitted during a transition from $n$ to $n-1$ state.

    $f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\left(\frac{2n-1}{n^2{\left(n-1\right)}^2}\right)$                                                                   (VII)

When $n\to \infty$, expression (VII) reduces to simpler form.

    $\begin{array}{c}f=\left(\frac{2{\pi }^2{m}_e{k}_e^2{e}^4}{h^3}\right)\left(\frac{2}{n^3}\right)\\ =\frac{4{\pi }^2{m}_e{k}_e^2{e}^4}{h^3{n}^3}\end{array}$                                                                                   (VIII)

Write the expression for the classical frequency of electron.

    $f=\frac{v}{2\pi r}$                                                                                                          (IX)

Here, $v$ is the speed of electron, and $r$ is the radius of the circular path followed by electron.

Write the expression for velocity of electron.

    $v=\sqrt{\frac{k_e{e}^2}{m_{e}r}}$                                                                                                   (X)

Here, $m_e$ is the mass of electron.

Use expression (X) in (IX).

    $\begin{array}{c}f=\frac{1}{2\pi r}\sqrt{\frac{k_e{e}^2}{m_{e}r}}\\ =\sqrt{\frac{k_e{e}^2}{4{\pi }^2{m}_e{r}^3}}\end{array}$                                                                                                 (XI)

Rewrite expression (XI) in terms of the radius of the $n^{th}$ orbit of the atom.

    $f=\sqrt{\frac{k_e{e}^2}{4{\pi }^2{m}_e{r}_n^3}}$                                                                                                  (XII)

Here, $r_n$ is the radius of the $n^{th}$ orbit of the atom.

Write the expression for the radius of the $n^{th}$ orbit of the atom.

    $r_n=\frac{n^2{h}^2}{4{\pi }^2{m}_e{k}_e{e}^2}$                                                                                                     (XIII)

Here, $n$ is the principle quantum number corresponding to the $n^{th}$ orbit of the atom, and $h$ is the Planck’s constant.

Use expression (XIII) in (XII) and solve for $f$.

    $\begin{array}{c}f=\sqrt{\frac{k_e{e}^2}{4{\pi }^2{m}_e}{\left(\frac{4{\pi }^2{m}_e{k}_e{e}^2}{n^2{h}^2}\right)}^3}\\ =\sqrt{\frac{{\left(4{\pi }^2\right)}^2{m}_e^2{k}_e^4{e}^8}{n^6{h}^6}}\\ =\frac{4{\pi }^2{m}_e{k}_e^2{e}^4}{h^3{n}^3}\end{array}$                                                                           (XIV)

The classical frequency of the electron undergoing a transition from $n$ to $n-1$ state is $\frac{4{\pi }^2{m}_e{k}_e^2{e}^4}{h^3{n}^3}$.

That is it is proven that as the value of $n$ approaches larger values, the frequency tends to classical limit. Also it is proven that as $n\to \infty$, the frequency varies as $\frac{1}{n^3}$.

Conclusion:

Therefore, it is shown that when $n\to \infty$, the expression for frequency varies as $\frac{1}{n^3}$ and also that it reduces to the classical limit.