Chapter 27, Problem 45E

(a)

To determine

The frequency emitted by hydrogen atom.

Answer to Problem 45E

The frequency emitted is $6.17\times {10}^{14}\text{Hz}$.

Explanation of Solution

Write the expression for wavelength of light.

    $\lambda =\frac{c}{\nu }$

Here, $\lambda$ is the wavelength, $c$ is the speed of light and $\nu$ is the frequency.

Rearrange the above expression for $\nu$.

    $\nu =\frac{c}{\lambda }$        (I)

Conclusion:

Substitute $3\times {10}^8{\text{ms}}^{-1}$ for $c$ and $486\text{nm}$ for $\lambda$ in equation (I).

    $\begin{array}{c}\nu =\frac{3\times {10}^8{\text{ms}}^{-1}}{486\text{nm}}\\ =\frac{3\times {10}^8{\text{ms}}^{-1}}{486\text{nm}}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\left(\frac{1\text{Hz}}{\text{1}{\text{s}}^{-1}}\right)\\ =6.17\times {10}^{14}\text{Hz}\end{array}$

Thus, the frequency of emitted light is $6.17\times {10}^{14}\text{Hz}$.

(b)

To determine

The energy lost due to emission of light.

Answer to Problem 45E

The energy lost is $2.32\text{eV}$.

Explanation of Solution

Write the expression for the energy.

    $E=h\nu$        (II)

Here, $E$ is the energy, $h$ is the Planck’s constant and $\nu$ is the frequency.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$ and $6.17\times {10}^{14}{\text{s}}^{-1}$ for $\nu$ in equation (II).

    $\begin{array}{c}E=\left(6.626\times {10}^{-34}\text{Js}\right)\left(6.17\times {10}^{14}{\text{s}}^{-1}\right)\\ =37.155\times {10}^{-20}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =2.32\text{eV}\end{array}$

Thus, the energy lost is $2.32\text{eV}$.

(c)

To determine

The final and initial level of atom.

Answer to Problem 45E

The final and initial levels are $2$ and $4$.

Explanation of Solution

Write the expression for energy.

    $E=\frac{Z^2{E}_0}{n_f^2}$                                                                                            

Here, $E$ is the energy lost, $Z$ is the atomic number, $E_0$ is the ground state energy of atom and $n_f$ is the number of final level.

Rearrange the above expression for $n_f^2$.

    $n_f^2=\frac{Z^2{E}_0}{E}$        (III)

Write the expression for the wavelength.

    $\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$                                                                                                  

Here, $\lambda$ is the wavelength, $R$ is the Rydberg’s constant, $n_f$ is the final level and $n_i$ is the initial level.

Rearrange the above expression for $n_i^2$.

    $\frac{1}{\lambda R}=\frac{1}{n_f^2}-\frac{1}{n_i^2}$

    $\frac{1}{n_i^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R}$        (IV)

Conclusion:

Substitute $2.32\text{eV}$ for $E$, $1$ for $Z$, $13.2\text{eV}$ for $E_0$ in equation (III).

  $\begin{array}{c}{n}_f^2=\frac{{\left(1\right)}^2\left(13.2\text{eV}\right)}{2.32\text{eV}}\\ =5.32\end{array}$

Solve the equation for $n_f$.

  $\begin{array}{c}{n}_f^2=5.32\\ n_f=2\end{array}$

Substitute $2$ for $n_f$, $486\text{nm}$ for $\lambda$ and $1.09\times {10}^7{\text{m}}^{-1}$ for $R$ in equation (IV).

    $\begin{array}{c}\frac{1}{n_i^2}=\frac{1}{{\left(2\right)}^2}-\frac{1}{\left(486\text{nm}\right)\left(1.09\times {10}^7\text{m}\right)}\\ =\frac{1}{{\left(2\right)}^2}-\frac{1}{\left(486\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)\left(1.09\times {10}^7{\text{m}}^{-1}\right)}\\ =\frac{1}{4}-\frac{1}{5.29}\\ =0.061\end{array}$

Solve the above expression for $n_i$.

    $\begin{array}{c}{n}_i^2=\frac{1}{0.061}\\ n_i^2=16.3\\ n_i=4\end{array}$

Thus, the final and initial levels are $2$ and $4$.