Chapter 42, Problem 80AP

(a)

To determine

Energy of electron in ground state and the first four excited state.

Answer to Problem 80AP

Energy of the electron in ground state is $\underset{\_}{-8.16\text{eV}}$, electron in first excited state is $\underset{\_}{-2.04\text{eV}}$, electron in second excited state is $\underset{\_}{-0.902\text{eV}}$, electron in third excited state is $\underset{\_}{-0.508\text{eV}}$, and electron in fourth excited state is $\underset{\_}{-0.325\text{eV}}$.

Explanation of Solution

Write the expression for the ground state energy of atom.

    $E_1=-\frac{hc}{{\lambda }_{\text{series limit}}}$                                                                                                (I)

Here, $E_1$ is the ground state energy of atom, $h$ is Planck’s constant, $c$ is the speed of light, and ${\lambda }_{\text{series limit}}$ is the wavelength.

Write the expression for difference in energy between first and ground state energy level.

    $\Delta E_{21}=\frac{hc}{{\lambda }_1}$                                                                                                        (II)

Here, $\Delta E_{21}$ is the energy difference between first and ground state energy level., and ${\lambda }_1$ is the wavelength of the light emitted during transition from first to ground state.

Write the expression to find $E_2$.

    $E_2=E_1+\Delta E_{21}$                                                                                                        (III)

Here, $E_2$ is the energy of first excited state.

Write the expression for the difference in energy between second excited level and ground state.

    $\Delta E_{31}=\frac{hc}{{\lambda }_2}$                                                                                                          (IV)

Here, $\Delta E_{31}$ is the energy difference between second and ground state energy level., and ${\lambda }_2$ is the wavelength of the light emitted during transition from second to ground state.

Write the expression to find energy of second excited level.

    $E_3=E_2+\Delta E_{31}$                                                                                                      (V)

Here, $E_3$ is the energy of second excited state.

Write the expression for the difference in energy between third excited level and ground state.

    $\Delta E_{41}=\frac{hc}{{\lambda }_3}$                                                                                                          (VI)

Here, $\Delta E_{41}$ is the energy difference between third and ground state energy level., and ${\lambda }_3$ is the wavelength of the light emitted during transition from third to ground state.

Write the expression to find energy of third excited level.

    $E_4=E_3+\Delta E_{41}$                                                                                                      (VII)

Here, $E_4$ is the energy of third excited state.

Write the expression for the difference in energy between fourth excited level and ground state.

    $\Delta E_{51}=\frac{hc}{{\lambda }_4}$                                                                                                        (VIII)

Here, $\Delta E_{51}$ is the energy difference between fourth and ground state energy level., and ${\lambda }_4$ is the wavelength of the light emitted during transition from fourth to ground state.

Write the expression to find energy of fourth excited level.

    $E_5=E_4+\Delta E_{51}$                                                                                                        (IX)

Here, $E_5$ is the energy of fourth excited state.

Conclusion:

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, and $152.0\text{nm}$ for ${\lambda }_{\text{series limit}}$ in expression (I) to find $E_1$.

    $\begin{array}{c}{E}_1=-\frac{1240\text{eV}\cdot \text{nm}}{152.0\text{nm}}\\ =-8.16\text{eV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, and $202.6\text{nm}$ for ${\lambda }_1$ in equation (II) to find $\Delta E_{21}$.

    $\begin{array}{c}\Delta E_{21}=\frac{1240\text{eV}\cdot \text{nm}}{\left(202.6\text{nm}\right)}\\ =6.12\text{eV}\end{array}$

Substitute $-8.16\text{eV}$ for $E_1$, and $6.12\text{eV}$ for $\Delta E_{21}$ in equation (III) to find $E_2$.

    $\begin{array}{c}{E}_2=-8.16\text{eV}+6.12\text{eV}\\ \text{=}-2.04\text{eV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, and $170.9\text{nm}$ for ${\lambda }_2$ in equation (IV) to find $\Delta E_{31}$.

    $\begin{array}{c}\Delta E_{31}=\frac{1240\text{eV}\cdot \text{nm}}{\left(170.9\text{nm}\right)}\\ =7.26\text{eV}\end{array}$

Substitute $-8.16\text{eV}$ for $E_1$, and $7.26\text{eV}$ for $\Delta E_{31}$ equation (V)to find $E_3$.

    $\begin{array}{c}{E}_3=-8.16\text{eV}+7.26\text{eV}\\ \text{=}-0.902\text{eV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, and $162.1\text{nm}$ for ${\lambda }_3$ in equation (VI) to find $\Delta E_{41}$.

    $\begin{array}{c}\Delta E_{41}=\frac{1240\text{eV}\cdot \text{nm}}{\left(162.1\text{nm}\right)}\\ =7.65\text{eV}\end{array}$

Substitute $-8.16\text{eV}$ for $E_1$, and $7.65\text{eV}$ for $\Delta E_{41}$ equation (VII)to find $E_4$.

    $\begin{array}{c}{E}_4=-8.16\text{eV}+7.65\text{eV}\\ \text{=}-0.508\text{eV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, and $158.3\text{nm}$ for ${\lambda }_4$ in equation (VIII) to find $\Delta E_{51}$.

    $\begin{array}{c}\Delta E_{51}=\frac{1240\text{eV}\cdot \text{nm}}{\left(158.3\text{nm}\right)}\\ =7.83\text{eV}\end{array}$

Substitute $-8.16\text{eV}$ for $E_1$, and $7.83\text{eV}$ for $\Delta E_{51}$ equation (IX)to find $E_5$.

    $\begin{array}{c}{E}_5=-8.16\text{eV}+7.83\text{eV}\\ \text{=}-0.325\text{eV}\end{array}$

Therefore, the energy of the electron in ground state is $\underset{\_}{-8.16\text{eV}}$, electron in first excited state is $\underset{\_}{-2.04\text{eV}}$, electron in second excited state is $\underset{\_}{-0.902\text{eV}}$, electron in third excited state is $\underset{\_}{-0.508\text{eV}}$, and electron in fourth excited state is $\underset{\_}{-0.325\text{eV}}$.

(b)

To determine

The wavelength of the first three lines and the short wavelength limit in the Balmer series.

Answer to Problem 80AP

The wavelength of the $\alpha$ line is $\underset{\_}{1090\text{nm}}$, $\beta$ line is $\underset{\_}{809\text{nm}}$, $\gamma$ line is $\underset{\_}{724\text{nm}}$, and the short wavelength limit is $\underset{\_}{609\text{nm}}$.

Explanation of Solution

In Balmer series transition takes place between $i^{th}$ level and second level.

Write the expression for the difference in energy between the energy levels for Balmer series.

    $E_i-E_2=\frac{hc}{\lambda }$                                                                                                   (X)

Here, $E_i$ is the energy corresponding to the $i^{th}$ energy level.

Rearrange expression (X) to find $\lambda$.

    $\lambda =\frac{hc}{E_i-E_2}$                                                                                                           (XI)

Put $i=3$ in equation (XI) to find expression for wavelength of $\alpha$ line.

    ${\lambda }_3=\frac{hc}{E_3-E_2}$                                                                                                      (XII)

Put $i=4$ in equation (XI) to find expression for wavelength of $\beta$ line.

    ${\lambda }_4=\frac{hc}{E_4-E_2}$                                                                                                      (XIII)

Put $i=5$ in equation (XI) to find expression for wavelength of $\gamma$ line.

    ${\lambda }_5=\frac{hc}{E_5-E_2}$                                                                                                      (XIV)

Put $i=\infty$ in equation (XI) to find expression for wavelength of shortest wave.

    ${\lambda }_{\infty }=\frac{hc}{E_{\infty }-E_2}$                                                                                                      (XV)

Conclusion:

Substitute $-0.0902\text{eV}$ for $E_3$, $1240\text{nm}\cdot \text{eV}$ for $hc$, and $-2.04\text{eV}$ for $E_2$ in expression (XII) to find ${\lambda }_3$.

    $\begin{array}{c}{\lambda }_3=\frac{1240\text{nm}\cdot \text{eV}}{\left(-0.902\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =1090\text{nm}\end{array}$

Substitute $-0.508\text{eV}$ for $E_4$, $1240\text{nm}\cdot \text{eV}$ for $hc$, and $-2.04\text{eV}$ for $E_2$ in expression (XIII) to find ${\lambda }_4$.

    $\begin{array}{c}{\lambda }_4=\frac{1240\text{nm}\cdot \text{eV}}{\left(-0.508\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =809\text{nm}\end{array}$

Substitute $-0.325\text{eV}$ for $E_5$, $1240\text{nm}\cdot \text{eV}$ for $hc$, and $-2.04\text{eV}$ for $E_2$ in expression (XIV) to find ${\lambda }_5$.

    $\begin{array}{c}{\lambda }_5=\frac{1240\text{nm}\cdot \text{eV}}{\left(-0.325\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =724\text{nm}\end{array}$

Substitute $\infty$ for $E_{\infty }$, $1240\text{nm}\cdot \text{eV}$ for $hc$, and $-2.04\text{eV}$ for $E_2$ in expression (XV) to find ${\lambda }_{\infty }$.

    $\begin{array}{c}{\lambda }_{\infty }=\frac{1240\text{nm}\cdot \text{eV}}{\left(\infty \right)-\left(-2.04\text{eV}\right)}\\ =609\text{nm}\end{array}$

Therefore, The wavelength of the $\alpha$ line is $\underset{\_}{1090\text{nm}}$, $\beta$ line is $\underset{\_}{809\text{nm}}$, that of $\gamma$ line is $\underset{\_}{724\text{nm}}$, and that of short wavelength is $\underset{\_}{609\text{nm}}$.

(c)

To determine

Show that the first four lines and the shortest wavelengths of Lyman series is equal to $60\%$ of the wavelength for Lyman series of a one electron system.

Answer to Problem 80AP

It is shown that the first four lines and the shortest wavelengths of Lyman series is equal to $60\%$ of the wavelength for Lyman series of a one electron system.

Explanation of Solution

Write the expression for the wavelength of Lyman series.

    $\frac{1}{\lambda }=R_H\left(1-\frac{1}{n^2}\right)$                                                                                          (XVI)

Here, $R_H$ is the Rydberg constant, and $n$ is the principle quantum number.

Rearrange expression (XVI) to find $\lambda$.

    $\lambda =\frac{1}{R_H\left(1-\frac{1}{n^2}\right)}$                                                                                             (XVII)

Conclusion:

For $\alpha$ series $n=2$. Put $2$ for $n$, $10973731.6{\text{m}}^{-1}$ for $R_H$ in expression (XVI) to find wavelength of $\alpha$ line.

    $\begin{array}{c}{\lambda }_{\alpha }=\frac{1}{10973731.6{\text{m}}^{-1}\left(1-\frac{1}{2^2}\right)}\\ =122\text{nm}\end{array}$

Put $3$ for $n$, $10973731.6{\text{m}}^{-1}$ for $R_H$ in expression (XVII) to find wavelength of $\beta$ line.

    $\begin{array}{c}{\lambda }_{\beta }=\frac{1}{10973731.6{\text{m}}^{-1}\left(1-\frac{1}{3^2}\right)}\\ =103\text{nm}\end{array}$

Put $4$ for $n$, $10973731.6{\text{m}}^{-1}$ for $R_H$ in expression (XVII) to find wavelength of $\gamma$ line.

    $\begin{array}{c}{\lambda }_{\gamma }=\frac{1}{10973731.6{\text{m}}^{-1}\left(1-\frac{1}{4^2}\right)}\\ =97.2\text{nm}\end{array}$

Put $5$ for $n$, $10973731.6{\text{m}}^{-1}$ for $R_H$ in expression (XVII) to find wavelength of $\delta$ line.

    $\begin{array}{c}{\lambda }_{\delta }=\frac{1}{10973731.6{\text{m}}^{-1}\left(1-\frac{1}{5^2}\right)}\\ =94.9\text{nm}\end{array}$

Put $\infty$ for $n$, $10973731.6{\text{m}}^{-1}$ for $R_H$ in expression (XVII) to find wavelength of short wavelength.

    $\begin{array}{c}{\lambda }_{\infty }=\frac{1}{10973731.6{\text{m}}^{-1}\left(1-\frac{1}{{\infty }^2}\right)}\\ =91.1\text{nm}\end{array}$

$60\%$ of the first spectral line of wavelength $202.6\text{nm}$ is $0.600\left(202.6\text{nm}\right)=122\text{nm}$.

$60\%$ of the second spectral line of wavelength $170.9\text{nm}$ is $0.600\left(170.9\text{nm}\right)=103\text{nm}$.

$60\%$ of the third spectral line of wavelength $162.1\text{nm}$ is $0.600\left(162.1\text{nm}\right)=97.3\text{nm}$.

$60\%$ of the fourth spectral line of wavelength $158.3\text{nm}$ is $0.600\left(158.3\text{nm}\right)=95.0\text{nm}$.

$60\%$ of the fifth spectral line of wavelength $2152.0\text{nm}$ is $0.600\left(150.0\text{nm}\right)=91.2\text{nm}$.

Comparing the $60\%$ value of wavelengths of spectral lines of the energy diagram in question, it is seen that the values are equal to the wavelength of $\alpha ,\beta ,\gamma ,\delta$, and the short wavelength.

Therefore, it is shown that the first four lines and the shortest wavelengths of Lyman series of Hydrogen atom is equal to $60\%$ of the wavelength for Lyman series of a one electron system.

(d)

To determine

The reason why the atom can be assumed to be Hydrogen.

Answer to Problem 80AP

The atom can be Hydrogen since the results of Doppler shifts is equal to the actual value of wavelength of Hydrogen spectrum.

Explanation of Solution

The wavelength of the spectral lines were measured by the Doppler shifting in the frequency and speed of motion of atom with respect to the Earth.

Write the expression for the ratio of original wavelength and the shifted wavelength of the atom.

    $\frac{\lambda }{{\lambda }^{\prime }}=\sqrt{\frac{c-v}{c+v}}$ (XVIII)

The ratio between the old wavelengths and Doppler shifted wavelength is equal to $0.600$.

    $\frac{\lambda }{{\lambda }^{\prime }}=\sqrt{\frac{c-v}{c+v}}=0.600$                                                                                  (XIX)

Square equation (XIX) and solve for $v$.

    $\begin{array}{c}\frac{c-v}{c+v}={\left(0.600\right)}^2\\ c-v=0.36c+0.36v\\ 0.64c=1.36v\\ v=0.471c\end{array}$

Thus if we multiply the shifted wavelength by $0.600$ we will get wavelengths that correspond to the wavelength of hydrogen spectrum. Here the source is moving with a speed of $0.471c$.

Conclusion:

Therefore, the atom can be Hydrogen since the results of Doppler shifts is equal to the actual value of wavelength of Hydrogen spectrum.