Chapter 27, Problem 64P

(a)

To determine

The wavelengths can absorbed by the $H$ atom in between the wavelength ranging $96\text{nm}$ to $110\text{nm}$.

Answer to Problem 64P

The wavelengths can absorbed by the $H$ atom in between the wavelength ranging $96\text{nm}$ to $110\text{nm}$ are $\underset{\_}{97.3\text{nm and }103\text{nm}}$.

Explanation of Solution

Write the expression for the absorbed energy associated from the transition between the energy levels.

    $E=\left|E_1\right|-\left|E_n\right|$        (I)

Here, $E$ is the absorbed energy associated from the transition between the energy levels, $E_1$ is the ground state energy, $E_n$ is the energy of the $n$ level.

Write the expression for the $E_n$.

    $E_n=\frac{E_1}{n^2}$        (II)

Here, $n$ is the $n^{\text{th}}$ energy level.

Write the expression for wavelength.

    $\lambda =\frac{hc}{E}$        (III)

Here, $\lambda$ is the wavelength, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum.

Use equation (II) and (I) in (III) to solve for absorbed wavelength associated from the transition between the energy levels.

    $\begin{array}{c}\lambda =\frac{hc}{\left|E_1\right|-\frac{\left|E_1\right|}{n^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{\left|E_1\right|-\frac{\left|E_1\right|}{n^2}}\end{array}$        (IV)

Conclusion:

Substitute $-13.6\text{eV}$ for $E_1$, $2$ for $n$ in equation (IV) to find the absorbed wavelength associated from the transition between the energy levels form $n=2$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\left|-13.6\text{eV}\right|-\frac{\left|-13.6\text{eV}\right|}{2^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{10.2\text{eV}}\\ =122\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $3$ for $n$ in equation (IV) to find the absorbed wavelength associated from the transition between the energy levels form $n=3$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\left|-13.6\text{eV}\right|-\frac{\left|-13.6\text{eV}\right|}{3^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{12.09\text{eV}}\\ =103\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $4$ for $n$ in equation (IV) to find the absorbed wavelength associated from the transition between the energy levels form $n=4$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\left|-13.6\text{eV}\right|-\frac{\left|-13.6\text{eV}\right|}{4^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{12.75\text{eV}}\\ =97.3\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $5$ for in equation (IV) to find the absorbed wavelength associated from the transition between the energy levels form $n=5$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\left|-13.6\text{eV}\right|-\frac{\left|-13.6\text{eV}\right|}{5^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{13.056\text{eV}}\\ =95.0\text{nm}\end{array}$

Therefore, the wavelengths can absorbed by the $H$ atom in between the wavelength ranging $96\text{nm}$ to $110\text{nm}$ are $\underset{\_}{97.3\text{nm and }103\text{nm}}$.

(b)

To determine

The recoil speed of the atom after absorption.

Answer to Problem 64P

The recoil speed of the atom after absorption are $\underset{\_}{4.07\text{m}/\text{s}\text{ for the }97.3\text{nm photon}}$ and $\underset{\_}{3.86\text{m}/\text{s}\text{ for the }103\text{nm photon}}$.

Explanation of Solution

Write the expression for the energy of the photon.

    $E_{\text{photn}}=pc$        (V)

Here, $E_{\text{photn}}$ is the energy of the photon, $p$ is the momentum.

Write the expression for $p$.

    $p=\frac{h}{\lambda }$        (VI)

Write the expression for the momentum associated with the photon.

    $p_{\text{photon}}=\frac{E_{\text{photon}}}{c}$        (VII)

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

    $p_{\text{photon}}=\frac{h}{\lambda }$        (VIII)

Write the expression for momentum associated with hydrogen atom.

    $p_H=m_{H}v$        (IX)

Here, $p_H$ is the momentum of hydrogen atom, $m_H$ is the mass of hydrogen atom, $v$ is the recoil velocity.

Equate equation (IX) and (VII) to solve for recoil velocity.

    $v=\frac{h}{m_H\lambda }$        (X)

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $97.3\text{nm}$ for $\lambda$, $1.674\times {10}^{-27}\text{kg}$ for $m_H$ in equation (X) to find the $v$ associated with $97.3\text{nm}$.

    $\begin{array}{c}v=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}{\left(1.674\times {10}^{-27}\text{kg}\right)\left(97.3\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =4.07\text{m}/\text{s}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $103\text{nm}$ for $\lambda$, $1.674\times {10}^{-27}\text{kg}$ for $m_H$ in equation (X) to find the $v$ associated with $103\text{nm}$.

    $\begin{array}{c}v=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}{\left(1.674\times {10}^{-27}\text{kg}\right)\left(103\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =3.86\text{m}/\text{s}\end{array}$

Therefore, the recoil speed of the atom after absorption are $\underset{\_}{4.07\text{m}/\text{s}\text{ for the }97.3\text{nm photon}}$ and $\underset{\_}{3.86\text{m}/\text{s}\text{ for the }103\text{nm photon}}$.

(c)

To determine

The wavelength of each photon emitted and classify its region in electromagnetic spectrum.

Answer to Problem 64P

The wavelength of photon emitted from $n=2$ to $n=1$ is $\underset{\_}{122\text{nm,}\text{UV}}$,  $n=2$ to $n=1$ is $\underset{\_}{103\text{nm,}\text{UV}}$,  $n=3$ to $n=1$ is $\underset{\_}{97.3\text{nm,}\text{UV}}$, $n=3$ to $n=2$ is $\underset{\_}{656\text{nm,}\text{Visible}\left(\text{red}\right)}$, $n=4$ to $n=2$ is $\underset{\_}{486\text{nm,}\text{Visible}\left(\text{blue}\right)}$, $n=4$ to $n=3$ is $\underset{\_}{\text{1}\text{.88}\times {\text{10}}^3\text{nm,}\text{IR}}$.

Explanation of Solution

Use equation (IV) to solve for the wavelength associated with the absorbed wavelength associated from the transition between the energy levels.

    $E=\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|E_1\right|}{n_f^2}-\frac{\left|E_1\right|}{n_i^2}}$        (XI)

Conclusion:

Substitute $-13.6\text{eV}$ for $E_1$, $2$ for $n_i$, $1$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=2$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{1^2}-\frac{\left|-13.6\text{eV}\right|}{2^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{10.2\text{eV}}\\ =122\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $3$ for $n_i$, $1$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=3$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{1^2}-\frac{\left|-13.6\text{eV}\right|}{3^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{12.09\text{eV}}\\ =103\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $4$ for $n_i$, $1$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=4$ to $n=1$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{1^2}-\frac{\left|-13.6\text{eV}\right|}{4^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{12.75\text{eV}}\\ =97.3\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $3$ for $n_i$, $2$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=3$ to $n=2$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{2^2}-\frac{\left|-13.6\text{eV}\right|}{3^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3.40\text{eV}-1.51\text{eV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.89\text{eV}}\\ =97.3\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $4$ for $n_i$, $2$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=4$ to $n=2$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{2^2}-\frac{\left|-13.6\text{eV}\right|}{4^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3.40\text{eV}-0.85\text{eV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{2.55\text{eV}}\\ =486\text{nm}\end{array}$

Substitute $-13.6\text{eV}$ for $E_1$, $4$ for $n_i$, $3$ for $n_f$ in equation (IX) to find the absorbed wavelength associated from the transition between the energy levels form $n=4$ to $n=3$.

    $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\frac{\left|-13.6\text{eV}\right|}{3^2}-\frac{\left|-13.6\text{eV}\right|}{4^2}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.51\text{eV}-0.85\text{eV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{0.661\text{eV}}\\ =1.88\times {10}^3\text{nm}\end{array}$

Therefore, the wavelength of photon emitted from $n=2$ to $n=1$ is $\underset{\_}{122\text{nm,}\text{UV}}$,  $n=2$ to $n=1$ is $\underset{\_}{103\text{nm,}\text{UV}}$,  $n=3$ to $n=1$ is $\underset{\_}{97.3\text{nm,}\text{UV}}$, $n=3$ to $n=2$ is $\underset{\_}{656\text{nm,}\text{Visible}\left(\text{red}\right)}$, $n=4$ to $n=2$ is $\underset{\_}{486\text{nm,}\text{Visible}\left(\text{blue}\right)}$, $n=4$ to $n=3$ is $\underset{\_}{\text{1}\text{.88}\times {\text{10}}^3\text{nm,}\text{IR}}$.