Chapter 27, Problem 64P

(a)

To determine

The wavelengths that can be absorbed by the hydrogen atom.

Answer to Problem 64P

The wavelengths that can be absorbed by the hydrogen atom are $\underset{\_}{97.3\text{nm and 103}\text{nm}}$.

Explanation of Solution

Given that the wavelength of the UV light is ranging from $96\text{nm}$ to $110\text{nm}$.

The photons that can be absorbed have energies equal to the energy difference between the energy levels of the hydrogen atom.

Write the expression for the difference in energy between the ground state and $n^{\text{th}}$ state of hydrogen atom.

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

Here, $\Delta E$ is the energy difference, $E_1$ is the ground state energy, and $E_n$ is the energy of the $n^{\text{th}}$ state.

Write the expression for the energy of the $n^{\text{th}}$ state in terms of the ground state energy.

  $\left|E_n\right|=\frac{\left|E_1\right|}{n^2}$                                                                                                          (II)

Use equation (II) in (I).

  $\begin{array}{c}\Delta E=\left|E_1\right|-\frac{\left|E_1\right|}{n^2}\\ =\left|E_1\right|\left(1-\frac{1}{n^2}\right)\end{array}$                                                                                             (III)

Write the expression for the absorbed wavelength corresponding to the energy level difference.

  $\lambda =\frac{hc}{\Delta E}$                                                                                                            (IV)

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

Use equation (III) in (IV).

  $\lambda =\frac{hc}{\left|E_1\right|\left(1-\frac{1}{n^2}\right)}$                                                                                                (V)

The value of $hc$ is $1240\text{eV}\cdot \text{nm}$. The ground state energy of hydrogen atom is $13.6\text{eV}$.

Conclusion:

Substitute $2$ for $n$, $1240\text{eV}\cdot \text{nm}$ for $hc$, and $13.6\text{eV}$ for $\left|E_1\right|$ in equation (V) to find the wavelength corresponding to the energy level transition from $n=1$ to $n=2$.

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

Substitute $3$ for $n$, $1240\text{eV}\cdot \text{nm}$ for $hc$, and $13.6\text{eV}$ for $\left|E_1\right|$ in equation (V) to find the wavelength corresponding to the energy level transition from $n=1$ to $n=3$.

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

Substitute $3$ for $n$, $1240\text{eV}\cdot \text{nm}$ for $hc$, and $13.6\text{eV}$ for $\left|E_1\right|$ in equation (V) to find the wavelength corresponding to the energy level transition from $n=1$ to $n=3$.

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

Substitute $4$ for $n$, $1240\text{eV}\cdot \text{nm}$ for $hc$, and $13.6\text{eV}$ for $\left|E_1\right|$ in equation (V) to find the wavelength corresponding to the energy level transition from $n=1$ to $n=4$.

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

Substitute $5$ for $n$, $1240\text{eV}\cdot \text{nm}$ for $hc$, and $13.6\text{eV}$ for $\left|E_1\right|$ in equation (V) to find the wavelength corresponding to the energy level transition from $n=1$ to $n=5$.

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\left(13.6\text{eV}\right)\left(1-\frac{1}{5^2}\right)}\\ =95.0\text{nm}\end{array}$

Therefore, the wavelengths that can be absorbed by the hydrogen atom are $\underset{\_}{97.3\text{nm and 103}\text{nm}}$.

(b)

To determine

The recoil speed of the atom after absorption of the UV photons corresponding to the absorbing wavelengths.

Answer to Problem 64P

The recoil speed of the atom after absorption of the UV photons corresponding to $97.3\text{nm}$ is $\underset{\_}{4.07\text{m/s}}$, and that corresponding to $103\text{nm}$ is $\underset{\_}{3.86\text{m/s}}$.

Explanation of Solution

According to the momentum conservation, the momentum of the photon must be equal to the momentum of the atom after absorbing the photon from rest.

Write the expression for the momentum of the photon.

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

Here, $p_{\text{photon}}$ is the momentum of the photon.

Write the expression for the momentum of the hydrogen atom after absorption of photon.

  $p_{\text{H}}=m_{\text{H}}v$                                                                                                      (VII)

Here, $p_{\text{H}}$ is the momentum of the hydrogen atom, $m_{\text{H}}$ is the mass of the hydrogen atom, and $v$ is the recoil speed of the atom.

Equate the right hand sides of equation (VI) and (VII) and solve for $v$.

  $\begin{array}{c}\frac{h}{\lambda }=m_{\text{H}}v\\ v=\frac{h}{m_{\text{H}}\lambda }\end{array}$                                                                                                  (VIII)

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.674\times {10}^{-27}\text{kg}$ for $m_{\text{H}}$, and $97.3\text{nm}$ for $\lambda$ on equation (VIII) to find the recoil velocity corresponding to $97.3\text{nm}$ photon absorption.

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

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.674\times {10}^{-27}\text{kg}$ for $m_{\text{H}}$, and $103\text{nm}$ for $\lambda$ on equation (VIII) to find the recoil velocity corresponding to $103\text{nm}$ photon absorption.

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

Therefore, the recoil speed of the atom after absorption of the UV photons corresponding to $97.3\text{nm}$ is $\underset{\_}{4.07\text{m/s}}$, and that corresponding to $103\text{nm}$ is $\underset{\_}{3.86\text{m/s}}$.

(c)

To determine

The number of ways in which the hydrogen atom returns to the ground state after absorption.

Answer to Problem 64P

The hydrogen atom returns to the ground state in two ways after absorption of $103\text{nm}$ wavelength and four ways after absorption of $97.3\text{nm}$ wavelength.

Explanation of Solution

The transition from the $n=3$ state to the ground state ($n=1$) can occur in two ways: $3$ to $2$, then $2$ to $1$; or $3$ to $1$. The transition from the $n=4$ state to the ground state can occur in four ways: $4$ to $3$, then $3$ to $2$, then $2$ to $1$; or $4$ to $3$, then $3$ to $1$; or $4$ to $2$, then $2$ to $1$; or $4$ to $1$. Therefore, in total there are six ways for the atom to return to the ground state: two ways after absorption of $103\text{nm}$ wavelength and four ways after absorption of $97.3\text{nm}$ wavelength.

Conclusion:

Therefore, the hydrogen atom returns to the ground state in two ways after absorption of $103\text{nm}$ wavelength and four ways after absorption of $97.3\text{nm}$ wavelength.