Chapter 27, Problem 89P

(a)

To determine

The recoil speed of the hydrogen atom after absorption of the ultraviolet photon.

Answer to Problem 89P

The recoil speed of the hydrogen atom is $4.1{\text{ms}}^{-1}$.

Explanation of Solution

The wavelength of the photon is $97\text{nm}$. The hydrogen atom is initially at rest.

Write the expression for momentum of the photon

  $p_{\nu }=\frac{h}{\lambda }$                                                                                                                  (I)

Here, $p_{\nu }$ is the momentum of the photon, $h$ is the Planck’s constant and $\lambda$ is the wavelength of the photon.

Write the expression for momentum

  $p_{\text{H}}=mv$                                                                                                            (II)

Here, $m$ is the mass of the hydrogen atom, $p_{\text{H}}$ is its momentum after absorbing the photon and $v$ is its recoil velocity.

Equate (I) and (II)

  $mv=\frac{h}{\lambda }$

Rearrange

  $v=\frac{h}{m\lambda }$                                                                                                            (III)

Substitute $97\text{nm}$ for $\lambda$, $1.674\times {10}^{-27}\text{kg}$ for $m$ and $6.636\times {10}^{-34}\text{Js}$ for $h$ in (III) to find $v$

  $\begin{array}{c}v=\frac{\left(6.636\times {10}^{-34}\text{Js}\right)}{\left(1.674\times {10}^{-27}\text{kg}\right)\left(97\text{nm}\right)}\\ =\frac{\left(6.636\times {10}^{-34}\text{Js}\right)}{\left(1.674\times {10}^{-27}\text{kg}\right)\left(97\times {10}^{-9}\text{m}\right)}\cdot \frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\\ =4.1{\text{ms}}^{-1}\end{array}$

Thus, the recoil speed of the hydrogen atom is $4.1{\text{ms}}^{-1}$.

(b)

To determine

The different possible transitions to reach ground state after being excited by absorbing a photon.

Answer to Problem 89P

The number of ways to return to ground state is $4$.

Explanation of Solution

Write the expression for energy of photon

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

Here, $E$ is the energy of the photon and $c$ is the speed of light.

Substitute $97\text{nm}$ for $\lambda$ and $1240\text{eV}\cdot \text{nm}$ for $hc$ in (IV) to find $E$

  $\begin{array}{c}E=\frac{1240\text{eV}\cdot \text{nm}}{97\text{nm}}\\ =12.8\text{eV}\end{array}$

Thus, the energy of each radiated photon is $3.69\times {10}^{-7}\text{eV}$.

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

  $E=\left|E_0\right|-\frac{\left|E_0\right|}{n^2}$                                                                                               (V)

 Here, $E$ is the difference in energy i.e. the energy of the emitted photon during transition, $E_0$ is the ground state energy and $n$ is the principal quantum number.

Rearrange for $n$

  $n=\sqrt{\frac{\left|E_0\right|}{E-\left|E_0\right|}}$                                                                                               (VI)

Substitute $13.6\text{eV}$ for $\left|E_0\right|$ and $12.8\text{eV}$ for $E$ in (VI) to find $n$

  $\begin{array}{c}n=\sqrt{\frac{13.6\text{eV}}{13.6\text{eV}-12.8\text{eV}}}\\ =4.12\\ \approx 4\end{array}$

Thus, the atom is excited to level $4$.

The possible transitions to return to ground state are $4\to 1$, $4\to 2\to 1$, $4\to 3\to 1$ and  $4\to 3\to 2\to 1$. Thus, there are $4$ different ways to return to the ground state with the emission of $6$ different photons.

(c)

To determine

The wavelength of the emitted photons and their classification.

Answer to Problem 89P

The wavelengths of the emitted photons are $97\text{nm}$, $486\text{nm}$, $1875\text{nm}$, $103\text{nm}$, $656\text{nm}$ and $122\text{nm}$, their classification is given in the table.

Explanation of Solution

Write the expression for wavelength of photon emitted during transition

  $\lambda =\frac{1}{R\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)}$                                                                                         (VII)

Here, $R$ is the Rydberg’s constant, $n_{\text{f}}$ is the final state and $n_{\text{i}}$ is the initial state.

Substitute $1$ for $n_{\text{f}}$, $4$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $4\to 1$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{4^2}\right)}\\ =97.2\times {10}^{-9}\text{m}\\ =97\text{nm}\end{array}$

Substitute $2$ for $n_{\text{f}}$, $4$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $4\to 2$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}\\ =486.1\times {10}^{-9}\text{m}\\ =486\text{nm}\end{array}$

Substitute $3$ for $n_{\text{f}}$, $4$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $4\to 3$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{4^2}\right)}\\ =1875\times {10}^{-9}\text{m}\\ =1875\text{nm}\end{array}$

Substitute $1$ for $n_{\text{f}}$, $3$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $3\to 1$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{3^2}\right)}\\ =102.5\times {10}^{-9}\text{m}\\ =103\text{nm}\end{array}$

Substitute $2$ for $n_{\text{f}}$, $3$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $3\to 2$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{4^2}\right)}\\ =656.3\times {10}^{-9}\text{m}\\ =656\text{nm}\end{array}$

Substitute $1$ for $n_{\text{f}}$, $2$ for $n_{\text{i}}$ and $1.097\times {10}^{-7}{\text{m}}^{-1}$ for $R$ in (VII) to find $\lambda$ for the transition $2\to 1$.

  $\begin{array}{c}\lambda =\frac{1}{\left(1.097\times {10}^{-7}{\text{m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}\\ =121.5\times {10}^{-9}\text{m}\\ =122\text{nm}\end{array}$

The wavelength range for UV light is $10\text{nm}<\lambda <400\text{nm}$, X-rays is $0.01\text{nm}<10\text{nm}$, visible light is $400\text{nm}\le \lambda \le 700\text{nm}$, IR radiation is $700\text{nm}<\lambda <{10}^6\text{nm}$.

The wavelength of the emitted photons and their classification is given in the table below:

Transition	$\lambda$ in $\text{nm}$	Class
$4\to 1$	97	UV
$4\to 2$	486	Visible
$4\to 3$	1875	IR
$3\to 1$	103	UV
$3\to 2$	656	Visible
$2\to 1$	122	UV