Chapter 27, Problem 53P

(a)

To determine

The energies of first four levels of doubly ionized lithium starting with $n=1$.

Answer to Problem 53P

The energies of first four levels of doubly ionized lithium starting with $n=1$ to $n=4$ are $-122\text{eV}$, $-30.6\text{eV}$, $-13.6\text{eV}$ and $-7.65\text{eV}$.

Explanation of Solution

Write an expression for the energy of level of doubly ionized lithium.

    $E_n=-\frac{Z^2}{n^2}{E}_0$                                                                                                            (I)

Here, $E_n$ is the energy of the level, $E_0$ is the rest energy, $Z$ is the atomic number and $n$ is the energy level.

Conclusion:

For $n=1$:

Substitute $3$ for $Z$, $13.6\text{eV}$ for $E_0$ and $1$ for $n$ in equation (I) to find $E_n$.

  $\begin{array}{c}{E}_1=-\frac{3^2}{1^2}\left(13.6\text{eV}\right)\\ =-9\left(13.6\text{eV}\right)\\ =-122\text{eV}\end{array}$

Here, $E_1$ is the energy at level $n=1$.

For $n=2$:

Substitute $3$ for $Z$, $13.6\text{eV}$ for $E_0$ and $2$ for $n$ in equation (I) to find $E_n$.

  $\begin{array}{c}{E}_2=-\frac{3^2}{2^2}\left(13.6\text{eV}\right)\\ =-\left(\frac{9}{4}\right)\left(13.6\text{eV}\right)\\ =-30.6\text{eV}\end{array}$

Here, $E_2$ is the energy at level $n=2$.

For $n=3$:

Substitute $3$ for $Z$, $13.6\text{eV}$ for $E_0$ and $3$ for $n$ in equation (I) to find $E_n$.

  $\begin{array}{c}{E}_3=-\frac{3^2}{3^2}\left(13.6\text{eV}\right)\\ =-\left(1\right)\left(13.6\text{eV}\right)\\ =-13.6\text{eV}\end{array}$

Here, $E_3$ is the energy at level $n=3$.

For $n=4$:

Substitute $3$ for $Z$, $13.6\text{eV}$ for $E_0$ and $4$ for $n$ in equation (I) to find $E_n$.

  $\begin{array}{c}{E}_4=-\frac{3^2}{4^2}\left(13.6\text{eV}\right)\\ =-\left(\frac{9}{16}\right)\left(13.6\text{eV}\right)\\ =-7.65\text{eV}\end{array}$

Here, $E_4$ is the energy at level $n=4$.

Thus, the energies of first four levels of doubly ionized lithium starting with $n=1$ to $n=4$ are $-122\text{eV}$, $-30.6\text{eV}$, $-13.6\text{eV}$ and $-7.65\text{eV}$.

(b)

To determine

The energies of the photons emitted or absorbed when the electron makes a transition between the levels.

Answer to Problem 53P

The energies of the photons emitted or absorbed when the electron makes a transition between the levels is given in table 1.

Explanation of Solution

Write an expression for energies of the photons emitted or absorbed when the electron makes a transition between the levels.

    $\Delta E=E_i-E_f$                                                                                                            (II)

Here, $E_f$ is the energy of the final level, $E_i$ is the energy of the initial level, $\Delta E$ is the energy of the photons emitted or absorbed when the electron makes a transition between the levels.

Conclusion:

For transition of photon from level $4\to 1$

Substitute $-7.65\text{eV}$ for $E_i$ and $-122\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-7.65\text{eV}-\left(-122\text{eV}\right)\\ =115\text{eV}\end{array}$

For transition of photon from level $4\to 2$

Substitute $-7.65\text{eV}$ for $E_i$ and $-30.6\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-7.65\text{eV}-\left(-30.6\text{eV}\right)\\ =23.0\text{eV}\end{array}$

For transition of photon from level $4\to 3$

Substitute $-7.65\text{eV}$ for $E_i$ and $-13.6\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-7.65\text{eV}-\left(-13.6\text{eV}\right)\\ =6.00\text{eV}\end{array}$

For transition of photon from level $3\to 1$

Substitute $-13.6\text{eV}$ for $E_i$ and $-122\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-13.6\text{eV}-\left(-122\text{eV}\right)\\ =109\text{eV}\end{array}$

For transition of photon from level $3\to 2$

Substitute $-13.6\text{eV}$ for $E_i$ and $-30.6\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-13.6\text{eV}-\left(-30.6\text{eV}\right)\\ =17.0\text{eV}\end{array}$

For transition of photon from level $2\to 1$

Substitute $-30.6\text{eV}$ for $E_i$ and $-122\text{eV}$ for $E_f$ in equation (II) to find $\Delta E$.

    $\begin{array}{c}\Delta E=-30.6\text{eV}-\left(-122\text{eV}\right)\\ =92.0\text{eV}\end{array}$

Thus, the energies of the photons emitted or absorbed when the electron makes a transition between the levels is given in table 1.

Initial level	Final level	Energy of photon (eV)
4	1	115
4	2	23.0
4	3	6.00
3	1	109
3	2	17.0
2	1	92.0

Table 1

(c)

To determine

The presence of photons in the visible part of EM.

Answer to Problem 53P

No photons will be present in visible range.

Explanation of Solution

Write an expression for the maximum energy of the photon in visible range.

    $E_{\max }=\frac{hc}{{\lambda }_{\min }}$                                                                                                            (III)

Here, $E_{\max }$ is the maximum energy of the photon in visible range, $h$ is the Plank’s constant, $c$ is the speed of light and ${\lambda }_{\min }$ is the minimum wavelength of the visible range.

Write an expression for the minimum energy of the photon in visible range.

    $E_{\min }=\frac{hc}{{\lambda }_{\max }}$                                                                                                            (IV)

Here, $E_{\min }$ is the minimum energy of the photon in visible range and ${\lambda }_{\max }$ is the maximum wavelength of the visible range.

The wavelength of the visible spectrum varies from $400\text{nm}$ to $700\text{nm}$.

Conclusion:

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$ and $400\text{nm}$ for ${\lambda }_{\min }$ in equation (III) to find $E_{\max }$.

    $\begin{array}{c}{E}_{\max }=\frac{1240\text{eV}\cdot \text{nm}}{400\text{nm}}\\ =3.10\text{eV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$ and $700\text{nm}$ for ${\lambda }_{\max }$ in equation (IV) to find $E_{\min }$.

    $\begin{array}{c}{E}_{\min }=\frac{1240\text{eV}\cdot \text{nm}}{700\text{nm}}\\ =1.77\text{eV}\end{array}$

The lowest energy of the photon absorbed or emitted during the transition is $6.00\text{eV}$. The energy of photon is greater than the maximum energy possible for a photon in visible region. Thus, there will not be any photon in visible region.