Chapter 3, Problem 3.128QP

Four atomic energy levels of an atom are shown here. When an electron in an excited atom moves from (d), (d) ⇄ (c), (c) ⇄ (b), and (b) ⇄ (a), a photon of light is emitted each time. The wavelengths of the various photons are 575 nm, 162 nm, and 131 nm. Determine the energies of each photon emitted and match each emission to the appropriate wavelength knowing the energy levels are drawn to scale.

Interpretation Introduction

Interpretation:

The energies of each photon emitted having the wavelengths of the various photons as $\text{575 nm}$, $\text{162 nm}$ and $\text{131 nm}$ when an electron in an excited atom moves from (d), $\left(\text{d}\right)\rightleftarrows \left(\text{c}\right)$, $\left(\text{c}\right)\rightleftarrows \left(\text{b}\right)$ and $\left(\text{b}\right)\rightleftarrows \left(\text{a}\right)$ in four atomic energy levels of an atom should be calculated using the relation between speed, wavelength and frequency of a wave and Planck’s quantum theory. Each emission should also be matched to the appropriate wavelength knowing the energy levels.

Concept Introduction:

A wave is a disturbance or variation that travels through a medium transporting energy without transporting matter.  The wavelength is the distance between identical points on successive waves.  The frequency is the number of waves that pass through any particular point in 1 second.

Figure 1

The speed, wavelength and frequency of a wave are related by the equation: $\text{c }=\text{ λν}$ where $\text{λ and ν}$ are expressed in meters ($\text{m}$) and reciprocal seconds (${\text{s}}^{-1}$) respectively.

Planck’s quantum theory

• Different atoms and molecules can emit or absorb energy in discreet quantities only.  The smallest amount of energy that can be emitted or absorbed in the form of electromagnetic radiation is known as quantum.
• The energy of the radiation absorbed or emitted is directly proportional to the frequency of the radiation.  The energy of radiation is expressed in terms of frequency as, $\text{E }=\text{ hν}$ where, $\text{E}$ = energy of the radiation; $\text{h}$ = Planck’s constant ($6.626\times {10}^{–34}\text{ Js}$); $\text{ν}$ = frequency of radiation.

To find: The energies of each photon emitted having the wavelengths of the various photons as $\text{575 nm}$, $\text{162 nm}$ and $\text{131 nm}$ when an electron in an excited atom moves from (d), $\left(\text{d}\right)\stackrel{}{\rightleftarrows }\left(\text{c}\right)$, $\left(\text{c}\right)\stackrel{}{\rightleftarrows }\left(\text{b}\right)$ and $\left(\text{b}\right)\stackrel{}{\rightleftarrows }\left(\text{a}\right)$ in four atomic energy levels of an atom and match each emission to the appropriate wavelength knowing the energy levels.

Answer to Problem 3.128QP

The energies of the photons emitted having the wavelength of $\text{575 nm}$ is $\text{3}{\text{.46 × 10}}^{-\text{19}}\text{ J}$; The energies of the photons emitted having the wavelength of $\text{162 nm}$ is $\text{1}{\text{.23 × 10}}^{-\text{18}}\text{ J}$; The energies of the photons emitted having the wavelength of $\text{131 nm}$ is $\text{1}{\text{.52 × 10}}^{-\text{18}}\text{ J}$.

Each emission to the appropriate wavelength knowing the energy levels is given as follows:

Transition $\left(\text{c}\right)\to \left(\text{b}\right)$ has the largest difference in energy levels between the initial and final states.  Transition $\left(\text{c}\right)\to \left(\text{b}\right)$ emits the photon 2 with the highest energy ($\text{E }=\text{ 1}{\text{.52 × 10}}^{\text{–18}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{131 nm}$.

Transition $\left(\text{b}\right)\to \left(\text{a}\right)$ has the second largest difference in energy levels between the initial and final states. Transition $\left(\text{b}\right)\to \left(\text{a}\right)$ emits the photon 3 with the second highest energy ($\text{E }=\text{ 1}{\text{.23 × 10}}^{\text{–18}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{162 nm}$.

Transition $\left(\text{d}\right)\to \left(\text{c}\right)$ has the smallest difference in energy levels between the initial and final states.  Transition $\left(\text{d}\right)\to \left(\text{c}\right)$ emits the photon 1 with the lowest energy ($\text{E }=\text{ 3}{\text{.46 × 10}}^{\text{–19}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{575 nm}$.

Explanation of Solution

The speed, wavelength and frequency of a wave are related by the equation: $\text{c }=\text{ λν}$ where $\text{λ and ν}$ are expressed in meters ($\text{m}$) and reciprocal seconds (${\text{s}}^{-1}$) respectively.  Here, wavelength is given; energy is to be calculated.  Hence, rearranging the equation for getting frequency is

$\text{ν }=\frac{\text{c}}{\text{λ}}$

where wavelength, $\text{λ}$ is $\text{575 nm}$; the speed of light, $\text{c}$ is $\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$

The speed of light is used in $\text{m/s}$.  Hence, wavelength, $\text{λ}$ in $\text{nm}$ is converted into that in $\text{m}$.  We know that, $\text{1 nm }={\text{ 1 × 10}}^{-\text{9}}\text{ m}$.

$\begin{array}{l}\text{λ }=\text{ 575 nm }\times \frac{{\text{1 × 10}}^{-\text{9}}\text{ m}}{\text{1 nm}}\\ \text{λ }=\text{ 5}{\text{.75 × 10}}^{-\text{7}}\text{ m}\end{array}$

The frequency having the wavelength $\text{575 nm}$ is

$\begin{array}{l}\text{ν }=\frac{\text{c}}{\text{λ}}\\ \text{ν }=\frac{\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}}{\text{5}{\text{.75 × 10}}^{-\text{7}}\text{ m}}\\ \text{ν }=\text{ 5}{\text{.22 × 10}}^{\text{14}}{\text{ s}}^{-\text{1}}\end{array}$

In the same manner, the frequency having the wavelength $\text{162 nm}$ is

$\begin{array}{l}\text{λ }=\text{ 162 nm }\times \frac{{\text{1 × 10}}^{-\text{9}}\text{ m}}{\text{1 nm}}\\ \text{λ }=\text{ 1}{\text{.62 × 10}}^{-\text{7}}\text{ m}\\ \text{ν }=\frac{\text{c}}{\text{λ}}\\ \text{ν }=\frac{\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}}{\text{1}{\text{.62 × 10}}^{-\text{7}}\text{ m}}\\ \text{ν }=\text{ 1}{\text{.85 × 10}}^{\text{15}}{\text{ s}}^{-\text{1}}\end{array}$

In the same manner, the frequency having the wavelength $\text{131 nm}$ is

$\begin{array}{l}\text{λ }=\text{ 1}\text{.31 nm }\times \frac{{\text{1 × 10}}^{-\text{9}}\text{ m}}{\text{1 nm}}\\ \text{λ }=\text{ 1}{\text{.31 × 10}}^{-\text{7}}\text{ m}\\ \text{ν }=\frac{\text{c}}{\text{λ}}\\ \text{ν }=\frac{\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}}{\text{1}{\text{.31 × 10}}^{-\text{7}}\text{ m}}\\ \text{ν }=\text{ 2}{\text{.29 × 10}}^{\text{15}}{\text{ s}}^{-\text{1}}\end{array}$

The energy of radiation is expressed in terms of frequency as, $\text{E }=\text{ hν}$ where, $\text{E}$ = energy of the radiation; $\text{h}$ = Planck’s constant ($6.626\times {10}^{–34}\text{ Js}$); $\text{ν}$ = frequency of radiation.  Substitute the given values which has a wavelength of $\text{575 nm}$,

$\begin{array}{l}\text{E }=\text{ hν}\\ \text{E }=\text{ (6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(5}{\text{.22 × 10}}^{\text{14}}\text{ /s)}\\ \text{E }=\text{ 3}{\text{.46 × 10}}^{-\text{19}}\text{ J}\end{array}$

Therefore, the energy of the photon which has a wavelength of $\text{575 nm}$ is $\text{3}{\text{.46 × 10}}^{-\text{19}}\text{ J}$.

In the similar way, the energy of the photon which has a wavelength of $\text{162 nm}$ is

$\begin{array}{l}\text{E }=\text{ hν}\\ \text{E }=\text{ (6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(1}{\text{.85 × 10}}^{\text{15}}\text{ /s)}\\ \text{E }=\text{ 1}{\text{.23 × 10}}^{-\text{18}}\text{ J}\end{array}$

In the similar way, the energy of the photon which has a wavelength of $\text{131 nm}$ is

$\begin{array}{l}\text{E }=\text{ hν}\\ \text{E }=\text{ (6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(2}{\text{.29 × 10}}^{\text{15}}\text{ /s)}\\ \text{E }=\text{ 1}{\text{.52 × 10}}^{-\text{18}}\text{ J}\end{array}$

To match the emissions to the appropriate wavelengths, the quantity of energy depends on the difference in energy levels between the initial and final states. Since the energy levels are drawn to scale, the photon with the highest energy will have the greatest difference between energy levels.

Transition $\left(\text{c}\right)\to \left(\text{b}\right)$ has the largest difference in energy levels between the initial and final states.  Transition $\left(\text{c}\right)\to \left(\text{b}\right)$ emits the photon 2 with the highest energy ($\text{E }=\text{ 1}{\text{.52 × 10}}^{\text{–18}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{131 nm}$.

Transition $\left(\text{b}\right)\to \left(\text{a}\right)$ has the second largest difference in energy levels between the initial and final states. Transition $\left(\text{b}\right)\to \left(\text{a}\right)$ emits the photon 3 with the second highest energy ($\text{E }=\text{ 1}{\text{.23 × 10}}^{\text{–18}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{162 nm}$.

Transition $\left(\text{d}\right)\to \left(\text{c}\right)$ has the smallest difference in energy levels between the initial and final states.  Transition $\left(\text{d}\right)\to \left(\text{c}\right)$ emits the photon 1 with the lowest energy ($\text{E }=\text{ 3}{\text{.46 × 10}}^{\text{–19}}\text{ J}$).  This energy is produced by the photon with a wavelength of $\text{575 nm}$.

Conclusion

The energies of each photon emitted having the wavelengths of the various photons as $\text{575 nm}$, $\text{162 nm}$ and $\text{131 nm}$ when an electron in an excited atom moves from (d), $\left(\text{d}\right)\stackrel{}{\rightleftarrows }\left(\text{c}\right)$, $\left(\text{c}\right)\stackrel{}{\rightleftarrows }\left(\text{b}\right)$ and $\left(\text{b}\right)\stackrel{}{\rightleftarrows }\left(\text{a}\right)$ in four atomic energy levels of an atom are calculated using the relation between speed, wavelength and frequency of a wave and Planck’s quantum theory. Each emission is matched to the appropriate wavelength knowing the energy levels.