Unit One

Day 2: Quantum Theory

If you have not yet worked through the Introduction, please do so before beginning this section. As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Sections 5-2a and 5-3, and/or Chapter 1.3-1.5 and Chapter 2.1 in the Additional Reading Materials section.

D2.1 The Photoelectric Effect

Activity 1: Preparation—Photoelectric Effect

In your course notebook, make a heading for Photoelectric Effect. After the heading write down what you recall about the photoelectric effect from courses you have already taken. If you remember having a question about the photoelectric effect or if there is anything you remember being puzzled about, write that down as well. We will ask you to refer back to what you have written when you complete this section.

Planck’s quantum theory was able to predict accurately the distribution of wavelengths emitted by a blackbody at various temperatures. However, Planck found it difficult to justify his assumption that vibration energies had to be multiples of a minimum energy—a quantum. When Albert Einstein used Planck’s quantum hypothesis to explain a different phenomenon, the photoelectric effect, the validity of quantum theory became clearer.

Exercise 1: Quanta and Laser Light (Review)

Before doing any calculation or looking at the hint, write in your class notebook an explanation of how you plan to work out the problem. Do all the steps in the calculation in your notebook. Once you have arrived at an answer, then submit your results below and click the “Check” button to see if it is correct. If one or more parts of your answer is incorrect, go over your work in your notebook carefully and check for errors. “Retry” with your new answer.  Look at the hint (click on it to expand for view) only after you have attempted to answer the question at least twice.
Determine how many quanta of radiation are emitted by a 5.0 mW red (λ = 632.8 nm) laser pointer each second. It is useful to know that 1 W = 1 J s-1.
Number of quanta per second = [Number] × 10[exponent] s-1

When electromagnetic radiation shines on a metal, such as sodium, electrons can be emitted and an electric current (a flow of electrons) can occur. This is called the photoelectric effect. The effect is complicated: for some wavelengths no electrons are emitted, but at other wavelengths electrons are emitted.

Exercise 2: Photoelectric Effect Simulation

Watch this photoelectric effect animation, where sodium is already selected, to answer the questions below. Write down your observations as you watch the animation, and then answer the questions in your course notebook.

  1. For which colors of visible light are electrons emitted by sodium? Determine the maximum wavelength at which electrons are emitted.
  2. At a wavelength where electrons are emitted, describe the shape of a graph of number of electrons emitted vs. light intensity.
  3. Determine the shape of a graph of electron energy vs. light frequency. How does this graph differ from the electrons vs. intensity graph?
  4. Based on your observations, draw a rough graph with labeled axes to show how the number of electrons emitted varies with wavelength of light. How would the graph change if the intensity of light increased?

If you would like to experiment with the simulation further, download and save the simulation program, go to the location where you saved it, and double-click on the file name (photoelectric_en.jar) to install and run it. You need to have Java installed for this to work.

Additional Practice

  1. Assume that the intensity of light corresponds with the amplitude of light waves and try to explain the observations you made using the simulation.
    1. Above a maximum wavelength no electrons are emitted
    2. Electron current increases with light intensity
    3. Above a threshold frequency electron energy increases with frequency of light
  2. Assume that light comes in quanta and the energy of a quantum of light is [latex]h\nu[/latex] (or [latex]\dfrac{hc}{\lambda}[/latex]). Does this idea enable all three observations to be explained? Write an explanation in your own words.
  3. Use this quantum assumption (number 2) to calculate the minimum energy required to cause an electron to be emitted from the surface of a sample of sodium.
  4. Suggest additional experiments you can do using the simulation to help confirm your response to question 2. Do those experiments and report the results.

Activity 2: Summary of Photoelectric Effect Results

In your course notebook, write a brief summary of the results you obtained from experimenting with the photoelectric effect simulation. Also write a summary of how your experimental results can be interpreted based on the idea that light consists of quanta. Compare what you wrote with the summary below [also available at this link] and, if necessary, revise what you wrote.

Exercise 3: Photoelectric Effect

Based on Einstein’s explanation of the photoelectric effect, decide whether each statement below is true or false. If a statement is true, but only under certain conditions, write down the conditions required to make it true. If a statement is false, rewrite the statement to make it true.

Additional Practice

It takes 5.86 × 10-19 J to free one electron from the surface of magnesium metal. Can the diode from a Blu-Ray player (405 nm light) generate a photoelectric effect from magnesium?

Activity 3: Wrap-up—Photoelectric Effect

Refer to what you wrote (including things that puzzled you) when you made the heading Photoelectric Effect in your class notebook. Revise what you wrote based on what you have just learned. Write a summary so that it will be a good study aid when you review for an exam. If you still have questions, ask them on Piazza.

D2.2 Line Spectra

Activity 4: Preparation—Atomic Spectra and Bohr Theory

In your course notebook, make a heading for Atomic Spectra and the Bohr Theory. After the heading write down what you remember about atomic spectra and the Bohr theory from courses you have already taken. If there is anything you remember being puzzled about, write that down as well. We will ask you to refer back to what you have written when you complete this section.

When solids, liquids, or high-pressure gases are heated sufficiently, they radiate light, producing a continuous spectrum in which there is significant intensity across a broad range of wavelengths. These continuous spectra can often be approximated by blackbody radiation curves.

Heating a gaseous element at low pressure, or passing an electric current through the gas, produces a line spectrum, which consists of very sharp peaks (lines) at a few specific wavelengths and no emission whatsoever at the rest of the wavelengths. (The colors of “neon” signs are produced by passing electric current through low-pressure gases.) Each element displays its own characteristic set of lines, as shown in Figure 1. For example, when electricity passes through a tube containing hydrogen gas at low pressure, the H2 molecules are broken apart into separate H atoms and we see a blue-pink color. Passing the blue-pink light through a prism produces the line spectrum in Figure 1: the blue-pink color consists of four discrete visible wavelengths.

An image is shown with 5 rows. Across the top and bottom of the image is a scale that begins at 4000 angstroms at the left and extends to 740 angstroms at the far right. The top row is a continuous band of the visible spectrum, showing the colors from violet at the far left through indigo, blue, green, yellow, orange, and red at the far right. The second row, labeled, “N a,” shows the emission spectrum for the element sodium, which includes two narrow vertical bands in the blue range, two narrow bands in the yellow-green range, two narrow bands in the yellow range, and one narrow band in the orange range. The third row, labeled, “H,” shows the emission spectrum for hydrogen. This spectrum shows single bands in the violet, indigo, blue, and orange regions. The fourth row, labeled, “C a,” shows the emission spectrum for calcium. This spectrum shows bands in the following colors and frequencies; one violet, five indigo, one blue, two green, two yellow-green, one yellow, two yellow-orange, one orange, and one red. The fifth row, labeled, “H g,” shows the emission spectrum for mercury. This spectrum shows bands in the following colors and frequencies; two violet, one indigo, two blue, one green, two yellow, two orange, and one orange-red. It is important to note that each of the color bands for the emission spectra of the elements matches to a specific wavelength of light. Extending a vertical line from the bands to the scale above or below the diagram will match the band to a specific measurement on the scale.
Figure 1. Compare the two types of emission spectra: continuous spectrum of white light (top) and the line spectra of the light from excited sodium, hydrogen, calcium, and mercury atoms.

In 1885, Johann Balmer discovered an equation that related the four visible wavelengths of light emitted by hydrogen atoms to the integers 3, 4, 5, and 6. For the nth line:

[latex]\dfrac{1}{\lambda} = k\left(\dfrac{1}{4} - \dfrac{1}{n^2}\right),\;\; n = 3, 4, 5, 6; k\;\text{is a constant}[/latex]

Other discrete lines for the hydrogen atom were found in the ultraviolet and infrared regions. Johannes Rydberg generalized Balmer’s work and developed an empirical formula that predicted all hydrogen emission lines, not just visible ones, where, n1 and n2 are integers, n2 < n1, and R is the Rydberg constant (1.097 × 107 m−1).

[latex]\dfrac{1}{\lambda} = R_{\infty} \left(\dfrac{1}{n^2_2} - \dfrac{1}{n^2_1}\right)[/latex]

Because the wavelengths of hydrogen emission lines were measured to very high accuracy, the Rydberg constant could be determined very precisely. That a simple formula involving integers could account for such precise measurements seemed astounding at the time.

Exercise 4: Hydrogen Atom Emissions

 Use the Rydberg equation to calculate the wavelength of light emitted by a hydrogen atom when n2 = 3 and n1 = 7.

Additional Practice

Calculate the energy of one photon with the wavelength you just calculated.
Photon energy = [number] × 10[exponent] J

D2.3 Energy Levels: The Bohr Model

Why do emission spectra for low-pressure gases consist of sharp peaks (lines) at certain wavelengths and zero emission everywhere else? How can the Rydberg equation, a simple equation involving integers, calculate the exact wavelengths of such emissions? These questions can be answered if we assume that an electron in a hydrogen atom can have only certain specific energies but cannot have any other energy values—that is, if we assume that there are specific energy levels for an electron in a hydrogen atom.

In 1913, Niels Bohr used Planck’s and Einstein’s ideas that energy is quantized and photon energies are proportional to their frequency to explain hydrogen-atom energy levels. The Bohr model assumed that in a hydrogen atom the electron moved around the nucleus in specific orbits, each with a different distance from the nucleus. Based on Coulomb’s law, the farther the electron is from the nucleus, the higher its energy is, so each orbit corresponded to a different energy. Instead of allowing for all possible values for the orbit energies, Bohr assumed that only certain energies could occur; that is, the energies were quantized. Bohr found that the orbit energies were inversely proportional to the square of an integer:

[latex]E_n = - \dfrac{k}{n^2},\;\; n = 1, 2, 3, \dots[/latex]

The proportionality constant k (k = 2.179 × 10–18 J) can be calculated from fundamental values such as the charge and mass of an electron and Planck’s constant.


Exercise 5: Hydrogen-Atom Energies

Calculate the energy of a hydrogen atom when the electron

  1. has n = 7;  E7 = [number] × 10[exponent] J
  2. has n = 3E3 = [number] × 10[exponent] J

Bohr further assumed that the atom would only emit or absorb a photon if the electron moved from one orbit to a different orbit. The energy absorbed or emitted corresponded to a difference between the orbit energies:

[latex]\Delta E = E_\text{f} - E_\text{i}[/latex]

where Ei and Ef are the initial and final orbit energies.

A positive value of [latex]\Delta E[/latex] means that the atom’s energy increases, corresponding to absorption of a photon with the photon energy added to the atom’s initial energy. A negative [latex]\Delta E[/latex] indicates that the atoms has less energy after the change—the energy is lost through emission of a photon. The energy of the photon is the absolute value of [latex]\Delta E[/latex], [latex]|\Delta E|[/latex]; the sign of [latex]\Delta E[/latex] indicates photon emission (negative) or absorption (positive). Combining the two preceding equations gives:

[latex]|\Delta E| = \left|-k\left(\dfrac{1}{n^2_f} - \dfrac{1}{n^2_i}\right)\right| = \dfrac{hc}{\lambda}[/latex]

or

[latex]\dfrac{1}{\lambda} = \left|\dfrac{-k}{hc} \left(\dfrac{1}{n^2_f} - \dfrac{1}{n^2_i}\right)\right|[/latex]

which is identical to the Rydberg equation with [latex]R_{\infty} = –\dfrac{k}{hc}[/latex].

When Bohr calculated the Rydberg constant,

[latex]R_{\infty} = \dfrac{2.179\;\times\;10^{–18}\;\text{J}}{(6.626\;\times\;10^{–34}\;\text{Js})(2.998\;\times\;10^{8}\;\text{m/s})}[/latex]

and compared it with the experimentally accepted value, he got excellent agreement. That is, the Bohr model was able to calculate the wavelengths of hydrogen-atom emission lines very accurately. This excellent agreement with experimental results implied that the Bohr model should be taken seriously.


Exercise 6: Hydrogen-Atom Electronic Transition

Calculate

  1. the change in energy of a hydrogen atom when an electron transitions from n = 7 to n = 3.
    • ΔE = [number] × 10[exponent] J
  2. the wavelength of the photon that results from this change.

Additional Practice
Is the photon emitted or absorbed? Compare the photon wavelength with the wavelength calculated using the Rydberg equation.

D2.4 Hydrogen-Atom Energy Levels

The lowest few energy levels for a hydrogen atom are shown in Figure 2.

An atom is most stable when it has the lowest possible energy. Thus, the electron in a hydrogen atom usually has n = 1, in which case the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to a higher n value (higher energy) and the atom is now in an excited electronic state (or simply an excited state).

If a photon is absorbed by an atom, the energy of the photon moves the electron from a lower energy level up to a more excited one. When an atom is in an excited state the extra energy can be released as one quantum if the electron returns to its ground state (say, from n = 5 to n = 1), or it can be released as two or more smaller quanta if the electron falls to an intermediate state then to the ground state (say, from n = 5 to n = 4, emitting one photon, then to n = 1, emitting a second photon).

he figure includes a diagram representing the relative energy levels determined by the quantum numbers of the hydrogen atom. An upward pointing arrow at the left of the diagram is labeled, “E.” Long, horizontal line segments are placed to the right of the arrow. Labels to the right of each line segment are arranged in columns with the headings, “n” and “Energy.” The bottom horizontal line segment is labeled, “1” and “-2.179 aj.” At a level approximately three-quarters of the distance to the top a horizontal line segment is labeled “2” and “-0.545 aJ.” At a level approximately seven-eighths the distance from the bottom a green horizontal line segment is labeled “3” and “-0.242 aJ.” Just a short distance above this segment a line segment is labeled “4” and “-0.136 aJ.” Just above this segment a horizontal line segment is labeled “5” and “-0.097 aJ.” Just a short distance above this segment a horizontal line segment is labeled “infinity” and “0.0 aJ.” Arrows are drawn to depict energies of photons absorbed, as shown by upward pointing arrows on the left, or released as shown by downward pointing arrows on the right side of the diagram between the horizontal line segments. The label, “Electron moves to higher energy as light is absorbed,” is placed beneath the upward pointing arrows. Similarly, the label, “Electron moves to lower energy as light is emitted,” appears beneath the downward pointing arrows. Moving left to right across the diagram, arrows extend from one horizontal line segment to the next in the following order: 1 to 2, 1 to 3, 1 to 4, 1 to 5, 1 to infinity, 2 to 3, 2 to 4, and 2 to 5. The arrows originating from the segment labeled 1 are grouped together by close placement of the arrows. Similarly, the downward arrows follow in this sequence; infinity to a, 5 to 1, 4 to 1, 3 to 1, 2 to 1, 5 to 2, 4 to 2, and 3 to 2. Arrows are again grouped by close placement according to the horizontal segment at which the arrows end. Three of the downward pointing arrows are colored. The one from 5 to 2 is violet; the one from 4 to 2 is cyan; the one from 3 to 2 is red. These colors correspond to the colors of the visible lines in the H atom spectrum at 434.1 nm, 486.2 nm, and 656.4 nm.
Figure 2. The horizontal lines show the relative energy of orbits in the Bohr model of the hydrogen atom, and the vertical arrows depict the energy of photons absorbed (left) or emitted (right) as electrons move between these orbits.

As an electron’s energy increases (as n increases), the electron is found at greater distances from the nucleus (the radius, r, of the Bohr orbit increases). This is a corollary of Coulomb’s law: if the electron is farther from the nucleus, the electrostatic attraction between electron and nucleus decreases, and the electron is held less tightly in the atom. Note that as n gets larger and the electron gets farther from the nucleus, energies get closer to zero, and so the limits [latex]n \longrightarrow \infty[/latex] and [latex]r \longrightarrow \infty[/latex] imply that E = 0 corresponds to ionization of the H atom—complete removal of the electron from the nucleus. Thus, for hydrogen in the ground state n = 1, the ionization energy is:

[latex]\Delta E = E_{n \rightarrow \infty} - E_1= 0 – \dfrac{-k}{1^2} = k = 2.179\;\times\;10^{–18}\;\text{J}[/latex]

D2.5 Bohr Model Summary

Bohr’s model of the hydrogen atom includes several important features of all models used to describe the distribution of electrons in an atom:

  • The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed values and used to characterize the arrangement of electrons in an atom.
  • An electron’s energy increases with increasing distance from the nucleus.
  • The discrete energies (lines) in the spectra of the elements result from electronic transitions among quantized energy levels.

Bohr’s model works very well, as long as there is only a single electron outside an atomic nucleus. For example, it can also be applied to the single-electron ions He+, Li2+, Be3+, and so forth, which differ from hydrogen only in their nuclear charges. Such single-electron ions are collectively referred to as hydrogen-like ions. The energy expression for hydrogen-like atoms/ions is a generalization of the hydrogen-atom energy, in which Z is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and k has a value of 2.179 × 10–18 J.

[latex]E_n = \dfrac{Z^2\;k}{n^2}[/latex]

Unfortunately, neither Bohr nor anyone else was able to extend his theory to the next simplest atom, He, which has two electrons or to the rest of the atoms in the periodic table. The only line spectra that could be explained using the Bohr model were for species with a single electron. The quantum ideas that accounted for blackbody radiation, the photoelectric effect, and the line spectra of hydrogen-like atom had to be extended further, which involved a quantum mechanical model of the atom.

Activity 5: Wrap-up—Atomic Spectra and Bohr Theory

Review what you wrote about atomic spectra and the Bohr theory at the beginning of this section. Update the information based on what you have learned. Write a summary that will be a good study aid when you review for an exam. If you still have things that puzzle you, ask about them on Piazza.

Podia Question

In the photoelectric effect animation, electrons emitted from the metal surface moved faster (had greater energy) the shorter the wavelength of the incident light was. The graph at the right expresses this quantitatively, plotting electron energy vs frequency of light for a particular metal. For each question below show all mathematical work and explain briefly and clearly, using scientifically appropriate language, how you arrive at each result.

  • Write an equation that relates electron energy, Eelec, threshold energy, Ethresh (the minimum energy required to eject an electron), and photon energy, Ephot.
  • Based on the equation you wrote, figure out the value of the slope of the graph where it is non-zero. (Base your answer on the equation you wrote, not on the graph.)
  • For light with wavelength 400 nm the electron energy is found to be 5.76 × 10−20 J. Calculate the minimum energy required to eject an electron from the surface of the metal.

Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.

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Chemistry 109 Copyright © by John Moore; Jia Zhou; and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.