Quantized Energy Levels and Atomic Spectra Study Guide

Learning Objectives & Prerequisites

  • Prerequisites:

    • Required reading prior to beginning activity: Sections 12.3–12.4 in Chemical Principles.

    • Course context: CHEM 1311.

  • Core Learning Objectives:

    • Explain the physical meaning of energy quantization.

    • Explain how atomic emission spectra relate to quantized energy levels in atoms.

    • Calculate discrete energy levels, energy level changes (ΔE\Delta E), and corresponding photon emission wavelengths for single-electron atomic systems.

Atomic Spectra: Continuous vs. Line Spectra

  • Continuous Spectra:

    • Generated when white light (such as light from the sun or an incandescent lightbulb) passes through a prism.

    • Produces an unbroken band containing every wavelength of visible light, resembling a rainbow.

  • Atomic Emission Spectra (Line Spectra):

    • Generated when a pure gaseous element is heated or electrically excited.

    • A common commercial example is the vibrant red emission produced by neon gas in open/closed neon signs.

    • Emits light containing only specific, discrete wavelengths rather than a continuous spectrum.

    • Represented as sharp vertical lines on a spectrum; each line corresponds to a distinct wavelength emitted by the element.

    • Dark spaces located between spectral lines represent all omitted or non-emitted wavelengths.

    • The line spectrum of every element is unique, functioning as a spectral fingerprint for identification.

    • Spectral patterns are a direct physical consequence of an element's distinct electronic structure.

Quantization of Energy in Single-Electron Systems

  • Principle of Quantization:

    • The energy of an electron within an atom is quantized, meaning it can only possess specific, allowed energy values.

    • Intermediate energy values between allowed states are strictly forbidden.

    • Allowed energy states are termed energy levels and are categorized using quantum numbers.

  • Single-Electron Species:

    • Applies to species containing exactly one electron, such as hydrogen (H\text{H}), singly ionized helium (He+\text{He}^+), and doubly ionized lithium (Li2+\text{Li}^{2+}).

    • Energy levels are identified using positive integers starting with n=1n = 1 for the lowest-energy, most stable state.

    • The integer nn is designated as the principal quantum number.

  • Spatial Distribution relative to the Nucleus:

    • Electrons occupying lower energy levels (smaller nn) reside physically closer to the atomic nucleus.

    • Electrons occupying higher energy levels (larger nn) reside farther from the nucleus.

The energy levels of a hydrogen atom and their relationship to the atomic emission spectrum.

Mechanism of Photons and Energy Transitions

  • Energy Level Transitions:

    • For an electron to move between energy levels, it must absorb or release an amount of energy equal to the exact mathematical difference between its initial and destination levels.

    • Energy transfer occurs via the absorption or emission of light particles known as photons.

  • Conservation of Energy Requirements:

    • The energy carried by a photon (Ephoton=hν=hcλE_{\text{photon}} = h \nu = \frac{hc}{\lambda}) must strictly equal the magnitude of energy change (∣ΔE∣|\Delta E|) between the initial and final electron energy levels.

  • Origin of Atomic Line Spectra:

    • Heating or energizing an element promotes electrons into higher, excited energy levels.

    • Electrons cannot remain indefinitely in higher excited levels and eventually relax back down to lower energy levels.

    • Relaxation causes the electron to release the exact difference in energy as an emitted photon.

    • Because energy levels are quantized, emitted photons are restricted to discrete energies and wavelengths, yielding the observed lines in atomic line spectra.

A change between two discrete energy levels emits a photon of light.

Quantitative Formulas for Single-Electron Systems

  • Energy Level Calculation Formula:

    • For any single-electron atom or ion, the energy in joules (J\text{J}) of a given energy level nn is given by:     En=(−2.178×10−18 J)(Z2n2)E_n = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{Z^2}{n^2} \right)

    • Definitions of variables:

    • EnE_n: Energy of the electron at principal quantum level nn (measured in joules, J\text{J}).

    • ZZ: Atomic number of the species (Z=1Z = 1 for hydrogen H\text{H}, Z=2Z = 2 for He+\text{He}^+, Z=3Z = 3 for Li2+\text{Li}^{2+}).

    • nn: Principal quantum number (n=1,2,3,…n = 1, 2, 3, \dots).

  • Ground State vs. Excited State Definitions:

    • Ground State: The lowest possible energy level (n=1n = 1).

    • Excited State: Any higher energy level (n>1n > 1) attained by absorbing energy.

  • Transition Energy Change Formula:

    • The net change in energy when an electron transitions between states is:     ΔE=Efinal−Einitial\Delta E = E_{\text{final}} - E_{\text{initial}}

  • Sign Conventions for Energy Flow:

    • Negative ΔE\Delta E (ΔE<0\Delta E < 0): Indicates energy is released/emitted by the atom as a photon (relaxation from higher nn to lower nn).

    • Positive ΔE\Delta E (ΔE>0\Delta E > 0): Indicates energy must be absorbed by the atom to promote the electron (excitation from lower nn to higher nn).

    • Photon Energy: Photon energy EphotonE_{\text{photon}} is inherently positive (Ephoton=∣ΔE∣E_{\text{photon}} = |\Delta E|).

Model 1: Worked Step-by-Step Calculation of Energy Change (\Delta E)

  • Scenario: Calculation of ΔE\Delta E for an electron transition in a hydrogen atom (Z=1Z = 1) moving from n=5n = 5 to n=2n = 2.

Model 1 calculation table for an electron transition in hydrogen from n=5 to n=2.

  • Step 1: Identify initial and final energy levels

    • Initial level (where electron begins): n=5n = 5

    • Final level (where electron ends): n=2n = 2

  • Step 2: Calculate energy at initial level (n=5n = 5)

    • E5=(−2.178×10−18 J)(1252)E_5 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{1^2}{5^2} \right)

    • E5=(−2.178×10−18 J)(125)=−8.712×10−20 JE_5 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{1}{25} \right) = -8.712 \times 10^{-20}\,\text{J}

  • Step 3: Calculate energy at final level (n=2n = 2)

    • E_2 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{1^2}{2^2} \n\right)

    • E2=(−2.178×10−18 J)(14)=−5.445×10−19 JE_2 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{1}{4} \right) = -5.445 \times 10^{-19}\,\text{J}

  • Step 4: Calculate total energy change (ΔE=E2−E5\Delta E = E_2 - E_5)

    • ΔE=(−5.445×10−19 J)−(−8.712×10−20 J)=−4.574×10−19 J\Delta E = (-5.445 \times 10^{-19}\,\text{J}) - (-8.712 \times 10^{-20}\,\text{J}) = -4.574 \times 10^{-19}\,\text{J}

  • Sign Analysis:

    • The negative sign indicates energy release during relaxation.

    • The reverse transition (n=2→n=5n = 2 \rightarrow n = 5) requires energy absorption equal to +4.574×10−19 J+4.574 \times 10^{-19}\,\text{J}.

Key Questions, Conceptual Exercises, and Practice Problems

  • Key Questions and Solutions:

    • Question 1: What feature of an atomic spectrum indicates that the energy emitted by heating an element is not continuous?

    • Answer: The spectrum consists of distinct, individual lines separated by blank spaces (omitted wavelengths), rather than a continuous rainbow band.

    • Question 2: Complete the following sentences:

    • a. Electrons can jump to higher energy levels when they absorb a photon.

    • b. Electrons can drop to lower energy levels when they emit a photon.

    • Question 3: What is the equation that gives the possible energies (EnE_n) of an electron in the hydrogen atom?

    • Answer: En=(−2.178×10−18 J)(12n2)=−2.178×10−18 Jn2E_n = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{1^2}{n^2} \right) = \frac{-2.178 \times 10^{-18}\,\text{J}}{n^2}

    • Question 4: How can the wavelength of a photon be calculated from its energy?

    • Answer: Using the equation Ephoton=hcλE_{\text{photon}} = \frac{hc}{\lambda}, rearranged to solve for wavelength: λ=hcEphoton\lambda = \frac{hc}{E_{\text{photon}}}, where h=6.626×10−34 J sh = 6.626 \times 10^{-34}\,\text{J\,s} and c=2.998×108 m/sc = 2.998 \times 10^8\,\text{m/s}.

  • Exercises and Detailed Solutions:

    • Exercise 1: Consider the transition that occurs when the electron in a hydrogen atom drops from n=2n = 2 to n=1n = 1, compared to the transition from n=3n = 3 to n=1n = 1

    • Part a: Which transition causes the larger change in the energy of the hydrogen atom?

      • Answer: The transition from n=3n = 3 to n=1n = 1 causes a larger change in energy because the energy spacing between n=3n = 3 and n=1n = 1 is greater than between n=2n = 2 and n=1n = 1

    • Part b: Which transition will produce light with a longer wavelength?

      • Answer: The transition from n=2n = 2 to n=1n = 1 produces light with a longer wavelength. Smaller energy transition (ΔE\Delta E) corresponds to a photon of lower frequency and longer wavelength (λ∝1ΔE\lambda \propto \frac{1}{\Delta E}).

    • Exercise 2: Consider a He+\text{He}^+ ion (Z=2Z = 2) with an electron in the n=3n = 3 level.

    • Part a: Calculate the energy of the n=3n = 3 level.

      • Calculation: E3=(−2.178×10−18 J)(2232)=(−2.178×10−18 J)(49)=−9.680×10−19 JE_3 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{2^2}{3^2} \right) = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{4}{9} \right) = -9.680 \times 10^{-19}\,\text{J}

    • Part b: The lowest energy emission from n=3n = 3 will move the electron to which level?

      • Answer: The n=2n = 2 level. The lowest energy emission corresponds to the smallest energy difference dropping down, which is to the adjacent lower level (n=2n = 2).

    • Part c: Calculate the wavelength (in nm\text{nm}) of the photon associated with the lowest energy emission from the n=3n = 3 level.

      • Calculation Steps:

      • Energy at n=2n = 2 level for He+\text{He}^+: E2=(−2.178×10−18 J)(2222)=−2.178×10−18 JE_2 = (-2.178 \times 10^{-18}\,\text{J}) \left( \frac{2^2}{2^2} \right) = -2.178 \times 10^{-18}\,\text{J}

      • Energy difference: ΔE=E2−E3=(−2.178×10−18 J)−(−9.680×10−19 J)=−1.210×10−18 J\Delta E = E_2 - E_3 = (-2.178 \times 10^{-18}\,\text{J}) - (-9.680 \times 10^{-19}\,\text{J}) = -1.210 \times 10^{-18}\,\text{J}

      • Photon energy: Ephoton=1.210×10−18 JE_{\text{photon}} = 1.210 \times 10^{-18}\,\text{J}

      • Wavelength: λ=hcEphoton=(6.626×10−34 J s)(2.998×108 m/s)1.210×10−18 J=1.641×10−7 m=164.1 nm\lambda = \frac{hc}{E_{\text{photon}}} = \frac{(6.626 \times 10^{-34}\,\text{J\,s})(2.998 \times 10^8\,\text{m/s})}{1.210 \times 10^{-18}\,\text{J}} = 1.641 \times 10^{-7}\,\text{m} = 164.1\,\text{nm}

  • Practice Problems and Detailed Solutions:

    • Problem 1: In a hydrogen atom (Z=1Z = 1), what energy level is associated with an energy of −1.36×10−19 J-1.36 \times 10^{-19}\,\text{J}?

    • Calculation:

      • En=−2.178×10−18 Jn2=−1.36×10−19 JE_n = \frac{-2.178 \times 10^{-18}\,\text{J}}{n^2} = -1.36 \times 10^{-19}\,\text{J}

      • n2=−2.178×10−18 J−1.36×10−19 J≈16.01n^2 = \frac{-2.178 \times 10^{-18}\,\text{J}}{-1.36 \times 10^{-19}\,\text{J}} \approx 16.01

      • n=16.01=4n = \sqrt{16.01} = 4

      • Answer: The energy level is n=4n = 4

    • Problem 2: One of the visible lines in the hydrogen emission spectrum corresponds to the n=6n = 6 to n=2n = 2 electron transition. What color light is this transition?

    • Calculation:

      • E6=−2.178×10−18 J62=−6.050×10−20 JE_6 = \frac{-2.178 \times 10^{-18}\,\text{J}}{6^2} = -6.050 \times 10^{-20}\,\text{J}

      • E2=−2.178×10−18 J22=−5.445×10−19 JE_2 = \frac{-2.178 \times 10^{-18}\,\text{J}}{2^2} = -5.445 \times 10^{-19}\,\text{J}

      • ΔE=E2−E6=(−5.445×10−19 J)−(−6.050×10−20 J)=−4.840×10−19 J\Delta E = E_2 - E_6 = (-5.445 \times 10^{-19}\,\text{J}) - (-6.050 \times 10^{-20}\,\text{J}) = -4.840 \times 10^{-19}\,\text{J}

      • Ephoton=4.840×10−19 JE_{\text{photon}} = 4.840 \times 10^{-19}\,\text{J}

      • λ=hcEphoton=(6.626×10−34 J s)(2.998×108 m/s)4.840×10−19 J=4.10×10−7 m=410 nm\lambda = \frac{hc}{E_{\text{photon}}} = \frac{(6.626 \times 10^{-34}\,\text{J\,s})(2.998 \times 10^8\,\text{m/s})}{4.840 \times 10^{-19}\,\text{J}} = 4.10 \times 10^{-7}\,\text{m} = 410\,\text{nm}

      • Answer: A wavelength of 410 nm410\,\text{nm} corresponds to violet (or blue-violet) visible light.

    • Problem 3: An electron in the n=7n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light with a wavelength of 397 nm397\,\text{nm}. What is the value of nn for the level to which the electron relaxed?

    • Calculation:

      • Ephoton=hcλ=(6.626×10−34 J s)(2.998×108 m/s)397×10−9 m=5.003×10−19 JE_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}\,\text{J\,s})(2.998 \times 10^8\,\text{m/s})}{397 \times 10^{-9}\,\text{m}} = 5.003 \times 10^{-19}\,\text{J}

      • Because energy is emitted, ΔE=−5.003×10−19 J\Delta E = -5.003 \times 10^{-19}\,\text{J}

      • E7=−2.178×10−18 J72=−4.445×10−20 JE_7 = \frac{-2.178 \times 10^{-18}\,\text{J}}{7^2} = -4.445 \times 10^{-20}\,\text{J}

      • Efinal=E7+ΔE=(−4.445×10−20 J)+(−5.003×10−19 J)=−5.447×10−19 JE_{\text{final}} = E_7 + \Delta E = (-4.445 \times 10^{-20}\,\text{J}) + (-5.003 \times 10^{-19}\,\text{J}) = -5.447 \times 10^{-19}\,\text{J}

      • Efinal=−2.178×10−18 Jnfinal2=−5.447×10−19 JE_{\text{final}} = \frac{-2.178 \times 10^{-18}\,\text{J}}{n_{\text{final}}^2} = -5.447 \times 10^{-19}\,\text{J}

      • nfinal2=−2.178×10−18 J−5.447×10−19 J≈3.998n_{\text{final}}^2 = \frac{-2.178 \times 10^{-18}\,\text{J}}{-5.447 \times 10^{-19}\,\text{J}} \approx 3.998

      • nfinal=4=2n_{\text{final}} = \sqrt{4} = 2

      • Answer: The final relaxed energy level is n=2n = 2

Academic Literature References

  • Zumdahl, S., DeCoste, D., Chemical Principles; Cengage Learning, Inc.: Boston, MA, 2017; p. 445-451.

  • Hanson, D.; Goodwin, J.; Phillips, M. Foundations of Chemistry; Pacific Crest: Hampton, p. 142-146.

  • Timberlake, K.; Timberlake, W. Basic Chemistry; Prentice Hall: U.S.A., 2011; p. 125-128.