Quantum Theory and Electronic Structure — Bohr Model, Determine Photon Energy in Joules
Question
Using the Bohr model:
- Determine the change in energy (in joules) when the electron in a Li2+ ion moves from the n = 2 to the n = 1 orbit.
- What is the energy (in joules) of the photon produced?
Show/Hide Answer
[latex]\Delta \mathrm{E}=-1.47 \times 10^{-17} \mathrm{J}[/latex]
[latex]\mathrm{E}_{\text {photon }}=1.47 \times 10^{-17} \mathrm{~J}[/latex]
Refer to Section 2.2: Atomic Spectroscopy and The Bohr Model (1).
Strategy Map
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Check out the strategy map.
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| Strategy Map Steps |
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1. Identify the information (data and/or variables) given the question.
Show/Hide Hintni=2, nf=1 (initial and final energy levels). |
2. Identify what the question is asking for; what variable would represent it in a formula (E, λ, n, etc.).
Show/Hide HintYou are looking for ΔE (total energy change). |
3. Choose a formula that allows you to plug in your given information. Fill in any known constant values.
Show/Hide HintTo check if you are using the correct formula, ask yourself:
[latex]\Delta \mathrm{E}=-2.18 \times 10^{-18} \mathrm{J}\left(\frac{\mathrm{Z}^2}{\mathrm{n_f}^2}-\frac{\mathrm{Z}^2}{\mathrm{n_i}^2}\right)[/latex] [latex]\begin{aligned} &\mathrm{n_i} = 2\\ &\mathrm{n_f} = 1\\ &\Delta \mathrm{E}=? \end{aligned}[/latex] Rydberg’s Constant (RH)=2.18 x 10-18 J Z = Atomic Number = 3 Note that you can find the atomic number on the periodic table. Most questions consider ‘hydrogen’s electron’ because hydrogen has an atomic number of Z=1. The formula is often simplified to: [latex]\Delta \mathrm{E}=\mathrm{-R_H}\left(\frac{1}{\mathrm{n_f}^2}-\frac{1}{\mathrm{n_i}^2}\right)[/latex] |
4. Calculate the final answer. Do not forget about sign, units and significant figures.
Show/Hide Watch Out!Be sure to follow typical math rules such as order of operations. Recall the Order of Operations. Refer to Section 1.3: The Order of Operations (2). Show/Hide HintStudents commonly make mistakes when filling in the formula. When using nonhydrogen atoms, students often forget to include the atomic number (Z) in the numerator of both fractions. In this example, the atomic number is for Li2+ is 3. |
Solution
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Check out this solution.
Show/Hide Solution
1. Change in energy (in joules) when the electron in a Li2+ ion moves from the n = 2 to the n = 1 orbit:
[latex]\begin{gathered} \Delta \mathrm{E=-R_H}\left(\frac{\mathrm{Z^2}}{\mathrm{n_f^2}}-\frac{\mathrm{Z^2}}{\mathrm{n_i^2}}\right) \\\\ \Delta \mathrm{E=-2.18} \times 10^{-18} \mathrm{J}\left(\frac{\mathrm{Z^2}}{\mathrm{n_f^2}}-\frac{\mathrm{Z^2}}{\mathrm{n_i^2}}\right) \\\\ \Delta \mathrm{E=-2.18} \times 10^{-18} \mathrm{J}\left(\frac{3^2}{1^2}-\frac{3^2}{2^2}\right) \\\\ \Delta \mathrm{E=-2.18} \times 10^{-18} \mathrm{J}\left(\frac{9}{1}-\frac{9}{4}\right) \\\\ \Delta \mathrm{E=-2.18} \times 10^{-18} \mathrm{J}\left(\frac{27}{4}\right) \\\\ \Delta \mathrm{E=-2.18} \times 10^{-18} \mathrm{J}(6.25) \\\\ \Delta \mathrm{E=-1.4715} \times 10^{-17} \mathrm{J} \\\\ \Delta \mathrm{E=-1.47} \times 10^{-17} \mathrm{~J} \end{gathered}[/latex]
2. Energy (in joules) of the photon produced:
Energy of the photon = Ephoton = |ΔEelectron|
[latex]\begin{gathered} \Delta \mathrm{E}_{\text {electron }}=-1.47 \times 10^{-17}\mathrm{~J} \end{gathered}[/latex]
[latex]\begin{gathered} \mathrm{E}_{\text {photon }}=1.47 \times 10^{-17}\mathrm{~J} \end{gathered}[/latex]
Guided Solution
The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
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| Guided Solution Ideas |
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This question is a calculation problem where you use the Rydberg Equation to calculate the energy of the transition of the electron.
Show/Hide Resource
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| Using the Bohr model,
1. Determine the change in energy (in joules) when the electron in a Li2+ ion moves from the n = 2 to the n = 1 orbit. Show/Hide Don’t Forget!change in energy = ΔE 2. What is the energy (in joules) of the photon produced? Show/Hide Don’t Forget!Ephoton = |ΔE| |
Summarize what the question is asking.
Show/Hide Think About This!You are looking for the change in energy of the Li2+ electron when it transitions from n=2 to n=1. The energy change is exactly the same (quantized) amount as what is emitted (i.e., as a photon). |
Based on your understanding of the related chemistry concepts, predict what you expect the answer to be.
Show/Hide Think About This!Since the Li2+ ion is going from a higher energy orbital (n=2) to a lower energy orbital (n=1), you can expect the answer to be negative. When energy is emitted, it is released or lost from the system; ΔE is a negative value. When energy is absorbed, it is added to the system; ΔE is a positive value. |
| Complete Solution |
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Recall the Rydberg equation. Show/Hide Don’t Forget!The Rydberg equation allows you to calculate the overall change in an atom’s energy, given its initial(ni) and final(nf) energy values. [latex]\Delta \mathrm{E=-2.18 \times 10^{-18} J\left(\frac{Z^2}{n_f^2}-\frac{Z^2}{n_i^2}\right)}[/latex] |
Input your data provided by the question.
Show/Hide Watch Out!Students commonly make mistakes here. When using nonhydrogen atoms, students often forget to include the atomic number. [latex]\Delta \mathrm{E=-2.18 \times 10^{-18} J\left(\frac{3^2}{1^2}-\frac{3^2}{2^2}\right)}[/latex]
Square the numerator and denominator of both fractions. |
| Start by solving inside the brackets.
[latex]\Delta \mathrm{E=-2.18 \times 10^{-18} J\left(\frac{9}{1}-\frac{9}{4}\right)}[/latex] Combine the fractions by subtracting them. [latex]\Delta \mathrm{E=-2.18 \times 10^{-18} J\left(\frac{27}{4}\right)}[/latex] Optional: Convert your fraction before the multiplication step. |
| Multiply your constant RH and the number inside the brackets.
[latex]\begin{aligned} & \Delta \mathrm{E}=-1.4715 \times 10^{-17}\mathrm{ J} \\ & \Delta \mathrm{E}=-1.47 \times 10^{-17} \mathrm{J} \end{aligned}[/latex] Notice that there is a negative sign in front of the constant value. |
Recall the relationship between the energy change for the transition of an electron to the energy of the photon that is absorbed or emitted during that change.
Show/Hide Think About This!The energy of a photon must exactly match the energy change resulting from the transition of the electron from one energy level (orbit) to another (orbit). A photon is a quantized packet of energy. When an electron transitions from a low to a high energy level, the electron must absorb energy (ΔEelectron = positive) from a photon with the same amount of energy required for that transition. When an electron transitions from a high to a low energy level, the electron must release energy (ΔEelectron = negative). The energy lost from the electron is emitted as a photon with the same amount of energy released during that transition. Energy of the photon = Ephoton = |ΔEelectron| [latex]\Delta \mathrm{E}_{\text {electron }}=-1.47 \times 10^{-17}\mathrm{~J}[/latex] therefore, [latex]\mathrm{E}_{\text {photon }}=1.47 \times 10^{-17}\mathrm{~J}[/latex] |
Check Your Work
Comparing the bond energy values, we see that the product bond (2 H-Br) have a greater magnitude than the reactant bonds (H-H and Br-Br), so we would expect this reaction to have an exothermic ΔH.
Show/Hide Watch Out!
Make sure you add up all bonds and consider the reaction stoichiometry.
Does your answer make chemical sense?
Show/Hide Answer
When any reaction occurs, energy will be absorbed and released; however, the overall sum can be endothermic (requires more energy than it releases) or exothermic (releases more energy than it requires). You can numerically observe this using the enthalpy change from the bond energies equation.
PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (3).
- Question 2.E.26 from LibreTexts PASS Chemistry Book CHEM 1500 (4) is used under a CC BY-NC-SA 4.0 license.
- Question 2.E.26 was adapted from question 6.2.11 from LibreTexts Chemistry 1e (OpenSTAX) (5), which is under a CC BY 4.0 license.
- Question 6.2.11 is question 11 from OpenStax Chemistry 2e (6), which is under a CC BY 4.0 license. Access for free at https://openstax.org/books/chemistry-2e/pages/1-introduction.
References
1. LibreTexts. 2.2: Atomic Spectroscopy and The Bohr Model. In CHEM1500: Chemical Bonding and Organic Chemistry. LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/CHEM1500%3A_Chemical_Bonding_and_Organic_Chemistry/02%3A_Quantum_Theory_and_Electronic_Structure_of_Atoms/2.02%3A_Atomic_Spectroscopy_and_The_Bohr_Model.
2. Elhitti, S.; Bonanome, M.; Carley, H.; Tradler, T.; Zhou, L. 1.3: Order of Operations. In Arithmetic and Algebra (ElHitti, Bonanome, Carley, Tradler, and Zhou). LibreTexts. 2021. https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Arithmetic_and_Algebra_(ElHitti_Bonanome_Carley_Tradler_and_Zhou)/01%3A_Chapters/1.03%3A_The_Order_of_Operations#:~:text=’PE(MD)(AS),together%20from%20left%20to%20right).
3. Blackstock, L.; Brewer, S.; Jensen, A. PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.
4. Blackstock, L.; Brewer, S.; Jensen, A. 2.2: Question 2.E.26 PASS – Bohr Model, Quantized Energy Change. In PASS Chemistry Book CHEM 1500. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500/02%3A_Quantum_Theory_and_Electronic_Structure/2.02%3A_Question_2.E.26_PASS_-_Bohr_Model_quantized_energy_change.
5. OpenStax. 6.E: Electronic Structure and Periodic Properties (Exercises). In Chemistry 1e (OpenSTAX). LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX).
6. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 6 Exercises. In Chemistry 2e; OpenStax, 2019. https://openstax.org/books/chemistry-2e/pages/6-exercises.
