Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

9. Quantum Mechanics

The Energy of Light

General Chemistry

9. Quantum Mechanics

The Energy of Light

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Multiple Choice

An electron in a hydrogen atom relaxes to the n = 4 level, emitting light at 138 THz. What is the initial energy level of the electron before relaxation?

n = 5

n = 8

n = 7

n = 6

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Verified step by step guidance

Start by understanding the concept of electron transitions in a hydrogen atom. When an electron moves from a higher energy level to a lower one, it emits energy in the form of light. The frequency of this light is related to the energy difference between the two levels.

Use the formula for the frequency of emitted light during an electron transition: \( \nu = \frac{R_H}{h} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( \nu \) is the frequency, \( R_H \) is the Rydberg constant, \( h \) is Planck's constant, \( n_1 \) is the initial energy level, and \( n_2 \) is the final energy level.

Given that the frequency of the emitted light is 138 THz (or \( 138 \times 10^{12} \) Hz) and the final energy level \( n_2 \) is 4, substitute these values into the formula to solve for \( n_1 \).

Rearrange the formula to solve for \( n_1 \): \( \frac{1}{n_1^2} = \frac{1}{n_2^2} + \frac{h \cdot \nu}{R_H} \). Calculate the right-hand side using the given frequency and constants.

Compare the calculated \( \frac{1}{n_1^2} \) with the possible values for \( n_1 \) (5, 6, 7, 8) to determine which initial energy level corresponds to the given frequency of emitted light.