University:
Carleton College
Course:
CHEM 122 | Introduction to Chemistry
Academic year:

2022
Views:

83

Pages:

8
Author:

AbrariHermoo

Practice Problems: Wavelength, Frequency, Energy Content of One Quantum of Light Examples: A certain photon of light has a wavelength of 422 nm. What is the frequency of the light? Given: c = 3.00 x 10^8 m/s, λ = 422 nm = 422 x 10^-9 m Solution: c = λv v = c / λ = (3.00 x 10^8 m/s) / (422 x 10^-9 m) = 7.11 x 10^14 Hz What is the energy of a quantum of light from part 1? Given: h = 6.626 x 10^-34 J s Solution: E = hv = (6.626 x 10^-34 J s) * (7.11 x 10^14 Hz) = 4.71 x 10^-19 J Problems: What is the energy of a quantum of light with a frequency of 7.39 x 10^14 Hz? Solution: E = hv = (6.626 x 10^-34 J s) * (7.39 x 10^14 Hz) = 4.90 x 10^-19 J What is the wavelength of the quantum of light in question 1? Solution: c = λv λ = c / v = (3.00 x 10^8 m/s) / (7.39 x 10^14 Hz) = 4.06 x 10^-7 m = 406 nm A certain red light has a wavelength of 680 nm. What is the frequency of the light? Solution: c = λv v = c / λ = (3.00 x 10^8 m/s) / (680 x 10^-9 m) = 4.41 x 10^14 Hz 4. What is the energy of a quantum of light from question 3? Given: Frequency (v) = 4.41 x 10^14 Hz, Planck's constant (h) = 6.626 x 10^-34 J s Formula: E = hv Calculation: E = (6.626 x 10^-34 J s) * (4.41 x 10^14 Hz) = 2.92 x 10^-19 J 5. A certain blue light has a frequency of 6.91 x 10^14 Hz. What is the wavelength of the light? Given: Frequency (v) = 6.91 x 10^14 Hz, Speed of light (c) = 3.0 x 10^8 m/s Formula: c = λv Calculation: λ = c / v = (3.0 x 10^8 m/s) / (6.91 x 10^14 Hz) = 4.34 x 10^-7 m = 434 nm 6. What is the energy of a quantum of light from question 5? Given: Frequency (v) = 6.91 x 10^14 Hz, Planck's constant (h) = 6.626 x 10^-34 J s Formula: E = hv Calculation: E = (6.626 x 10^-34 J s) * (6.91 x 10^14 Hz) = 4.58 x 10^-19 J 7. The energy for a quantum of light is 2.84 x 10^-19 J. What is the wavelength of this light? Given: Energy (E) = 2.84 x 10^-19 J, Planck's constant (h) = 6.626 x 10^-34 J s, Speed of light (c) = 3.0 x 10^8 m/s Formula: E = hv and c = λv Calculation: First, find the frequency: v = E / h = (2.84 x 10^-19 J) / (6.626 x 10^-34 J s) = 4.29 x 10^14 Hz Then, find the wavelength: λ = c / v = (3.0 x 10^8 m/s) / (4.29 x 10^14 Hz) = 6.99 x 10^-7 m = 699 nm Information: Energy of Sublevels Rach sublevel bes a different amount of enrgy. For example, orbitals in the 3p subleval have more energy than orbitals in the 2p sublavel. The following is a list of the sublevels from lowest to highest energy: 18, 28, 2p, 38, 3p, 4a, 3d, 4p, 5a, 4d , 5р , ба, 44, 54, бр. 78, 51, 6d... To help you, here is the above list with the orbitals included. Recall that each blank represents an cubital: 18 , 28, 2p 38, 3, 4, 3, 4p, 58, 4, 5, 6 , 41 Note that d and f sublevels, appear to be out of place. This is because they have extra high energies. For example, the 3d sublevel has a higaor energy than a 4s sublevel and the 4f sublevel has a higher energy, dun the 6s sublevel. When electrons occupy orbitals, they try to have the lowest amount of energy possible. (This is called the Aufben Principle.) An electron will enter a 2s orbital only after the is sublevel is filled up and an electron will enter a 3d orbital only after the 4s sublevel is filled. Recall that only two electrons can fit in each orbitel. (This is called the Pauli Exclusion Principle.) When two electrons occupy the same orbital they must spin in opposite directions one clockwise and the other counterclockwise. Critical Thinking Questions 1. a) How many electrons would an atom need to have before it can begin filling the 3s subleval? 10 b) What is the first element that has enough electrons to have one in the 3s sublevel? (Use your periodic table.) 2. Na a) How many electrons would an atom need to have before it can begin filling the 3d sublevel? 20 b) What is the first elament that has enough electrons to begin placing electrons in the 3d sublevel? Sc