,CHEM 103 Module 6 | Practice Q&A | 2026/2027 | Chemistry |
Portage Learning | 100% PASS
1. What is the energy of a photon with a frequency of 5.0 × 10¹⁴ s⁻¹? (h =
6.626 × 10⁻³⁴ J·s)
A) 3.31 × 10⁻¹⁹ J
B) 6.63 × 10⁻¹⁹ J
C) 1.33 × 10⁻¹⁸ J
D) 3.31 × 10⁻²⁰ J
Correct Answer: 3.31 × 10⁻¹⁹ J
Rationale: The energy of a photon is calculated using E = hν. Substituting the
values: E = (6.626 × 10⁻³⁴ J·s)(5.0 × 10¹⁴ s⁻¹) = 3.31 × 10⁻¹⁹ J. This equation
is fundamental to quantum theory.
2. What is the wavelength of a photon with a frequency of 6.0 × 10¹⁴ s⁻¹? (c
= 3.00 × 10⁸ m/s)
A) 5.0 × 10⁻⁷ m
B) 1.8 × 10²³ m
C) 2.0 × 10⁶ m
D) 5.0 × 10⁻⁹ m
Correct Answer: 5.0 × 10⁻⁷ m
Rationale: Wavelength and frequency are related by c = λν. Solving for λ: λ =
c/ν = (3.00 × 10⁸ m/s)/(6.0 × 10¹⁴ s⁻¹) = 5.0 × 10⁻⁷ m. This wavelength is in
the visible region of the electromagnetic spectrum.
3. As the wavelength of electromagnetic radiation increases, the frequency:
,A) Increases
B) Decreases
C) Remains constant
D) Doubles
Correct Answer: Decreases
Rationale: Wavelength and frequency are inversely proportional (c = λν). As
wavelength increases, frequency decreases. This relationship is fundamental
to understanding the electromagnetic spectrum.
4. As the frequency of electromagnetic radiation increases, the energy:
A) Increases
B) Decreases
C) Remains constant
D) Doubles
Correct Answer: Increases
Rationale: Energy is directly proportional to frequency (E = hν). Higher
frequency radiation (such as ultraviolet) has more energy than lower
frequency radiation (such as infrared).
5. What is the principal quantum number (n) for an electron in the 3d orbital?
A) 1
B) 2
C) 3
D) 4
, Correct Answer: 3
Rationale: The principal quantum number (n) indicates the energy level or
shell. For the 3d orbital, n = 3. The number before the letter indicates the
principal quantum number.
6. What is the angular momentum quantum number (l) for an electron in an s
orbital?
A) 0
B) 1
C) 2
D) 3
Correct Answer: 0
Rationale: The angular momentum quantum number (l) defines the subshell
shape. For s orbitals, l = 0. p orbitals have l = 1, d orbitals have l = 2, and f
orbitals have l = 3.
7. What is the angular momentum quantum number (l) for an electron in a p
orbital?
A) 0
B) 1
C) 2
D) 3
Correct Answer: 1
Portage Learning | 100% PASS
1. What is the energy of a photon with a frequency of 5.0 × 10¹⁴ s⁻¹? (h =
6.626 × 10⁻³⁴ J·s)
A) 3.31 × 10⁻¹⁹ J
B) 6.63 × 10⁻¹⁹ J
C) 1.33 × 10⁻¹⁸ J
D) 3.31 × 10⁻²⁰ J
Correct Answer: 3.31 × 10⁻¹⁹ J
Rationale: The energy of a photon is calculated using E = hν. Substituting the
values: E = (6.626 × 10⁻³⁴ J·s)(5.0 × 10¹⁴ s⁻¹) = 3.31 × 10⁻¹⁹ J. This equation
is fundamental to quantum theory.
2. What is the wavelength of a photon with a frequency of 6.0 × 10¹⁴ s⁻¹? (c
= 3.00 × 10⁸ m/s)
A) 5.0 × 10⁻⁷ m
B) 1.8 × 10²³ m
C) 2.0 × 10⁶ m
D) 5.0 × 10⁻⁹ m
Correct Answer: 5.0 × 10⁻⁷ m
Rationale: Wavelength and frequency are related by c = λν. Solving for λ: λ =
c/ν = (3.00 × 10⁸ m/s)/(6.0 × 10¹⁴ s⁻¹) = 5.0 × 10⁻⁷ m. This wavelength is in
the visible region of the electromagnetic spectrum.
3. As the wavelength of electromagnetic radiation increases, the frequency:
,A) Increases
B) Decreases
C) Remains constant
D) Doubles
Correct Answer: Decreases
Rationale: Wavelength and frequency are inversely proportional (c = λν). As
wavelength increases, frequency decreases. This relationship is fundamental
to understanding the electromagnetic spectrum.
4. As the frequency of electromagnetic radiation increases, the energy:
A) Increases
B) Decreases
C) Remains constant
D) Doubles
Correct Answer: Increases
Rationale: Energy is directly proportional to frequency (E = hν). Higher
frequency radiation (such as ultraviolet) has more energy than lower
frequency radiation (such as infrared).
5. What is the principal quantum number (n) for an electron in the 3d orbital?
A) 1
B) 2
C) 3
D) 4
, Correct Answer: 3
Rationale: The principal quantum number (n) indicates the energy level or
shell. For the 3d orbital, n = 3. The number before the letter indicates the
principal quantum number.
6. What is the angular momentum quantum number (l) for an electron in an s
orbital?
A) 0
B) 1
C) 2
D) 3
Correct Answer: 0
Rationale: The angular momentum quantum number (l) defines the subshell
shape. For s orbitals, l = 0. p orbitals have l = 1, d orbitals have l = 2, and f
orbitals have l = 3.
7. What is the angular momentum quantum number (l) for an electron in a p
orbital?
A) 0
B) 1
C) 2
D) 3
Correct Answer: 1
