Question 1

The greater the energy of a photon,the
A)longer the wavelength and the higher the frequency.
B)longer the wavelength and the lower the frequency.
C)shorter the wavelength and the higher the frequency.
D)shorter the wavelength and the lower the frequency.

Accepted Answer

According to the equation E=hf (where E is the energy of the photon, h is Planck's constant, and f is the frequency), as the energy of the photon increases, the frequency must also increase to maintain the constant value of Planck's constant. Since c (the speed of light) is also constant, this means that the wavelength must decrease. Therefore, the correct choice is C: shorter wavelength and higher frequency.

Question 2

What is a quantum of light called?
A)the amplitude
B)the frequency
C)a photon
D)the wavelength

Accepted Answer

A quantum of light is called a photon.

Question 3

According to the Balmer-Rydberg equation,the transition from n = 6 to m = 2 results in a spectral line having which color?
A)blue
B)blue-green
C)indigo
D)red

Accepted Answer

C

Question 4

A quantized variable
A)can be continuously varied.
B)can only assume certain values.
C)consists of photons.
D)is extremely small.

Accepted Answer

A quantized variable can only assume certain specific, discrete values, rather than being continuously varied.

Question 5

Arrange the following spectral regions in order of increasing wavelength: infrared,microwave,ultraviolet,visible.
A)microwave < infrared < visible < ultraviolet
B)microwave < visible < infrared < ultraviolet
C)ultraviolet < infrared < visible < microwave
D)ultraviolet < visible < infrared < microwave

Accepted Answer

The correct order from shortest to longest wavelength is ultraviolet, visible, infrared, and then microwave. Ultraviolet light has shorter wavelengths than visible light, which is followed by infrared with longer wavelengths, and microwaves have the longest wavelengths among the options provided.

Question 6

Arrange the following spectral regions in order of increasing energy: infrared,microwave,ultraviolet,visible.
A)microwave < infrared < visible < ultraviolet
B)microwave < visible < infrared < ultraviolet
C)ultraviolet < infrared < visible < microwave
D)ultraviolet < visible < infrared < microwave

Accepted Answer

The correct order from lowest to highest energy is microwave < infrared < visible < ultraviolet. Energy increases from radio waves to microwaves to infrared to visible light to ultraviolet to X-rays to gamma rays in the electromagnetic spectrum.

Question 7

Which of the following is not quantized?
A)the charge on a monatomic ion
B)the distance between two objects
C)the population of the United States
D)the static charge on a balloon rubbed with wool

Accepted Answer

The distance between two objects is not quantized because it can vary continuously and does not come in discrete units. Charge, population numbers, and static charges are quantized, meaning they can only take on certain discrete values.

Question 8

What is the frequency of a helium-neon laser light with a wavelength of 632.8 nm? The speed of light is 3.00 × 10⁸ m/s. A)4.74 × 10¹⁴ s^-1 B)4.74 × 10⁵ s^-1 C)2.11 × 10^-15 s^-1 D)1.58 × 10^-15 s^-1

Accepted Answer

The frequency (f) can be calculated using the formula f = c/λ, where c is the speed of light and λ is the wavelength. Substituting the given values, f = (3.00 × 10^8 m/s) / (632.8 × 10^-9 m) = 4.74 × 10^14 s^-1.

Question 9

The absorption of a photon of wavelength 4.67 × 10^-6 m is necessary for a CO molecule to pass from the lowest vibrational energy level to the next highest vibrational level.If this higher vibrational level has an energy of 6.41 × 10^-20 J,what is the energy of the lowest vibrational level? (h = 6.626 × 10^-34 J ∙ s) A)1.60 × 10^-20 J B)2.15 × 10^-20 J C)3.20 × 10^-20 J D)4.26 × 10^-20 J

Accepted Answer

The answer of The absorption of a photon of wavelength...

Question 10

What is the first ionization energy for a hydrogen atom in the ground state? The Rydberg constant is . A)7.27 × 10^-36 J B)1.63 × 10^-27 J C)2.18 × 10^-18 J D)0.00823 J

Accepted Answer

The first ionization energy of a hydrogen atom can be calculated using the expression:
IE1 = -RH/n^2
Where RH is the Rydberg constant, n is the principal quantum number of the excited state, and IE1 is the ionization energy.
For a hydrogen atom in the ground state (n=1), the ionization energy is:
IE1 = -RH/n^2 = -11.0 x 10^6 m^-1
To convert this to joules, we multiply by the Planck constant (h) and the speed of light (c):
IE1= -11.0 x 10^6 m^-1 x h x c = 2.18 x 10^-18 J
Therefore, the correct answer is C) 2.18 × 10^-18 J.

Question 11

The intensity of a beam of light is related to its
A)frequency.
B)relative number of photons.
C)speed.
D)wavelength.

Accepted Answer

The intensity of a beam of light is related to the relative number of photons, not its other properties like frequency, speed or wavelength. A higher number of photons per unit of time equates to a higher intensity of light.

Question 12

The absorption of light of frequency 1.16 × 10¹¹ Hz is required for CO molecules to go from the lowest rotational energy level to the next highest rotational energy level.Determine the energy for this transition in kJ/mol.(h = 6.626 × 10^-34 J ∙ s) A)7.69 × 10^-23 kJ/mol B)0.0463 kJ/mol C)46.3 kJ/mol D)949 kJ/mol

Accepted Answer

The energy of the transition can be found using the equation:

ΔE = hν

where h is Planck's constant and ν is the frequency of the absorbed light.

Substituting the given values, we get:

ΔE = (6.626 × 10^-34 J ∙ s) × (1.16 × 10^11 Hz)

= 7.69 × 10^-24 J

To convert this to kJ/mol, we need to divide by Avogadro's number (6.022 × 10^23) and multiply by 10^-3:

ΔE = (7.69 × 10^-24 J/atom) / (6.022 × 10^23 atoms/mol) × 10^-3 kJ/J

= 0.0463 kJ/mol

Therefore, the correct answer is B.

Question 13

For hydrogen,what is the wavelength of the photon emitted when an electron drops from a 3d orbital to a 2p orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10^-2 nm^-1. A)656.3 nm B)486.2 nm C)364.6 nm D)2.057 × 10^-3 nm

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy difference between the 3d and 2p orbitals in hydrogen is given by the Rydberg formula: E = -13.6 eV (1/n1^2 - 1/n2^2), where n1 and n2 are the principal quantum numbers of the orbitals. Plugging in the values for n1 = 3 and n2 = 2, we get E = -1.51 eV. Converting this to joules, we get E = -2.42 × 10^-19 J. Using the equation E = hc/λ, we can solve for λ: λ = hc/E = (6.626 × 10^-34 J s)(3 × 10^8 m/s) / (-2.42 × 10^-19 J) = 656.3 nm.

Question 14

According to the Balmer-Rydberg equation,electromagnetic radiation with wavelength λ = 486.1 nm will be absorbed when an electron undergoes which of the following transitions?
A)m = 2 → n = 3
B)m = 2 → n = 4
C)n = 3 → m = 2
D)n = 4 → m = 2

Accepted Answer

The answer of According to the Balmer-Rydberg equation,electromagnetic radiation with...

Question 15

Molecular vibrational energy transitions are observed in the infrared,molecular rotational transitions in the microwave,and electronic transitions in the ultraviolet-visible range.Which transitions require the most energy and which the least energy?
A)Electronic transitions require the least energy and vibrational transitions the most.
B)Rotational transitions require the least energy and electronic transitions the most.
C)Vibrational transitions require the least energy and electronic transitions the most.
D)Vibrational transitions require the least energy and rotational transitions the most.

Accepted Answer

Electronic transitions require the most energy as they involve the movement of electrons between energy levels. Vibrational transitions require intermediate energy as they involve the stretching and bending of chemical bonds. Rotational transitions require the least energy as they involve the rotation of the entire molecule around its axis. Therefore, option B is the correct answer.

Question 16

Light can be made to have a higher intensity by raising its
A)amplitude.
B)energy.
C)frequency.
D)wavelength.

Accepted Answer

The intensity of light is directly related to the square of its amplitude. Increasing the amplitude of a light wave increases the energy it carries per unit area, thereby increasing its intensity. Frequency and wavelength adjustments change the color or type of light but not directly its intensity.

Question 17

The Balmer-Rydberg equation can be extended to ions with only one electron,such as He⁺.In that case it has the form: 1/λ = Z²^R(1/m² - 1/n²),where Z is the atomic number.What is the energy of the photon required to promote an electron in He⁺ from a 1s orbital to a 2p orbital? A)(3/4)hcR B)3hcR C)4hcR D)12hcR

Accepted Answer

The answer of The Balmer-Rydberg equation can be extended to...

Question 18

According to the Balmer-Rydberg equation,which transition results in the emission of a photon in the ultraviolet region of the electromagnetic radiation spectrum?
A)m = 1 → n = 2
B)m = 3 → n = 4
C)m = 2 → n = 1
D)m = 4 → n = 3

Accepted Answer

The Balmer-Rydberg equation predicts the wavelength of photons emitted or absorbed when electrons in an atom transition between energy levels. Transitions to the n=2 level (from higher levels) fall within the Balmer series, visible in the visible spectrum, not ultraviolet. However, the question's phrasing seems to intend to ask about transitions that result in ultraviolet emissions, which are typically associated with the Lyman series (transitions to n=1). None of the provided transitions directly correspond to the Lyman series or explicitly result in ultraviolet emission based on the standard understanding of the Balmer-Rydberg equation and known series in the hydrogen spectrum. Therefore, it appears there might be a misunderstanding in the question's formulation as none of the options directly result in ultraviolet emission based on the typical assignments to the Balmer or Lyman series.

Question 19

Photochemists use electromagnetic radiation to initiate chemical reactions,often by providing the energy required to break bonds within a molecule.Lowering which of the following will result in electromagnetic radiation having more energy per photon?
A)amplitude
B)frequency
C)intensity
D)wavelength

Accepted Answer

The energy of a photon is inversely proportional to its wavelength, meaning that shorter wavelengths correspond to higher energy per photon.

Question 20

According to the Balmer-Rydberg equation,electromagnetic radiation with the shortest wavelength will be emitted when an electron undergoes which of the following transitions?
A)m = 1 → n = 2
B)m = 2 → n = 3
C)n = 2 → m = 1
D)n = 3 → m = 2

Accepted Answer

The Balmer-Rydberg equation predicts the wavelength of light resulting from an electron moving between energy levels in a hydrogen atom. The shortest wavelength corresponds to the transition with the greatest energy difference. The transition from a higher energy level to a lower one (n = 2 → m = 1) releases the most energy, resulting in the shortest wavelength of light.

Quiz 5: Periodicity and the Electronic Structure of Atoms

The greater the energy of a photon,the

What is a quantum of light called?

According to the Balmer-Rydberg equation,the transition from n = 6 to m = 2 results in a spectral line having which color?

A quantized variable

Arrange the following spectral regions in order of increasing wavelength: infrared,microwave,ultraviolet,visible.

Arrange the following spectral regions in order of increasing energy: infrared,microwave,ultraviolet,visible.

Which of the following is not quantized?

What is the frequency of a helium-neon laser light with a wavelength of 632.8 nm? The speed of light is 3.00 × 10⁸ m/s.

What is the first ionization energy for a hydrogen atom in the ground state? The Rydberg constant is .

The intensity of a beam of light is related to its

The absorption of light of frequency 1.16 × 10¹¹ Hz is required for CO molecules to go from the lowest rotational energy level to the next highest rotational energy level.Determine the energy for this transition in kJ/mol.(h = 6.626 × 10^-34 J ∙ s)

For hydrogen,what is the wavelength of the photon emitted when an electron drops from a 3d orbital to a 2p orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10^-2 nm^-1.

According to the Balmer-Rydberg equation,electromagnetic radiation with wavelength λ = 486.1 nm will be absorbed when an electron undergoes which of the following transitions?

Molecular vibrational energy transitions are observed in the infrared,molecular rotational transitions in the microwave,and electronic transitions in the ultraviolet-visible range.Which transitions require the most energy and which the least energy?

Light can be made to have a higher intensity by raising its

The Balmer-Rydberg equation can be extended to ions with only one electron,such as He⁺.In that case it has the form: 1/λ = Z²^R(1/m² - 1/n²) ,where Z is the atomic number.What is the energy of the photon required to promote an electron in He⁺ from a 1s orbital to a 2p orbital?

According to the Balmer-Rydberg equation,which transition results in the emission of a photon in the ultraviolet region of the electromagnetic radiation spectrum?

Photochemists use electromagnetic radiation to initiate chemical reactions,often by providing the energy required to break bonds within a molecule.Lowering which of the following will result in electromagnetic radiation having more energy per photon?

According to the Balmer-Rydberg equation,electromagnetic radiation with the shortest wavelength will be emitted when an electron undergoes which of the following transitions?