AP Chemistry Quiz: Properties Of Photons

What this quiz covers

This quiz focuses on Properties Of Photons, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Photon Y has greater energy and shorter wavelength than Photon X (correct answer)
2. Photon X has greater energy because it has the lower frequency
3. Photon Y has greater wavelength because it has the higher frequency
4. Photon X and Photon Y have the same energy because both are photons
5. Photon energy depends on intensity, so frequency alone is insufficient

Explanation: This question assesses knowledge of properties of photons, particularly the relationship between frequency and energy. Photon energy is directly proportional to frequency (E = hν), so Photon Y with frequency 7.5×10¹⁴ s⁻¹ has greater energy than Photon X with frequency 5.0×10¹⁴ s⁻¹. Since c = λν, wavelength is inversely proportional to frequency, meaning Photon Y also has a shorter wavelength than Photon X. Choice C is incorrect because it claims higher frequency corresponds to greater wavelength, when the relationship is actually inverse. When comparing photons, always remember that higher frequency means both higher energy and shorter wavelength.

Question 2

1. UV light has lower frequency, so it delivers more energy per photon
2. UV light has higher frequency, so each photon has greater energy (correct answer)
3. Visible light has a longer wavelength, so it must have higher photon energy
4. Visible light cannot transfer energy because only UV light is electromagnetic radiation
5. The intensity of visible light is always lower than the intensity of UV light

Explanation: This question examines properties of photons in the context of the photoelectric effect. UV light has a higher frequency than visible light (since it has shorter wavelength), and photon energy is directly proportional to frequency (E = hν). Therefore, each UV photon carries more energy than a visible light photon. The photoelectric effect requires each photon to have sufficient energy to overcome the work function of the metal, explaining why UV light is effective while lower-frequency visible light is not. Choice A incorrectly states that UV light has lower frequency, when UV actually has higher frequency than visible light. Remember that higher frequency (shorter wavelength) photons carry more energy per photon.

Question 3

1. half the frequency and half the energy
2. half the frequency and twice the energy
3. twice the frequency and twice the energy (correct answer)
4. twice the wavelength and twice the energy
5. the same energy because the speed of light is constant

Explanation: This problem examines properties of photons when wavelength changes. Radiation N has wavelength λ/2, which is half the wavelength of Radiation M. Since frequency and wavelength are inversely related (c = λν), halving the wavelength doubles the frequency. Additionally, photon energy is inversely proportional to wavelength (E = hc/λ), so halving the wavelength doubles the energy. Choice D incorrectly claims that shorter wavelength corresponds to higher energy while also claiming twice the wavelength, which is contradictory. When wavelength is halved, both frequency and energy are doubled.

Question 4

1. Laser B delivers more energy per second because each photon has greater energy
2. Laser A delivers more energy per second because each photon has greater energy (correct answer)
3. Both deliver the same energy per second because the number of photons per second is the same
4. Laser B photons have higher frequency because they have longer wavelength
5. Laser A photons have lower energy because they have shorter wavelength

Explanation: This problem tests properties of photons in comparing laser outputs. Since photon energy E = hc/λ is inversely proportional to wavelength, Laser A (532 nm) emits photons with greater energy than Laser B (1064 nm). If both lasers emit the same number of photons per second, Laser A delivers more total energy per second because each of its photons carries more energy. Choice E incorrectly claims that shorter wavelength corresponds to lower energy, which contradicts the inverse relationship between wavelength and photon energy. When comparing light sources, remember that shorter wavelength means higher energy per photon.

Question 5

1. half the energy and twice the wavelength.
2. twice the energy and half the wavelength. (correct answer)
3. twice the energy and twice the wavelength.
4. the same energy because intensity determines photon energy.
5. half the energy and half the wavelength.

Explanation: This problem tests understanding of properties of photons when frequency changes. Photon energy is directly proportional to frequency (E = hf), so doubling the frequency doubles the photon energy. Since wavelength and frequency are inversely related (c = fλ), doubling the frequency halves the wavelength. Therefore, photons from the second source (frequency 2ν) have twice the energy and half the wavelength compared to the first source (frequency ν). Choice C incorrectly suggests that both energy and wavelength would double, but this violates the inverse relationship between frequency and wavelength. The key principle is that frequency and energy change proportionally, while wavelength changes inversely.

Question 6

1. Photon X has shorter wavelength because it has lower frequency.
2. Photons X and Y have the same energy because both frequencies are in the same order of magnitude.
3. Photon X has greater energy because it has lower frequency.
4. Photon Y has greater energy and longer wavelength than photon X.
5. Photon Y has greater energy and shorter wavelength than photon X. (correct answer)

Explanation: This question assesses knowledge of properties of photons, particularly how frequency relates to energy and wavelength. Photon energy is directly proportional to frequency (E = hf), so Radiation Y with frequency 6.0 × 10¹⁴ s⁻¹ has twice the energy of Radiation X with frequency 3.0 × 10¹⁴ s⁻¹. Since wavelength and frequency are inversely related (c = fλ), higher frequency means shorter wavelength, so photon Y also has shorter wavelength than photon X. Choice B incorrectly states that higher energy corresponds to longer wavelength, which violates the inverse relationship. Remember that higher frequency always means both higher energy and shorter wavelength for photons.

Question 7

1. Photon A has greater energy because its wavelength is larger.
2. Photon B has greater energy because its wavelength is shorter. (correct answer)
3. Photon A has greater energy because it is in the infrared region.
4. Photons A and B have the same energy because both are electromagnetic radiation.
5. Photon B has lower energy because it is visible light.

Explanation: This question examines properties of photons by comparing different wavelengths. Radiation A has wavelength 2.0 μm (2000 nm) while Radiation B has wavelength 500 nm. Since photon energy is inversely proportional to wavelength (E = hc/λ), the photon with shorter wavelength has higher energy. Photon B, with wavelength 500 nm, has four times the energy of Photon A with wavelength 2000 nm. Choice A incorrectly claims that larger wavelength means greater energy, which reverses the actual relationship. Remember that for photons, shorter wavelength always means higher energy, regardless of which region of the electromagnetic spectrum they occupy.

Question 8

1. Radiation with the same frequency but higher intensity.
2. Radiation with a shorter wavelength (higher frequency). (correct answer)
3. Radiation with a lower intensity but the same wavelength.
4. Radiation with a longer wavelength (lower frequency).
5. Any radiation, as long as enough photons are absorbed.

Explanation: This problem involves properties of photons and energy transitions. To promote an electron to a higher energy level, a photon must have energy equal to the energy difference (ΔE) between levels. Since ΔE₂ > ΔE₁, Transition 2 requires a photon with greater energy than Transition 1. Because photon energy is inversely proportional to wavelength (E = hc/λ), higher energy photons have shorter wavelengths and higher frequencies. Choice A incorrectly suggests longer wavelength radiation would work, but longer wavelengths have lower energy photons. The strategy is to match photon energy to the required transition energy: larger transitions need shorter wavelength (higher frequency) radiation.

Question 9

1. 700 nm, because longer wavelength means higher photon energy.
2. 400 nm, because it is in the visible range.
3. 250 nm, because it has the shortest wavelength. (correct answer)
4. All three, because equal intensity means equal energy per photon.
5. 700 nm, because lower frequency photons transfer energy more efficiently.

Explanation: This question assesses understanding of the properties of photons. Shorter wavelengths correspond to higher energy because E = hc/λ shows an inverse relationship. Frequency is also inversely related to wavelength, with higher frequency meaning higher energy. The 250 nm source has the shortest wavelength among 250 nm, 400 nm, and 700 nm, so its photons have the greatest energy. A tempting distractor is choice A, which wrongly states 700 nm has higher energy due to longer wavelength, but longer wavelengths actually mean lower energy. Remember, shorter wavelength corresponds to higher energy.

Question 10

1. Both photons have the same energy because they are both absorbed by hydrogen.
2. The photon for the $n=1\to n=2$ transition, because lower $n$ always means higher photon energy.
3. The photon for the $n=1\to n=2$ transition, because it involves a smaller wavelength.
4. The photon for the $n=1\to n=3$ transition, because it corresponds to a larger energy change. (correct answer)
5. The photon for the $n=1\to n=3$ transition, because it has a longer wavelength and therefore higher energy.

Explanation: This question assesses understanding of the properties of photons. Photon energy is inversely related to wavelength because E = hc/λ, meaning shorter wavelengths have higher energy. Frequency relates inversely to wavelength but directly to energy via E = hν. The n=1 to n=3 transition involves a larger energy change than n=1 to n=2, resulting in a higher-energy photon. A tempting distractor is choice E, which incorrectly states that longer wavelength means higher energy, but longer wavelengths actually indicate lower energy. Remember, larger energy changes produce higher-energy photons with shorter wavelengths.

Question 11

1. Photons from Source 2 have greater energy because they have a longer wavelength.
2. Photons from Source 1 have greater energy because they have a shorter wavelength. (correct answer)
3. Photons from Source 1 and Source 2 have the same energy because both are forms of electromagnetic radiation.
4. Photons from Source 2 have greater energy because red light is more intense than blue light.
5. Photons from Source 1 have lower frequency because they have a shorter wavelength.

Explanation: This question assesses understanding of the properties of photons. The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength, as described by the equation E = hc/λ, where h is Planck's constant and c is the speed of light. This means that as the wavelength decreases, the frequency increases, leading to higher photon energy. For the two sources, Source 1 with 450 nm has a shorter wavelength than Source 2 with 650 nm, so its photons have higher energy. A tempting distractor is choice A, which incorrectly states that longer wavelength means greater energy, but actually, longer wavelengths correspond to lower energy photons. Remember, shorter wavelength corresponds to higher energy.

Question 12

1. Photon S has three times the frequency of Photon R.
2. Photon S has one-third the frequency of Photon R. (correct answer)
3. Photon S and Photon R have the same frequency because they travel at the same speed.
4. Photon R has one-third the frequency of Photon S.
5. Frequency cannot be compared without knowing the intensity of each photon.

Explanation: This question assesses understanding of the properties of photons. Frequency is inversely proportional to wavelength (ν = c/λ), so a longer wavelength means lower frequency. Energy follows similarly, being lower for longer wavelengths. Photon S with λ_S = 3λ_R has three times the wavelength, so one-third the frequency of Photon R. A tempting distractor is choice A, which incorrectly claims Photon S has three times the frequency, but the inverse relationship shows it is one-third. Remember, longer wavelength corresponds to lower frequency.

Question 13

1. Microwave photons have higher energy because microwaves are used to heat food.
2. Ultraviolet photons have higher energy because they have higher frequency. (correct answer)
3. Microwave photons have higher energy because they have longer wavelength.
4. Ultraviolet and microwave photons have the same energy because both travel at the speed of light.
5. Microwave photons have higher energy because they are more intense in typical ovens.

Explanation: This problem involves comparing properties of photons from different regions of the electromagnetic spectrum. Ultraviolet radiation has much shorter wavelengths (higher frequencies) than microwave radiation. Since photon energy is proportional to frequency (E = hf) and inversely proportional to wavelength, UV photons have significantly higher energy than microwave photons. Choice A incorrectly assumes microwave photons have higher energy because microwaves heat food, but heating efficiency depends on water molecule absorption, not photon energy. The strategy is to use the electromagnetic spectrum ordering: as you go from radio waves through microwaves, infrared, visible, UV, to X-rays, photon energy increases.

Question 14

1. lower energy and lower frequency
2. higher energy and higher frequency (correct answer)
3. higher energy and lower frequency
4. lower energy and higher frequency
5. the same energy because the speed of light is constant

Explanation: This problem examines properties of photons by comparing different wavelengths. Since photon energy E = hc/λ, energy is inversely proportional to wavelength. The 300 nm photons have half the wavelength of 600 nm photons, so they have twice the energy. Additionally, since c = λν, frequency is also inversely proportional to wavelength, so 300 nm photons have twice the frequency of 600 nm photons. Choice C incorrectly suggests that shorter wavelength corresponds to lower frequency, when the opposite is true. To solve photon comparison problems, remember that shorter wavelength always means both higher energy and higher frequency.

Question 15

1. Source 2, because 900 nm photons have a longer wavelength
2. Source 1, because shorter-wavelength photons have greater energy (correct answer)
3. Source 2, because longer-wavelength photons have greater frequency
4. Both sources emit photons with the same energy because both are light
5. The source with greater intensity emits higher-energy photons, so it cannot be determined from wavelength

Explanation: This question tests understanding of properties of photons, specifically the relationship between wavelength and energy. 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 wavelength. Since energy is inversely proportional to wavelength, shorter wavelengths correspond to higher energy photons. Source 1 with 450 nm wavelength emits photons with greater energy than Source 2 with 900 nm wavelength. Choice C is incorrect because it wrongly claims that longer wavelengths have greater frequency, when actually frequency and wavelength are inversely related (c = λν). Remember: shorter wavelength always means higher energy and higher frequency for photons.

Question 16

1. Increase the wavelength of the photon
2. Decrease the frequency of the photon
3. Increase the frequency of the photon (correct answer)
4. Decrease the intensity of the radiation
5. Use the same frequency but increase the amplitude so each photon has more energy

Explanation: This question tests understanding of properties of photons in atomic transitions. To bridge a larger energy gap with a single photon, the photon must have more energy. Since photon energy is directly proportional to frequency (E = hν) and inversely proportional to wavelength (E = hc/λ), increasing the frequency of the photon will increase its energy. Choice A is incorrect because increasing wavelength would decrease photon energy, making it less capable of bridging large energy gaps. To increase the energy of individual photons, always increase frequency or decrease wavelength.

Question 17

1. greater energy because Beam A has greater intensity.
2. greater frequency because Beam A is brighter.
3. the same energy because photon energy depends on wavelength (or frequency), not intensity. (correct answer)
4. lower energy because Beam A contains more photons.
5. the same energy only if Beam A and Beam B have the same power.

Explanation: This problem tests understanding of properties of photons versus beam intensity. For monochromatic light of a given wavelength, all photons have the same energy (E = hc/λ), regardless of beam intensity or brightness. Intensity refers to the number of photons per unit time per unit area, not the energy of individual photons. A brighter beam simply contains more photons of the same energy, not photons with different energies. Choice A incorrectly conflates intensity with photon energy, but these are independent properties. Remember that photon energy depends only on wavelength (or frequency), while intensity depends on the number of photons.

Question 18

1. Photon energy decreases and photon frequency decreases. (correct answer)
2. Photon energy stays constant but intensity decreases.
3. Photon energy increases and photon frequency decreases.
4. Photon energy increases and photon frequency increases.
5. Photon energy decreases and photon frequency increases.

Explanation: This question assesses understanding of the properties of photons. Increasing wavelength from 300 nm to 600 nm doubles the wavelength, which halves the frequency since ν = c/λ. Energy is inversely proportional to wavelength (E = hc/λ), so longer wavelengths result in lower energy per photon. Consequently, both photon energy and frequency decrease with this change. A tempting distractor is choice B, which incorrectly states that energy and frequency increase, but they actually decrease as wavelength increases. Remember, longer wavelength corresponds to lower energy and frequency.

Question 19

1. Radiation Q photons have twice the energy of Radiation P photons. (correct answer)
2. Radiation Q photons have half the energy of Radiation P photons.
3. Radiation Q photons have the same energy as Radiation P photons because both travel at the speed of light.
4. Radiation Q photons have twice the energy only if Radiation Q is more intense.
5. Radiation Q photons have lower energy because higher frequency corresponds to longer wavelength.

Explanation: This question assesses understanding of the properties of photons. Energy is directly proportional to frequency (E = hν), so doubling the frequency doubles the energy. Wavelength is inversely related to frequency (λ = c/ν), meaning higher frequency corresponds to shorter wavelength and higher energy. Radiation Q with frequency 2ν_P has twice the energy per photon compared to Radiation P. A tempting distractor is choice B, which incorrectly states half the energy, but the direct proportionality shows it is twice. Remember, higher frequency corresponds to higher energy.

Question 20

1. A UV photon has lower energy because UV light is less intense than microwaves in an oven.
2. A microwave photon has greater energy because it has a shorter wavelength than UV.
3. Both photons have the same energy because both are electromagnetic radiation.
4. A microwave photon has greater energy because microwaves are used to heat food.
5. A UV photon has greater energy because UV radiation has higher frequency. (correct answer)

Explanation: This question assesses understanding of the properties of photons. UV radiation has a shorter wavelength than microwave radiation, leading to higher frequency since ν = c/λ. Energy is directly proportional to frequency (E = hν) and inversely to wavelength. Thus, a UV photon has greater energy than a microwave photon due to its higher frequency. A tempting distractor is choice C, which wrongly states microwave has shorter wavelength and greater energy, but microwaves have longer wavelengths and lower energy. Remember, higher frequency corresponds to higher energy.