Properties of Photons - AP Chemistry

A source emits monochromatic light at frequency $\nu$ . A second source emits monochromatic light at frequency $2\nu$ . Compared with photons from the first source, photons from the second source have
Answer: twice the energy and half the wavelength. 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.
A student compares two monochromatic light sources used in spectroscopy. Source 1 emits light with wavelength 450 nm, and Source 2 emits light with wavelength 650 nm. Which statement correctly compares the photons emitted by the two sources?
Answer: Photons from Source 1 have greater energy because they have a shorter wavelength. 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.
A student compares two monochromatic beams of electromagnetic radiation, each delivering the same total power to a surface. Beam 1 has wavelength $450\ \text{nm}$ , and Beam 2 has wavelength $900\ \text{nm}$ . Which statement about the photons in the two beams is correct?
Answer: Photons in Beam 1 have greater energy because shorter wavelength corresponds to higher frequency. This question tests understanding of properties of photons, specifically the relationship between wavelength and energy. The energy of a photon is given by E = hf = hc/λ, where h is Planck's constant, f is frequency, c is the speed of light, and λ is wavelength. Since energy is inversely proportional to wavelength, Beam 1 with wavelength 450 nm has photons with higher energy than Beam 2 with wavelength 900 nm (which is twice as long). Choice C incorrectly assumes that equal power means equal photon energy, but power depends on both photon energy and the number of photons per second. The key strategy is to remember that shorter wavelength always corresponds to higher photon energy.
A sample is irradiated with electromagnetic radiation. Radiation P has frequency $\nu_P$ , and Radiation Q has frequency $2\nu_P$ . How does the energy per photon of Radiation Q compare to that of Radiation P?
Answer: Radiation Q photons have twice the energy of Radiation P photons. 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.
A sample is irradiated separately with three types of electromagnetic radiation. The wavelengths are listed. Which radiation produces the highest-energy photons?
Answer: Radiation with wavelength 400 nm This question evaluates understanding of properties of photons across different wavelengths. Photon energy is inversely proportional to wavelength (E = hc/λ), so the shortest wavelength produces the highest-energy photons. Among the three options, 400 nm is the shortest wavelength, making it the radiation that produces the highest-energy photons. Choice D is incorrect because it suggests all electromagnetic radiation produces photons of equal energy, ignoring the fundamental wavelength-energy relationship. To identify the highest-energy photons, always look for the shortest wavelength or highest frequency.
An atom absorbs a photon and an electron is promoted from a lower energy level to a higher energy level. Which change in the absorbed photon would increase the energy gap that can be bridged by a single photon?
Answer: Increase the frequency of the photon 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.
Light from two sources is compared. Source A emits photons of wavelength $1.0\ \mu\text{m}$ and Source B emits photons of wavelength $500\ \text{nm}$ . Which statement correctly compares the frequencies of the photons?
Answer: Source B photons have higher frequency because they have a shorter wavelength. This question assesses understanding of the properties of photons. Frequency is inversely proportional to wavelength, as ν = c/λ, so shorter wavelengths yield higher frequencies. Energy follows this pattern, being directly proportional to frequency (E = hν) and inversely to wavelength. Source B with 500 nm has a shorter wavelength than Source A with 1.0 μm (1000 nm), so its photons have higher frequency. A tempting distractor is choice D, which wrongly suggests shorter wavelengths mean lower frequency, but the inverse relationship shows the opposite. Remember, shorter wavelength corresponds to higher frequency.
Two photons, R and S, travel through a vacuum. Photon R has wavelength $\lambda_R$ and Photon S has wavelength $\lambda_S$ , where $\lambda_S = 3\lambda_R$ . Which statement correctly compares the photon frequencies?
Answer: Photon S has one-third the frequency of Photon R. 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.
A student observes that Radiation A has a wavelength of $2.0\ \mu\text{m}$ and Radiation B has a wavelength of $500\ \text{nm}$ . For individual photons, which statement is correct?
Answer: Photon B has greater energy because its wavelength is shorter. 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.
A student records the following wavelengths for three lasers: Laser P: $400\ \text{nm}$ , Laser Q: $600\ \text{nm}$ , Laser R: $800\ \text{nm}$ . Which ordering correctly ranks the photon energies from greatest to least?
Answer: $E_P > E_Q > E_R$ This question involves ranking photon energies based on properties of photons. Since photon energy is inversely proportional to wavelength (E = hc/λ), shorter wavelengths correspond to higher energies. Laser P (400 nm) has the shortest wavelength and thus highest photon energy, followed by Laser Q (600 nm), then Laser R (800 nm) with the longest wavelength and lowest energy. Choice D incorrectly assumes all laser photons have equal energy, but photon energy depends on wavelength/frequency, not on being produced by a laser. The strategy for ranking photon energies is simple: arrange wavelengths from shortest to longest to get energies from highest to lowest.
A student uses three different monochromatic light sources to eject electrons from a metal surface (photoelectric effect). The wavelengths are 250 nm, 400 nm, and 700 nm. Assuming all three sources have the same intensity, which light source produces photons with the greatest energy?
Answer: 250 nm, because it has the shortest wavelength. 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.
An electron in a hydrogen atom can undergo either transition I or transition II. Transition I emits a photon with frequency $\nu$ . Transition II emits a photon with frequency $2\nu$ . Compared with the photon from transition I, the photon from transition II has
Answer: twice the energy and half the wavelength This question tests understanding of properties of photons in atomic transitions. Since photon energy is directly proportional to frequency (E = hν), a photon with frequency 2ν has twice the energy of a photon with frequency ν. Using the relationship c = λν, we can determine that doubling the frequency halves the wavelength, so the photon from transition II has half the wavelength of the photon from transition I. Choice A incorrectly reverses these relationships, claiming that higher frequency corresponds to lower energy and longer wavelength. When analyzing photon emissions, remember that frequency and energy are directly proportional, while wavelength is inversely proportional to both.
A student observes that increasing the wavelength of electromagnetic radiation from 300 nm to 600 nm changes the photon properties. Which change must occur for individual photons?
Answer: Photon energy decreases and photon frequency decreases. 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.
Two lasers emit photons of different frequencies. Laser X emits light at $5.0\times10^{14}\ \text{s}^{-1}$ , and Laser Y emits light at $7.5\times10^{14}\ \text{s}^{-1}$ . Which statement is correct?
Answer: Laser Y emits photons with greater energy because it has the higher frequency. This question assesses understanding of the properties of photons. The energy of a photon is directly proportional to its frequency, following E = hν, where h is Planck's constant. Frequency and wavelength are inversely related via ν = c/λ, so higher frequency means shorter wavelength and thus higher energy. Laser Y has a higher frequency ( $7.5×10^14$ s⁻¹) than Laser X ( $5.0×10^14$ s⁻¹), so its photons have greater energy. A tempting distractor is choice E, which wrongly claims higher frequency means lower energy, but the opposite is true since energy increases with frequency. Remember, higher frequency corresponds to higher energy.
A student observes that ultraviolet (UV) light causes photoelectric emission from a metal surface, but visible light of lower frequency does not. Which statement best explains why UV light is effective while the visible light is not?
Answer: UV light has higher frequency, so each photon has greater energy 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.
A student compares two photons: Photon X has frequency $5.0\times10^{14}\ \text{s}^{-1}$ and Photon Y has frequency $7.5\times10^{14}\ \text{s}^{-1}$ . Which statement is correct?
Answer: Photon Y has greater energy and shorter wavelength than Photon X 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.
Two lasers emit monochromatic light. Laser A emits light of wavelength 532 nm. Laser B emits light of wavelength 1064 nm. If both lasers emit the same number of photons per second, which statement is correct?
Answer: Laser A delivers more energy per second because each photon has greater energy 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.
An atom emits a photon when an electron drops from a higher energy level to a lower energy level. Transition 1 emits light of wavelength 400 nm, and Transition 2 emits light of wavelength 800 nm. Which statement is correct?
Answer: Transition 1 emits a higher-energy photon because it has the shorter wavelength. This question assesses understanding of the properties of photons. Photon energy is inversely related to wavelength via E = hc/λ, so shorter wavelengths mean higher energy. Frequency is also inversely related to wavelength but directly to energy. Transition 1 at 400 nm has a shorter wavelength than Transition 2 at 800 nm, indicating higher energy. A tempting distractor is choice A, which wrongly claims longer wavelength means higher energy, but energy decreases with increasing wavelength. Remember, shorter wavelength corresponds to higher energy.
Two monochromatic light sources emit photons: Source 1 emits light with wavelength 450 nm, and Source 2 emits light with wavelength 900 nm. Which source emits photons with greater energy?
Answer: Source 1, because shorter-wavelength photons have greater energy 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.
A student compares two photons traveling in vacuum. Photon P has wavelength 500 nm, and Photon Q has wavelength 250 nm. Which statement is correct?
Answer: Photon Q has greater frequency and greater energy than Photon P This question tests properties of photons by comparing different wavelengths. Photon Q has wavelength 250 nm, which is half the wavelength of Photon P (500 nm). Since frequency is inversely proportional to wavelength (c = λν), Photon Q has twice the frequency of Photon P. Because photon energy is directly proportional to frequency (E = hν), Photon Q also has twice the energy of Photon P. Choice D incorrectly claims that longer wavelength corresponds to greater energy, when the relationship is actually inverse. Remember that shorter wavelength always means higher frequency and greater photon energy.

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