Properties of Photons
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AP Chemistry › Properties of Photons
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?
Photon Y has greater energy and shorter wavelength than Photon X
Photon X and Photon Y have the same energy because both are photons
Photon energy depends on intensity, so frequency alone is insufficient
Photon Y has greater wavelength because it has the higher frequency
Photon X has greater energy because it has the lower frequency
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.
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?
Visible light has a longer wavelength, so it must have higher photon energy
Visible light cannot transfer energy because only UV light is electromagnetic radiation
The intensity of visible light is always lower than the intensity of UV light
UV light has higher frequency, so each photon has greater energy
UV light has lower frequency, so it delivers more energy per photon
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.
A chemist uses two different types of radiation to probe molecular transitions. Radiation M has wavelength $\lambda$ and Radiation N has wavelength $\tfrac{1}{2}\lambda$. Compared with photons of Radiation M, photons of Radiation N have
the same energy because the speed of light is constant
half the frequency and twice the energy
twice the frequency and twice the energy
twice the wavelength and twice the energy
half the frequency and half the energy
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.
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?
Both deliver the same energy per second because the number of photons per second is the same
Laser A delivers more energy per second because each photon has greater energy
Laser B delivers more energy per second because each photon has greater energy
Laser B photons have higher frequency because they have longer wavelength
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.
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
the same energy because intensity determines photon energy.
half the energy and twice the wavelength.
twice the energy and half the wavelength.
twice the energy and twice the wavelength.
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.
Two types of electromagnetic radiation are described: Radiation X has frequency $3.0\times 10^{14}\ \text{s}^{-1}$, and Radiation Y has frequency $6.0\times 10^{14}\ \text{s}^{-1}$. Which comparison is correct for single photons of X and Y?
Photon Y has greater energy and longer wavelength than photon X.
Photon X has greater energy because it has lower frequency.
Photon Y has greater energy and shorter wavelength than photon X.
Photon X has shorter wavelength because it has lower frequency.
Photons X and Y have the same energy because both frequencies are in the same order of magnitude.
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.
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?
Photon A has greater energy because it is in the infrared region.
Photon A has greater energy because its wavelength is larger.
Photons A and B have the same energy because both are electromagnetic radiation.
Photon B has greater energy because its wavelength is shorter.
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.
A chemist uses light to promote electrons in a sample from a lower energy level to a higher energy level. Transition 1 requires $\Delta E_1$, and Transition 2 requires $\Delta E_2$, where $\Delta E_2 > \Delta E_1$. Which radiation would be required to induce Transition 2 rather than Transition 1?
Radiation with a shorter wavelength (higher frequency).
Radiation with the same frequency but higher intensity.
Any radiation, as long as enough photons are absorbed.
Radiation with a longer wavelength (lower frequency).
Radiation with a lower intensity but the same wavelength.
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.
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?
700 nm, because lower frequency photons transfer energy more efficiently.
700 nm, because longer wavelength means higher photon energy.
250 nm, because it has the shortest wavelength.
400 nm, because it is in the visible range.
All three, because equal intensity means equal energy per photon.
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.
A hydrogen atom absorbs a photon and an electron transitions from $n=1$ to $n=3$. In a separate experiment, a hydrogen atom absorbs a photon and an electron transitions from $n=1$ to $n=2$. Which absorbed photon has the greater energy?
The photon for the $n=1\to n=2$ transition, because it involves a smaller wavelength.
The photon for the $n=1\to n=3$ transition, because it has a longer wavelength and therefore higher energy.
The photon for the $n=1\to n=3$ transition, because it corresponds to a larger energy change.
The photon for the $n=1\to n=2$ transition, because lower $n$ always means higher photon energy.
Both photons have the same energy because they are both absorbed by hydrogen.
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.