Properties of Photons
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AP Chemistry › Properties of Photons
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
half the energy and twice the wavelength.
half the energy and half the wavelength.
twice the energy and twice the wavelength.
twice the energy and half the wavelength.
the same energy because intensity determines photon energy.
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.
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 B has lower energy because it is visible light.
Photon B has greater energy because its wavelength is shorter.
Photons A and B have the same energy because both are electromagnetic radiation.
Photon A has greater energy because its wavelength is larger.
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.
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 X has greater energy because it has lower frequency.
Photon X has shorter wavelength because it has lower frequency.
Photon Y has greater energy and shorter wavelength than photon X.
Photon Y has greater energy and longer wavelength than photon X.
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 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 longer wavelength (lower frequency).
Radiation with a shorter wavelength (higher frequency).
Any radiation, as long as enough photons are absorbed.
Radiation with the same frequency but higher intensity.
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 longer wavelength means higher photon energy.
400 nm, because it is in the visible range.
700 nm, because lower frequency photons transfer energy more efficiently.
250 nm, because it has the shortest wavelength.
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.
Both photons have the same energy because they are both absorbed by hydrogen.
The photon for the $n=1\to n=2$ transition, because lower $n$ always means higher photon energy.
The photon for the $n=1\to n=3$ transition, because it corresponds to a larger energy change.
The photon for the $n=1\to n=3$ 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.
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?
Photons from Source 2 have greater energy because they have a longer wavelength.
Photons from Source 1 have lower frequency because they have a shorter wavelength.
Photons from Source 1 have greater energy because they have a shorter wavelength.
Photons from Source 1 and Source 2 have the same energy because both are forms of electromagnetic radiation.
Photons from Source 2 have greater energy because red light is more intense than blue light.
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.
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?
Photon S has three times the frequency of Photon R.
Photon R has one-third the frequency of Photon S.
Photon S has one-third the frequency of Photon R.
Frequency cannot be compared without knowing the intensity of each photon.
Photon S and Photon R have the same frequency because they travel at the same speed.
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.
A student compares electromagnetic radiation in the microwave region and in the ultraviolet region. Which statement about individual photons is correct?
Ultraviolet and microwave photons have the same energy because both travel at the speed of light.
Ultraviolet photons have higher energy because they have higher frequency.
Microwave photons have higher energy because they have longer wavelength.
Microwave photons have higher energy because microwaves are used to heat food.
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.
Two beams of monochromatic light, Beam A and Beam B, have the same wavelength. Beam A appears brighter because it has greater intensity. Compared with Beam B, the photons in Beam A have
greater frequency because Beam A is brighter.
lower energy because Beam A contains more photons.
the same energy only if Beam A and Beam B have the same power.
the same energy because photon energy depends on wavelength (or frequency), not intensity.
greater energy because Beam A has greater intensity.
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.