MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Photoelectric Effect Line Spectra

20 questions · exam conditions

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4e Photoelectric Effect Line SpectraQuestion 1 of 20

A researcher studies a one-electron atom that emits a line spectrum similar to hydrogen. When the nucleus is replaced with a different isotope (same atomic number, different mass), the bright emission lines are observed at essentially the same wavelengths within the instrument's resolution. Which statement best accounts for this result using the principle of quantized energy levels?

Emission wavelengths are set primarily by electronic energy level differences, which depend on nuclear charge more than nuclear mass

Emission wavelengths depend on how many photons are emitted per second, which is unchanged by isotope substitution

Isotope substitution changes the work function, so emission lines should shift substantially but were missed due to low intensity

The unchanged wavelengths show that atomic emission is purely classical radiation from accelerating charges

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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Photoelectric Effect Line Spectra

Practice 4e Photoelectric Effect Line Spectra in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 4e Photoelectric Effect Line Spectra, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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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.

All questions

Question 1

Emission wavelengths are set primarily by electronic energy level differences, which depend on nuclear charge more than nuclear mass (correct answer)
Emission wavelengths depend on how many photons are emitted per second, which is unchanged by isotope substitution
Isotope substitution changes the work function, so emission lines should shift substantially but were missed due to low intensity
The unchanged wavelengths show that atomic emission is purely classical radiation from accelerating charges

Explanation: This question tests understanding of how atomic structure determines emission spectra. Emission line wavelengths are determined by the energy differences between electronic levels, which depend primarily on the nuclear charge (atomic number) that governs electron-nucleus attraction. When replacing the nucleus with a different isotope (same atomic number but different mass), the nuclear charge remains unchanged, so the electronic energy levels and their differences remain essentially the same. The correct answer A recognizes that electronic transitions depend on nuclear charge rather than nuclear mass. Option B incorrectly relates wavelength to photon emission rate rather than energy level differences. Option C wrongly suggests isotopes affect the work function, which is irrelevant to emission spectra. Option D incorrectly denies the quantum nature of atomic emission. The slight isotope shift that does exist is typically too small to detect with standard spectrometers.

Question 2

A researcher observes that a hydrogen discharge tube emits a line at 656 nm. When the tube is cooled, the line remains at 656 nm but becomes dimmer. Which explanation is most consistent with the core principle of line spectra?

Cooling reduces the number of atoms reaching excited states, but the energy gaps (and thus wavelengths) remain fixed (correct answer)
Cooling reduces the photon frequency, but wavelength stays constant because Planck's constant decreases
Cooling changes the quantized energy levels, but the observed wavelength remains due to detector calibration
Cooling makes emission more wave-like, so discrete lines should merge into a continuous spectrum

Explanation: This question tests understanding of atomic line spectra and how temperature affects emission intensity versus wavelength. Line spectra arise from electrons transitioning between discrete energy levels in atoms, with each transition producing a photon of specific wavelength determined by the energy difference between levels. When the hydrogen tube is cooled, fewer atoms have sufficient thermal energy to reach excited states through collisions, reducing the number of electrons available to make downward transitions that produce the 656 nm line. However, the energy levels themselves are intrinsic properties of hydrogen atoms and remain unchanged by temperature, so the wavelength of emitted photons stays constant at 656 nm. The correct answer recognizes that cooling affects the population of excited atoms (intensity) but not the quantized energy gaps (wavelength). Common distractors incorrectly suggest that temperature changes fundamental atomic properties or invoke incorrect physics like variable Planck's constant or wave-particle duality effects on spectral lines.

Question 3

A spectrometer measures emission from a hydrogen discharge tube and detects sharp lines. The same instrument measures emission from a heated tungsten filament and detects a broad continuous spectrum. Which statement best explains the difference in spectra?

A discharge tube primarily produces photons from discrete electronic transitions; a hot filament approximates blackbody radiation (correct answer)
A discharge tube produces continuous radiation because electrons accelerate; a filament produces lines because atoms are quantized
Both should be continuous, but the spectrometer resolves only a few wavelengths for hydrogen
Both should be line spectra, but tungsten lines overlap to appear continuous due to photon wave interference

Explanation: This question tests understanding of the difference between line spectra and continuous spectra. A hydrogen discharge tube contains isolated atoms that undergo specific electronic transitions between quantized energy levels, producing photons at discrete wavelengths - hence sharp emission lines. In contrast, a heated tungsten filament acts as a dense solid where atoms are closely packed and strongly interact. This creates a near-continuum of available energy states, and thermal vibrations produce a broad range of photon energies approximating blackbody radiation. The fundamental difference is between isolated atoms with well-defined quantum states (line spectra) and condensed matter with overlapping energy bands (continuous spectra). This distinction is crucial for understanding different light sources and their applications in spectroscopy.

Question 4

A metal is illuminated with two different monochromatic light sources. Source 1 has frequency $f_1$ and produces photoelectrons with stopping potential $V_{s1}$ . Source 2 has frequency $f_2>f_1$ and produces stopping potential $V_{s2}$ . Intensities are adjusted so that both sources produce the same photocurrent. Which relationship is expected if both frequencies exceed threshold? (Constants: $h$ , $e$ .)

$V_{s2}>V_{s1}$ because maximum electron kinetic energy increases with photon frequency (correct answer)
$V_{s2}=V_{s1}$ because equal photocurrent implies equal electron kinetic energy
$V_{s2}<V_{s1}$ because higher frequency implies fewer photons and thus lower electron energy
No stopping potential can be defined when photocurrents are equal

Explanation: This question tests understanding of how frequency affects stopping potential independent of photocurrent. The photoelectric effect establishes that stopping potential depends only on the maximum kinetic energy of emitted electrons, which is determined by photon frequency through Vs = (hf - φ)/e. Since f₂ > f₁, photons from source 2 have higher energy, resulting in electrons with greater maximum kinetic energy and thus requiring a larger stopping potential to stop them. The photocurrent (number of electrons per second) depends on the number of incident photons, which can be adjusted through intensity. Equal photocurrents simply mean equal numbers of electrons are emitted per second, but this doesn't affect the energy per electron. This demonstrates the independence of photon number (intensity) and photon energy (frequency) in the quantum model.

Question 5

In a photoelectric experiment, the stopping potential is measured for several frequencies above threshold. The student mistakenly concludes that because light is a wave, increasing frequency should increase the number of emitted electrons per second at fixed intensity. Which statement best corrects this conclusion using the photoelectric model?

At fixed intensity, increasing frequency increases photon energy, so fewer photons per second may strike the surface (correct answer)
At fixed intensity, increasing frequency increases wave amplitude, so more electrons are emitted per photon
Frequency affects only the work function, so emission rate must remain constant
Electron emission rate depends only on stopping potential, which is unrelated to intensity

Explanation: This question tests understanding of the relationship between photon flux and frequency at constant intensity. In the photoelectric effect, light intensity equals the number of photons per second times the energy per photon (I = nE = nhf). At fixed intensity, increasing frequency means each photon carries more energy, so fewer photons per second must arrive to maintain the same total power. This results in fewer electron emissions per second (lower photocurrent) even though each emitted electron has higher kinetic energy. This counterintuitive result highlights the particle nature of light - we're dealing with discrete photons, not continuous waves. The student's error stems from classical wave thinking where frequency and amplitude are independent, but in the photon model, higher frequency at fixed intensity necessarily means fewer photons.

Question 6

A hydrogen emission spectrum shows a line corresponding to a transition from a higher energy level to a lower one. If the atom instead undergoes a transition with a smaller energy difference, what change is expected in the emitted photon? (Constants: $E=hc/\lambda$ .)

The photon has lower energy and longer wavelength (correct answer)
The photon has lower energy and shorter wavelength
The photon has higher energy and longer wavelength
The photon wavelength is unchanged because it depends only on the element, not the transition

Explanation: This question tests understanding of how energy differences relate to photon properties in atomic transitions. When an electron transitions between energy levels, the emitted photon carries exactly the energy difference: E = E_initial - E_final. A smaller energy difference means the photon carries less energy. Since E = hf = hc/λ, lower photon energy corresponds to both lower frequency and longer wavelength. This inverse relationship between energy and wavelength is fundamental to spectroscopy. In hydrogen, transitions with smaller energy gaps (like those to n=3 versus n=2) produce longer wavelength, lower energy photons. This relationship is universal - it applies to all electromagnetic radiation, not just atomic emissions. The element determines which energy differences are possible, but the energy-wavelength relationship itself is a fundamental property of photons.

Question 7

In a photoelectric setup, electrons are emitted from a metal and collected at an anode, producing a current. The researcher increases the intensity of incident light while also decreasing the frequency slightly, keeping it still above threshold. Which combined effect is expected on (i) photocurrent and (ii) stopping potential? (Constants: $h$ , $e$ .)

(i) increases; (ii) decreases (correct answer)
(i) decreases; (ii) increases
(i) increases; (ii) increases
(i) unchanged; (ii) unchanged

Explanation: This question tests understanding of how simultaneous changes in intensity and frequency affect photoelectric measurements. Increasing intensity at constant frequency increases the number of incident photons, leading to more electron emissions and higher photocurrent. Separately, decreasing frequency (while staying above threshold) reduces photon energy, resulting in electrons with lower maximum kinetic energy and thus lower stopping potential via eVs = hf - φ. These effects are independent: intensity affects the number of electrons (current) while frequency affects their maximum energy (stopping potential). The combined result is increased photocurrent with decreased stopping potential. This independence of intensity and frequency effects is a key feature of the photon model, contrasting with classical wave predictions where amplitude and frequency would be interrelated.

Question 8

A spectrometer detects a set of discrete emission lines from a hydrogen discharge tube. The researcher replaces the tube with a different hydrogen tube at lower gas pressure and observes the same line wavelengths but narrower line widths. Which statement is most consistent with quantized emission and experimental broadening?

Allowed transition energies remain the same, while reduced collisions can decrease line broadening (correct answer)
Lower pressure changes hydrogen's energy levels, but the spectrometer rescales wavelengths to match
Lower pressure reduces photon energy, so wavelengths should shift to longer values
Narrower lines imply photons are less quantized at low pressure

Explanation: This question tests understanding of spectral line properties and broadening mechanisms. The wavelengths of emission lines are determined by energy differences between quantized atomic levels, which are intrinsic properties unaffected by gas pressure. Lower pressure reduces collision frequency between atoms, decreasing collision broadening - a mechanism that slightly spreads the observed wavelength distribution around the central value. With fewer collisions, lines appear narrower (more monochromatic) while maintaining the same center wavelengths. This observation confirms that the fundamental transition energies remain constant while only the statistical broadening changes. The principle that atomic energy levels are pressure-independent is crucial for spectroscopic analysis across different conditions. Line narrowing at low pressure is exploited in precision spectroscopy to better resolve closely spaced transitions.

Question 9

A researcher compares light emitted by a hydrogen discharge tube to light from a heated tungsten filament. The hydrogen source shows sharp lines at specific wavelengths, while the filament shows a broad continuous distribution. Which statement best accounts for this difference using energy quantization?

Hydrogen emission arises from quantized electronic transitions, whereas filament radiation comes from many closely spaced energy changes producing a continuum (correct answer)
Hydrogen produces lines because its photons travel as particles, while filament photons travel as waves
Filament light is continuous because its electrons have no threshold frequency, unlike hydrogen atoms
Hydrogen lines occur because higher intensity forces photons into discrete wavelengths, unlike the dim filament

Explanation: This question tests understanding of the difference between discrete line spectra and continuous spectra based on energy quantization. Hydrogen atoms have discrete, widely-spaced electronic energy levels, so electrons can only make specific transitions between these levels, producing photons of specific energies and thus discrete wavelengths (line spectrum). In contrast, a heated tungsten filament contains densely packed atoms where thermal vibrations create a near-continuum of possible energy states, allowing emission across a broad range of wavelengths (continuous spectrum). The correct answer A recognizes this fundamental difference between quantized transitions in isolated atoms versus the quasi-continuous energy distribution in condensed matter. Option B incorrectly attributes the difference to wave-particle duality rather than energy quantization. Option C wrongly introduces threshold frequency, which relates to the photoelectric effect, not emission. Option D incorrectly suggests intensity determines spectral type, when it's actually the nature of the energy states that matters.

Question 10

In a photoelectric experiment, two light sources illuminate the same metal surface: Source 1 has frequency $f_1$ slightly above the threshold $f_0$ ; Source 2 has frequency $f_2>f_1$ . The intensity of Source 1 is adjusted so that the measured photocurrent (number of emitted electrons per second) matches that of Source 2. Constants: $h=6.63\times 10^{-34}\ \text{J}\cdot\text{s}$ , $e=1.60\times 10^{-19}\ \text{C}$ . Which statement is most consistent with the photoelectric effect under these conditions?

The stopping potential is the same for both sources because the photocurrent is the same
Source 1 produces higher-energy electrons because it has higher intensity
Source 2 requires a larger stopping potential because higher-frequency photons produce higher maximum electron kinetic energy (correct answer)
Neither source can eject electrons because matching the photocurrent implies $f_1=f_2$

Explanation: This question tests understanding of how photon frequency affects electron kinetic energy in the photoelectric effect. The photoelectric effect equation KEmax = hf - φ shows that maximum electron kinetic energy depends only on photon frequency, not intensity. Since Source 2 has higher frequency (f₂ > f₁), it produces electrons with higher maximum kinetic energy, requiring a larger stopping potential to prevent current flow (eVs = KEmax). The correct answer C recognizes this relationship between frequency and stopping potential. Option A incorrectly assumes stopping potential depends on photocurrent (which relates to intensity), when it actually depends only on maximum electron energy. Option B confuses intensity with photon energy - higher intensity means more photons, not higher energy per photon. Option D makes the false claim that equal photocurrents require equal frequencies, when actually Source 1's higher intensity compensates for its lower frequency.

Question 11

A hydrogen discharge tube shows a prominent 656 nm line. The tube is then placed in an environment that slightly increases the energy spacing between electronic levels (e.g., due to an external perturbation that raises the energy of excited states relative to the ground state). Assuming the same transition is still allowed, what change is most expected for the emitted photon from that transition?

The wavelength increases because larger energy gaps produce lower-energy photons
The wavelength decreases because a larger energy gap corresponds to a higher-energy photon (correct answer)
The wavelength is unchanged because emission depends only on tube current, not energy levels
The emission becomes continuous rather than discrete because the perturbation destroys quantization

Explanation: This question tests understanding of how energy level spacing affects emission wavelengths in atomic spectra. The 656 nm line in hydrogen corresponds to a specific electronic transition with energy E = hc/λ. When an external perturbation increases the energy spacing between levels, the energy difference for this transition increases, requiring a higher-energy photon to be emitted. Since E = hc/λ, a higher-energy photon has shorter wavelength (energy and wavelength are inversely related). The correct answer B recognizes that larger energy gaps produce higher-energy photons with shorter wavelengths. Option A incorrectly states the inverse relationship between energy and wavelength. Option C wrongly suggests emission wavelengths are independent of energy levels. Option D incorrectly claims that perturbations destroy quantization, when in fact they simply shift the quantized levels. This principle explains phenomena like the Stark effect where external fields shift spectral lines.

Question 12

A vacuum photoelectric cell with a clean sodium cathode is illuminated with monochromatic light. The light frequency is varied while the intensity is held constant. Electron emission is first detected at $f_0 = 5.5\times10^{14}\ \text{Hz}$ . For $f>f_0$ , a stopping potential $V_s$ is measured. Constants: $h=6.63\times10^{-34}\ \text{J}\cdot\text{s}$ , $e=1.60\times10^{-19}\ \text{C}$ . Which outcome would be expected if the frequency is increased from $6.0\times10^{14}$ Hz to $7.0\times10^{14}$ Hz at the same intensity?

The stopping potential increases because the maximum electron kinetic energy increases with photon frequency above $f_0$ . (correct answer)
The stopping potential decreases because higher frequency photons are absorbed less efficiently by the metal.
Electron emission ceases because intensity, not frequency, determines whether electrons are emitted.
The stopping potential remains unchanged because only the intensity controls electron kinetic energy.

Explanation: The question tests understanding of the photoelectric effect and line spectra. The photoelectric effect involves the emission of electrons when light hits a material, highlighting the quantization of energy. In the scenario, increasing the light frequency while keeping intensity constant affects the stopping potential, illustrating key principles. The correct answer is aligned with the principle that maximum electron kinetic energy increases with photon frequency above the threshold. A common distractor fails because it incorrectly assumes that higher frequency reduces absorption efficiency. To apply this principle, consider how photon energy hf exceeds the work function, leading to higher K_max and thus higher V_s. Remember that energy quantization is crucial for predicting outcomes in photoelectric experiments.

Question 13

A photoelectric cell has threshold frequency $f_0=4.0\times10^{14}\ \text{Hz}$ . Light of frequency $3.5\times10^{14}$ Hz produces no current at any intensity tested. The experimenter then switches to $4.5\times10^{14}$ Hz at low intensity and detects a small current. Which outcome would be expected if the intensity at $4.5\times10^{14}$ Hz is increased while frequency is fixed?

The maximum kinetic energy of emitted electrons increases because more photons hit the surface per second.
The maximum kinetic energy decreases because higher intensity increases electron-electron repulsion.
The photocurrent increases because more photons per unit time eject more electrons, while $K_{\max}$ stays the same. (correct answer)
Electron emission stops because intensity changes the work function of the metal.

Explanation: The question tests understanding of the photoelectric effect and line spectra. The photoelectric effect involves the emission of electrons when light hits a material, highlighting the quantization of energy. In the scenario, increasing intensity at a frequency above threshold affects photocurrent, illustrating key principles. The correct answer is aligned with the principle that intensity increases photocurrent while K_max remains unchanged. A common distractor fails because it incorrectly assumes intensity affects K_max. To apply this principle, consider how more photons eject more electrons without changing per-photon energy. Remember that energy quantization is crucial for distinguishing intensity from frequency effects.

Question 14

A discharge tube experiment records two emission lines from hydrogen at 656 nm and 486 nm. The lab notes indicate both lines correspond to transitions ending at the same lower energy level. Based on quantized energy levels, which statement is most consistent with the relative photon energies?

The 656 nm line has higher photon energy because its wavelength is longer.
The 486 nm line has higher photon energy because shorter wavelength corresponds to higher frequency. (correct answer)
Both lines have the same photon energy because they end at the same lower level.
Photon energy depends on intensity, so the brighter line must have higher energy.

Explanation: The question tests understanding of the photoelectric effect and line spectra. Line spectra arise from quantized energy transitions in atoms, producing discrete wavelengths. In the scenario, comparing two emission lines ending at the same level assesses photon energies, illustrating key principles. The correct answer is aligned with the principle that shorter wavelength means higher photon energy via E = hc/λ. A common distractor fails because it incorrectly assumes longer wavelength has higher energy. To apply this principle, consider how upper level differences determine energy drops. Remember that energy quantization is crucial for relating wavelength to transition energy.

Question 15

In a hydrogen emission experiment, a student claims that increasing the discharge current should shift the wavelengths of all spectral lines because more electrons collide with atoms. The instructor observes that line positions stay fixed while brightness changes. Which conclusion is most consistent with quantized emission spectra?

Line wavelengths are determined by fixed energy differences between allowed states; collision rate mainly affects intensity. (correct answer)
Line wavelengths depend on the number of excited atoms, so higher current must shift the spectrum.
Higher current increases photon frequency directly, so wavelengths decrease continuously.
The spectrum is continuous but appears discrete due to limitations of the detector.

Explanation: The question tests understanding of the photoelectric effect and line spectra. Line spectra arise from quantized energy transitions in atoms, producing discrete wavelengths. In the scenario, increasing current brightens lines without shifting them, illustrating key principles. The correct answer is aligned with the principle that wavelengths are fixed by energy differences, while current affects intensity. A common distractor fails because it incorrectly assumes current shifts spectra. To apply this principle, consider how current influences excitation rates but not levels. Remember that energy quantization is crucial for fixed line positions.

Question 16

A spectrometer measures discrete emission lines from a hydrogen discharge tube. The student replaces hydrogen with a gas at much higher pressure in the same tube and notes that the lines broaden slightly but remain centered at the same wavelengths. Which interpretation is most consistent with the origin of line spectra?

Line centers are set by quantized electronic transitions; pressure can broaden lines without changing transition energies. (correct answer)
Line centers shift because pressure changes Planck's constant inside the tube.
Discrete lines occur only at low pressure; higher pressure forces a continuous spectrum by eliminating quantization.
Line centers are determined by the detector calibration, so pressure changes cannot affect any spectral features.

Explanation: The question tests understanding of the photoelectric effect and line spectra. Line spectra arise from quantized energy transitions in atoms, producing discrete wavelengths. In the scenario, higher pressure broadens lines without shifting centers, illustrating key principles. The correct answer is aligned with the principle that line centers are set by quantized transitions, with pressure affecting broadening. A common distractor fails because it incorrectly assumes pressure eliminates quantization. To apply this principle, distinguish between transition energies and environmental effects. Remember that energy quantization is crucial for core spectral features.

Question 17

A photoelectric experiment uses two different metals, X and Y, illuminated with the same monochromatic light of frequency $f$ . Metal X emits electrons; metal Y does not. No surface contamination is detected. Which statement is most consistent with this result?

Metal Y must have a higher threshold frequency (larger work function) than metal X. (correct answer)
Metal Y must have a lower work function, so electrons are emitted with too much kinetic energy to detect.
Metal Y does not emit because intensity is too low; frequency is irrelevant to emission.
Metal Y does not emit because photons behave purely as waves at this frequency.

Explanation: The question tests understanding of the photoelectric effect and line spectra. The photoelectric effect involves the emission of electrons when light hits a material, highlighting the quantization of energy. In the scenario, different metals respond differently to the same light, illustrating key principles. The correct answer is aligned with the principle that higher threshold frequency means larger work function. A common distractor fails because it incorrectly assumes intensity determines emission. To apply this principle, compare hf to φ for each metal. Remember that energy quantization is crucial for material-specific thresholds.

Question 18

In a hydrogen discharge tube, a new line appears in the ultraviolet when the accelerating voltage is increased, while the previously observed visible lines remain at the same wavelengths. Which explanation is most consistent with quantized energy levels?

Higher voltage allows access to higher excited states, enabling additional transitions with larger energy drops (shorter wavelengths). (correct answer)
Higher voltage increases the speed of light in the tube, shifting some photons into the ultraviolet.
Higher voltage increases photon intensity, which directly decreases wavelength for all emitted light.
The ultraviolet line indicates that energy levels are continuous at higher voltages.

Explanation: The question tests understanding of the photoelectric effect and line spectra. Line spectra arise from quantized energy transitions in atoms, producing discrete wavelengths. In the scenario, higher voltage adds a new ultraviolet line without shifting others, illustrating key principles. The correct answer is aligned with the principle that higher voltage accesses higher states, enabling shorter-wavelength transitions. A common distractor fails because it incorrectly assumes voltage shifts all wavelengths. To apply this principle, consider excitation energy and level access. Remember that energy quantization is crucial for discrete line appearance.

Question 19

A researcher compares two photoelectric measurements on the same metal surface. Trial 1 uses light of frequency $f=1.2f_0$ and low intensity; Trial 2 uses the same frequency but 10× higher intensity. The stopping potential is measured in both trials. Constants: $h=6.63\times10^{-34}\ \text{J}\cdot\text{s}$ , $e=1.60\times10^{-19}\ \text{C}$ . Which statement best reflects the principle illustrated by the photoelectric effect?

The stopping potential is higher in Trial 2 because higher intensity increases photon energy.
The stopping potential is the same in both trials because it depends on photon frequency, not intensity. (correct answer)
The stopping potential is lower in Trial 2 because more photons reduce the energy per photon.
No electrons are emitted in Trial 1 because intensity must exceed a threshold to eject electrons.

Explanation: The question tests understanding of the photoelectric effect and line spectra. The photoelectric effect involves the emission of electrons when light hits a material, highlighting the quantization of energy. In the scenario, comparing trials with same frequency but different intensities affects stopping potential, illustrating key principles. The correct answer is aligned with the principle that stopping potential depends on frequency, not intensity. A common distractor fails because it incorrectly assumes higher intensity increases photon energy. To apply this principle, consider how intensity affects photocurrent but not K_max. Remember that energy quantization is crucial for separating frequency and intensity effects.

Question 20

A photoelectric experiment measures stopping potential for multiple frequencies above threshold and finds that extrapolating to zero stopping potential gives $f_0$ . The student claims $f_0$ depends on the applied stopping voltage range used in the instrument. Which response is most consistent with the underlying principle?

$f_0$ is an intrinsic property of the metal surface (work function) and does not depend on the instrument's voltage range. (correct answer)
$f_0$ increases when the stopping voltage range is increased because electrons gain energy from the circuit.
$f_0$ depends only on intensity, so voltage range affects it indirectly through brightness.
$f_0$ varies randomly because photon energies are continuous and not fixed by frequency.

Explanation: The question tests understanding of the photoelectric effect and line spectra. The photoelectric effect involves the emission of electrons when light hits a material, highlighting the quantization of energy. In the scenario, threshold frequency is determined by extrapolation, illustrating key principles. The correct answer is aligned with the principle that f_0 is intrinsic to the metal's work function, independent of voltage range. A common distractor fails because it incorrectly assumes voltage affects f_0. To apply this principle, use the Einstein equation for consistency. Remember that energy quantization is crucial for material properties.