College Physics Quiz: The Photoelectric Effect

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The Photoelectric EffectQuestion 1 of 20

In a photoelectric effect experiment, light of frequency $f_1$ produces photoelectrons with maximum kinetic energy $KE_1$ . When the frequency is increased to $f_2 = 2f_1$ , the maximum kinetic energy becomes $KE_2$ . If the work function of the metal is $\phi$ , which expression correctly relates $KE_2$ to $KE_1$ ?

KE_2 = 2KE_1

KE_2 = KE_1 + hf_1

KE_2 = 2KE_1 + \phi

KE_2 = KE_1 + \phi

KE_2 = 4KE_1

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College Physics Quiz

College Physics Quiz: The Photoelectric Effect

Practice The Photoelectric Effect in College Physics 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 The Photoelectric Effect, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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.

All questions

Question 1

$KE_2 = 2KE_1$
$KE_2 = KE_1 + hf_1$ (correct answer)
$KE_2 = 2KE_1 + \phi$
$KE_2 = KE_1 + \phi$
$KE_2 = 4KE_1$

Explanation: When you encounter photoelectric effect problems, remember that Einstein's equation

KE_{max} = hf - \phi

governs the relationship between photon energy and electron kinetic energy. The work function

\phi

represents the minimum energy needed to remove an electron from the metal surface. Let's apply this equation to both scenarios. For the first frequency:

KE_1 = hf_1 - \phi

. For the doubled frequency:

KE_2 = h(2f_1) - \phi = 2hf_1 - \phi

. To find the relationship between

KE_2

and

KE_1

, we can rearrange the first equation:

hf_1 = KE_1 + \phi

. Substituting this into the second equation:

KE_2 = 2hf_1 - \phi = 2(KE_1 + \phi) - \phi = 2KE_1 + 2\phi - \phi = 2KE_1 + \phi

. Wait—let me recalculate more carefully. We have

KE_2 = 2hf_1 - \phi

and

KE_1 = hf_1 - \phi

, so

KE_2 - KE_1 = 2hf_1 - \phi - (hf_1 - \phi) = hf_1

. Therefore,

KE_2 = KE_1 + hf_1

, confirming answer B. Looking at the wrong answers: A assumes the kinetic energy simply doubles with frequency, ignoring that the work function creates a constant offset. C incorrectly adds the work function when deriving the difference. D confuses the work function with the photon energy difference. Remember this pattern: in photoelectric problems, when frequency changes, the kinetic energy difference equals the photon energy difference (

h \Delta f

), not a simple proportional relationship. The work function affects the absolute values but cancels out when finding differences.

Question 2

In a photoelectric effect setup, increasing the intensity of monochromatic light while keeping the frequency constant will:

increase the maximum kinetic energy of emitted photoelectrons and increase the photoelectric current
increase the maximum kinetic energy of emitted photoelectrons but leave the photoelectric current unchanged
leave the maximum kinetic energy unchanged but increase the photoelectric current (correct answer)
decrease the maximum kinetic energy but increase the photoelectric current proportionally
leave both the maximum kinetic energy and photoelectric current completely unchanged

Explanation: When you encounter photoelectric effect questions, focus on Einstein's key insight: light behaves as discrete packets of energy called photons, and the energy of each photon depends only on frequency, not intensity. The photoelectric equation is

KE_{max} = hf - \phi

, where

h

is Planck's constant,

f

is frequency, and

\phi

is the work function of the material. Notice that maximum kinetic energy depends only on frequency—intensity doesn't appear in this equation at all. Since the frequency remains constant in this problem, the maximum kinetic energy of emitted photoelectrons stays unchanged. However, intensity refers to the number of photons hitting the surface per unit time. More photons means more electrons can potentially be ejected (assuming the frequency is above the threshold). This increases the photoelectric current, which measures the flow of emitted electrons. Looking at the wrong answers: Choice A incorrectly suggests intensity affects maximum kinetic energy—this reflects pre-quantum thinking where more intense light was thought to carry more energy per particle. Choice B makes the same error about kinetic energy and wrongly claims current stays constant despite more photons arriving. Choice D incorrectly states that maximum kinetic energy decreases, which contradicts the photoelectric equation entirely. Remember this key distinction: frequency determines the energy per photon (affecting maximum kinetic energy), while intensity determines the number of photons (affecting current). This separation of energy and particle number is fundamental to quantum mechanics and frequently tested.

Question 3

In a photoelectric experiment, the threshold frequency for a particular metal is $f_0 = 5.0 \times 10^{14}$ Hz. If this metal is illuminated with light of frequency $f = 7.0 \times 10^{14}$ Hz, what is the ratio of the maximum kinetic energy of the photoelectrons to the work function of the metal?

$\frac{2.0}{5.0} = 0.40$ (correct answer)
$\frac{7.0}{5.0} = 1.40$
$\frac{5.0}{2.0} = 2.50$
$\frac{7.0}{2.0} = 3.50$
$\frac{12.0}{5.0} = 2.40$

Explanation: When you encounter photoelectric effect problems, focus on Einstein's equation: the energy of incoming photons equals the work function plus the maximum kinetic energy of ejected electrons. Einstein's photoelectric equation is

hf = \phi + KE_{max}

, where

hf

is the photon energy,

\phi

is the work function, and

KE_{max}

is the maximum kinetic energy of photoelectrons. The work function equals

hf_0

, where

f_0

is the threshold frequency. First, find the work function:

\phi = hf_0 = h(5.0 \times 10^{14})

. Next, calculate the maximum kinetic energy using

KE_{max} = hf - \phi = h(7.0 \times 10^{14}) - h(5.0 \times 10^{14}) = h(2.0 \times 10^{14})

. The ratio becomes:

\frac{KE_{max}}{\phi} = \frac{h(2.0 \times 10^{14})}{h(5.0 \times 10^{14})} = \frac{2.0}{5.0} = 0.40

Answer A is correct:

\frac{2.0}{5.0} = 0.40

. Answer B incorrectly uses the incident frequency as the numerator instead of the excess energy above threshold. Answer C flips the correct ratio, putting the work function in the numerator and kinetic energy in the denominator. Answer D combines both errors: using the wrong frequencies and inverting the ratio. Remember that in photoelectric problems, only the energy above the threshold frequency becomes kinetic energy of the ejected electrons. Always subtract the threshold frequency from the incident frequency to find the "excess" energy that converts to kinetic energy.

Question 4

A photoelectric experiment measures the stopping potential $V_s$ as a function of incident light frequency $f$ . The slope of the resulting $V_s$ vs. $f$ graph represents:

the work function of the metal divided by the elementary charge
Planck's constant divided by the elementary charge (correct answer)
the elementary charge divided by Planck's constant
the threshold frequency multiplied by Planck's constant
the reciprocal of the speed of light times Planck's constant

Explanation: When you encounter photoelectric effect problems involving graphs, focus on the fundamental equation that governs this phenomenon. The photoelectric equation is

E_{\text{photon}} = \phi + KE_{\max}

, where

E_{\text{photon}} = hf

is the photon energy,

\phi

is the work function, and

KE_{\max}

is the maximum kinetic energy of ejected electrons. The stopping potential

V_s

is the voltage needed to stop the most energetic photoelectrons, so

eV_s = KE_{\max}

. Substituting this into the photoelectric equation gives us

hf = \phi + eV_s

. Rearranging for stopping potential:

V_s = \frac{h}{e}f - \frac{\phi}{e}

. This is a linear equation in the form

y = mx + b

, where the slope is

\frac{h}{e}

(Planck's constant divided by elementary charge). This confirms answer B is correct. Let's examine why the other options are wrong: A represents the y-intercept (when

f = 0

), not the slope. C gives the reciprocal of the actual slope, which would have units of charge per energy rather than voltage per frequency. D has units of energy, not the voltage/frequency units required for this slope. Study tip: Remember that photoelectric effect graphs always have slope

\frac{h}{e}

when plotting stopping potential versus frequency. This ratio appears frequently in quantum physics problems, so memorize that

\frac{h}{e} \approx 4.14 \times 10^{-15}

V·s for quick sanity checks on your calculations.

Question 5

A metal surface has a work function of 2.4 eV. When illuminated with photons of energy 4.1 eV, photoelectrons are emitted. If a retarding potential of 1.2 V is applied, what happens to the photoelectrons?

All photoelectrons are stopped because the retarding potential exceeds their maximum kinetic energy
Only the most energetic photoelectrons can overcome the retarding potential and reach the collector (correct answer)
No photoelectrons can overcome the retarding potential because it equals their average kinetic energy
The photoelectrons gain additional energy from the retarding potential and move faster
All photoelectrons easily overcome the retarding potential since it is less than the photon energy

Explanation: This question tests the photoelectric effect and how retarding potentials affect photoelectron motion. When you see photoelectric problems involving retarding potentials, focus on calculating the maximum kinetic energy of emitted electrons and comparing it to the potential barrier. Using Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is:

KE_{max} = E_{photon} - \phi = 4.1 \text{ eV} - 2.4 \text{ eV} = 1.7 \text{ eV}

The retarding potential creates an energy barrier of 1.2 eV that photoelectrons must overcome to reach the collector. Since some photoelectrons have kinetic energies up to 1.7 eV (which exceeds 1.2 eV), the most energetic ones can overcome this barrier, while less energetic ones cannot. Choice A is incorrect because the retarding potential (1.2 V) is less than the maximum kinetic energy (1.7 eV), so not all electrons are stopped. Choice C misunderstands the physics—the retarding potential doesn't equal the average kinetic energy, and even if it did, the most energetic electrons would still pass through. Choice D fundamentally misunderstands retarding potentials, which oppose electron motion and reduce their kinetic energy, never increase it. Therefore, B correctly describes the outcome: only photoelectrons with sufficient initial kinetic energy (greater than 1.2 eV) can overcome the retarding potential. Study tip: Always calculate the maximum photoelectron kinetic energy first, then compare it to the retarding potential. Remember that photoelectrons have a distribution of energies from zero up to the maximum value.

Question 6

A photoelectric experiment uses a variable frequency light source. As the frequency increases from below the threshold frequency to well above it, which graph best describes how the maximum kinetic energy of photoelectrons varies with frequency?

A horizontal line at zero kinetic energy for all frequencies
A curve that starts at zero and increases exponentially with frequency
A straight line starting from the threshold frequency with positive slope (correct answer)
A parabolic curve starting from the threshold frequency
A straight line passing through the origin with positive slope

Explanation: When you encounter photoelectric effect questions, focus on Einstein's photoelectric equation: the maximum kinetic energy of emitted electrons depends linearly on photon frequency above a critical threshold. The photoelectric equation is

KE_{max} = hf - \phi

, where

h

is Planck's constant,

f

is frequency, and

\phi

is the work function (minimum energy needed to remove an electron). This equation reveals that maximum kinetic energy has a linear relationship with frequency, but only when

hf > \phi

. Below the threshold frequency

f_0 = \phi/h

, no electrons are emitted regardless of light intensity. Answer C correctly describes this behavior: a straight line starting from the threshold frequency with positive slope

h

. Below threshold, kinetic energy is zero (no photoelectron emission), and above threshold, kinetic energy increases linearly with frequency. Answer A is wrong because it ignores that photoelectrons are emitted with increasing kinetic energy once frequency exceeds the threshold. Answer B incorrectly suggests an exponential relationship, but the photoelectric effect demonstrates a strictly linear dependence on frequency above threshold. Answer D suggests a parabolic relationship, which has no basis in photoelectric theory—the relationship between kinetic energy and frequency is always linear, not quadratic. Remember this key pattern: photoelectric effect graphs always show a linear relationship above threshold with slope equal to Planck's constant. The x-intercept gives you the threshold frequency, and the y-intercept (when extrapolated) gives the negative work function.

Question 7

In a photoelectric effect demonstration, a zinc plate with work function 4.3 eV is illuminated with ultraviolet light of wavelength 250 nm. What is the maximum speed of the emitted photoelectrons? (Use $m_e = 9.11 \times 10^{-31}$ kg)

$4.2 \times 10^5$ m/s
$6.8 \times 10^5$ m/s
$8.1 \times 10^5$ m/s (correct answer)
$1.2 \times 10^6$ m/s
$2.1 \times 10^6$ m/s

Explanation: The photoelectric effect occurs when photons strike a metal surface and eject electrons. The key relationship is Einstein's photoelectric equation: the photon energy equals the work function plus the maximum kinetic energy of emitted electrons. Start by finding the photon energy using

E = \frac{hc}{\lambda}

. With

h = 6.63 \times 10^{-34}

J·s,

c = 3.00 \times 10^8

m/s, and

\lambda = 250 \times 10^{-9}

E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{250 \times 10^{-9}} = 7.96 \times 10^{-19} \text{ J}

Converting to eV:

E = \frac{7.96 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.97 \text{ eV}

The maximum kinetic energy of photoelectrons is:

KE_{max} = E_{photon} - W = 4.97 - 4.3 = 0.67 \text{ eV}

Converting back to joules:

KE_{max} = 0.67 \times 1.60 \times 10^{-19} = 1.07 \times 10^{-19} \text{ J}

Using

KE = \frac{1}{2}mv^2

v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(1.07 \times 10^{-19})}{9.11 \times 10^{-31}}} = 8.1 \times 10^5 \text{ m/s}

This confirms answer C. Answer A (

4.2 \times 10^5

m/s) results from using only half the correct kinetic energy. Answer B (

6.8 \times 10^5

m/s) comes from calculation errors in the energy conversion. Answer D (

1.2 \times 10^6

m/s) occurs when students forget to subtract the work function and use the full photon energy as kinetic energy. Remember: always subtract the work function from photon energy to find the kinetic energy available for electron motion.

Question 8

A photoelectric experiment is conducted with three different metals having work functions $\phi_1 = 2.1$ eV, $\phi_2 = 3.4$ eV, and $\phi_3 = 4.7$ eV. All three are illuminated with the same monochromatic light of energy 3.8 eV. Which statement correctly describes the photoelectric current from each metal?

All three metals produce the same photoelectric current since they receive the same light intensity
Metal 1 produces the highest current, metal 2 produces lower current, and metal 3 produces no current (correct answer)
Metal 3 produces the highest current because it has the highest work function
Only metal 1 produces photoelectrons because its work function is lowest
Metals 1 and 2 produce current, but metal 2's current is higher than metal 1's

Explanation: When you encounter photoelectric effect problems, focus on two key principles: whether photoelectrons are emitted at all (energy threshold), and how many are emitted (current magnitude). For photoelectron emission to occur, the photon energy must exceed the metal's work function. Here, the 3.8 eV photons can eject electrons from metals 1 and 2 (work functions 2.1 eV and 3.4 eV respectively) but not from metal 3 (4.7 eV work function) since

3.8 < 4.7

eV. The photoelectric current depends on how easily electrons are removed. Metal 1 has the lowest work function, so electrons are most easily ejected, producing the highest current. Metal 2 has a higher work function (3.4 eV), making electron ejection more difficult and resulting in lower current. Metal 3 produces zero current since

3.8 - 4.7 = -0.9

eV means no photoelectrons are emitted. Choice A incorrectly assumes equal currents despite different work functions - while light intensity is the same, the ease of electron ejection varies dramatically. Choice C wrongly suggests higher work functions produce more current, when actually they make electron emission harder. Choice D incorrectly claims only metal 1 works; metal 2 also produces photoelectrons since

3.8 > 3.4

eV, just fewer than metal 1. Study tip: In photoelectric problems, always check the energy threshold first (

E_{photon} > \phi

), then remember that lower work functions mean easier electron ejection and higher currents for the same incident light.

Question 9

A student claims that in the photoelectric effect, if you double both the frequency and intensity of the incident light, the photoelectric current will increase by a factor of four. Which part of this reasoning is incorrect?

Doubling the frequency does not affect the photoelectric current at all
Doubling the intensity does not affect the photoelectric current at all
The current increases by a factor of two, not four, when frequency is doubled
The current depends on the product of frequency and intensity, so it increases by a factor of four
The frequency affects current magnitude, but not proportionally to the frequency change (correct answer)

Explanation: When analyzing photoelectric effect problems, you need to understand how frequency and intensity each affect the current independently, not as a combined product. In the photoelectric effect, doubling the intensity doubles the number of photons hitting the surface, which doubles the number of emitted electrons and thus doubles the current. Doubling the frequency increases each photon's energy (

E = hf

), but doesn't change how many photons are present, so it doesn't affect the current magnitude - it only affects the maximum kinetic energy of the emitted electrons. If you double both frequency and intensity simultaneously, only the intensity doubling affects the current, so the current increases by a factor of two, not four. Looking at the wrong answers: Choice A incorrectly states that frequency has no effect on current - while frequency doesn't affect current magnitude, it must exceed the threshold frequency for any current to flow at all. Choice B wrongly claims intensity doesn't affect current, when in fact intensity directly determines the number of emitted electrons. Choice C correctly identifies that the current increases by a factor of two (not four), making this statement actually correct, not incorrect. Choice D reflects the student's original misconception that current depends on the product of frequency and intensity. Since the question asks which part of the reasoning is incorrect, and choice C actually contains correct reasoning, none of the given options A-D properly identify an incorrect part of the student's reasoning. Remember: In photoelectric problems, intensity affects the number of electrons (current), while frequency affects their individual energies.

Question 10

In a photoelectric effect experiment, visible light with frequency $f_1 = 5.5 \times 10^{14}$ Hz produces photoelectrons with maximum kinetic energy 0.8 eV. If the frequency is reduced to $f_2 = 4.5 \times 10^{14}$ Hz, what is the maximum kinetic energy of the photoelectrons?

0.4 eV (correct answer)
0.6 eV
Zero (no photoelectrons produced)
1.0 eV
Cannot be determined without knowing the work function

Explanation: When you encounter photoelectric effect problems, you're dealing with Einstein's equation:

KE_{max} = hf - \phi

, where

KE_{max}

is the maximum kinetic energy,

h

is Planck's constant,

f

is frequency, and

\phi

is the work function (a material property). First, use the initial conditions to find the work function. With

f_1 = 5.5 \times 10^{14}

Hz producing

KE_{max1} = 0.8

eV:

0.8 = hf_1 - \phi

Using

h = 4.14 \times 10^{-15}

eV·s:

0.8 = (4.14 \times 10^{-15})(5.5 \times 10^{14}) - \phi

0.8 = 2.28 - \phi

\phi = 1.48

eV Now for the lower frequency

f_2 = 4.5 \times 10^{14}

Hz:

KE_{max2} = (4.14 \times 10^{-15})(4.5 \times 10^{14}) - 1.48

KE_{max2} = 1.86 - 1.48 = 0.38

eV ≈ 0.4 eV Choice A (0.4 eV) is correct. Choice B (0.6 eV) likely comes from incorrectly assuming kinetic energy scales linearly with frequency without accounting for the work function threshold. Choice C (zero) would only occur if the photon energy fell below the work function, but

hf_2 = 1.86

eV still exceeds

\phi = 1.48

eV. Choice D (1.0 eV) might result from calculation errors or misapplying the frequency relationship. Remember: in photoelectric problems, always find the work function first using the given data, then apply Einstein's equation to new conditions. The work function is material-specific and doesn't change between measurements.

Question 11

A photoelectric experiment uses light with photon energy 4.2 eV incident on a metal surface. The stopping potential is measured to be 1.7 V. If the same metal is then illuminated with light of photon energy 5.1 eV, what will be the new stopping potential?

2.0 V
2.6 V (correct answer)
3.4 V
5.1 V
6.8 V

Explanation: When you encounter photoelectric effect problems, you're dealing with Einstein's equation: the energy of incoming photons equals the work function of the metal plus the maximum kinetic energy of ejected electrons. The stopping potential directly measures this maximum kinetic energy. Einstein's photoelectric equation is

E_{photon} = \phi + eV_s

, where

\phi

is the work function and

eV_s

is the kinetic energy corresponding to the stopping potential. From the first experiment:

4.2 \text{ eV} = \phi + 1.7 \text{ eV}

, so the work function

\phi = 2.5 \text{ eV}

. For the second experiment with 5.1 eV photons:

5.1 \text{ eV} = 2.5 \text{ eV} + eV_s

, giving us

V_s = 2.6 \text{ V}

. Looking at the wrong answers: (A) 2.0 V incorrectly assumes the work function is 3.1 eV, perhaps from careless arithmetic. (C) 3.4 V might come from forgetting to subtract the work function entirely, just subtracting the original stopping potential from the new photon energy. (D) 5.1 V represents the common misconception that stopping potential equals photon energy, ignoring that some energy goes into overcoming the work function. Study tip: In photoelectric problems, always find the work function first using the given data, then apply it to the new conditions. Remember that the work function is a property of the metal that doesn't change between experiments—only the photon energy and resulting stopping potential change.

Question 12

Which of the following observations would be impossible to explain using classical physics but is naturally explained by the photoelectric effect?

Increasing light intensity increases the number of emitted electrons
Electrons are emitted instantaneously when light hits the surface
The kinetic energy of emitted electrons depends on light frequency, not intensity (correct answer)
More energetic electrons are emitted when brighter light is used
Electrons can be emitted from metal surfaces when illuminated with visible light

Explanation: When you encounter photoelectric effect questions, you're dealing with one of the key experiments that revealed the particle nature of light and helped launch quantum mechanics. Classical physics predicted that light's energy should depend only on its intensity (brightness), but the photoelectric effect revealed something completely different. The correct answer is C because this observation directly contradicts classical predictions. According to classical wave theory, brighter light should give electrons more energy since it carries more total energy. However, experiments showed that electron kinetic energy depends entirely on light frequency (color), not intensity. This makes perfect sense in quantum mechanics: each photon carries energy

E = hf

, so higher frequency means more energetic photons that can give electrons more kinetic energy. Let's examine why the other options don't work: A is actually predicted by both classical and quantum physics - more intense light means more photons, so more electrons get emitted. B might seem quantum, but classical physics could potentially explain instantaneous emission through wave energy absorption. D is simply wrong - brighter light produces more electrons, not more energetic ones. The key insight is that classical physics treated light as a continuous wave that could transfer any amount of energy, while quantum mechanics revealed light comes in discrete energy packets (photons). This quantization explains why only frequency matters for electron energy. Remember this pattern: when you see photoelectric effect questions, look for observations that depend on frequency rather than intensity - these are the uniquely quantum mechanical features that stumped classical physicists.

Question 13

In a photoelectric experiment, monochromatic light produces photoelectrons with kinetic energies ranging from 0 to 2.3 eV. If the work function of the metal is 1.8 eV, what can be concluded about the incident photons?

The photons have energy 1.8 eV and the experiment has measurement errors
The photons have energy 2.3 eV and some energy is lost during emission
The photons have energy 4.1 eV and the range represents different emission mechanisms (correct answer)
The photons have varying energies between 1.8 eV and 4.1 eV
The photons have energy 0.5 eV above the threshold frequency

Explanation: When you encounter photoelectric effect problems, focus on Einstein's photoelectric equation: the energy of incident photons equals the work function plus the maximum kinetic energy of emitted electrons. The key insight here is understanding what "kinetic energies ranging from 0 to 2.3 eV" means. The maximum kinetic energy (2.3 eV) occurs when electrons are emitted from the surface with no energy loss. Using Einstein's equation:

E_{photon} = \phi + KE_{max} = 1.8 \text{ eV} + 2.3 \text{ eV} = 4.1 \text{ eV}

The range from 0 to 2.3 eV exists because electrons can lose energy through different mechanisms during emission - some lose energy through collisions with atoms, surface interactions, or inelastic scattering before escaping the metal. This explains why choice C is correct: photons have 4.1 eV energy, and the kinetic energy range represents different energy loss mechanisms during emission. Choice A incorrectly assumes measurement error and underestimates photon energy. Choice B miscalculates photon energy as only 2.3 eV, which wouldn't overcome the 1.8 eV work function to produce any photoelectrons. Choice D misunderstands monochromatic light - all photons have identical energy, not varying energies. Remember: in photoelectric problems, "monochromatic" means single photon energy, and kinetic energy ranges always result from energy loss mechanisms during electron emission, not from varying photon energies. Always use the maximum kinetic energy in your calculations.

Question 14

Two different metals, A and B, have work functions $\phi_A = 1.8$ eV and $\phi_B = 3.2$ eV respectively. Both are illuminated with the same monochromatic light of energy 2.5 eV. Which statement correctly describes the photoelectric effect for these metals?

Both metals will emit photoelectrons with the same maximum kinetic energy of 2.5 eV
Metal A will emit photoelectrons with maximum kinetic energy 0.7 eV, while metal B will not emit any photoelectrons (correct answer)
Metal A will emit photoelectrons with maximum kinetic energy 4.3 eV, while metal B will emit photoelectrons with maximum kinetic energy 5.7 eV
Both metals will emit photoelectrons, but metal B will produce more energetic electrons than metal A
Neither metal will emit photoelectrons because the photon energy is less than both work functions

Explanation: The photoelectric effect occurs when light provides enough energy to overcome a material's work function and eject electrons. The key equation is Einstein's photoelectric equation:

KE_{max} = h\nu - \phi

, where the maximum kinetic energy equals the photon energy minus the work function. For photoelectrons to be emitted, the photon energy must exceed the work function. Here, the light has energy 2.5 eV. For metal A with

\phi_A = 1.8

eV, since 2.5 eV > 1.8 eV, photoelectrons will be emitted with maximum kinetic energy:

KE_{max} = 2.5 - 1.8 = 0.7

eV. For metal B with

\phi_B = 3.2

eV, since 2.5 eV < 3.2 eV, the photon energy is insufficient to overcome the work function, so no photoelectrons are emitted. Answer A incorrectly assumes both metals emit electrons and that maximum kinetic energy equals photon energy, ignoring the work function entirely. Answer C adds the photon energy to the work function instead of subtracting it, showing a fundamental misunderstanding of the photoelectric equation. Answer D incorrectly assumes both metals will emit electrons—since metal B's work function exceeds the photon energy, it cannot emit any photoelectrons regardless of their potential energy. When approaching photoelectric effect problems, always first check if the photon energy exceeds each material's work function. Only then calculate kinetic energies using Einstein's equation. Remember: higher work function means it's harder to eject electrons, not that ejected electrons are more energetic.

Question 15

Two identical photoelectric experiments are set up, but Experiment A uses a light source with twice the power of Experiment B. Both use the same frequency of light and the same metal surface. How do the maximum kinetic energies and photoelectric currents compare?

Experiment A has twice the maximum kinetic energy and twice the photoelectric current
Experiment A has the same maximum kinetic energy but twice the photoelectric current (correct answer)
Experiment A has twice the maximum kinetic energy but the same photoelectric current
Both experiments have identical maximum kinetic energies and photoelectric currents
Experiment A has four times the maximum kinetic energy and twice the photoelectric current

Explanation: When you encounter photoelectric effect problems, focus on what determines each measured quantity. The photoelectric effect depends on both the frequency and intensity of light, but they affect different aspects of the phenomenon. The maximum kinetic energy of ejected photoelectrons is determined solely by the photon energy, which depends on frequency:

KE_{max} = hf - \phi

, where

h

is Planck's constant,

f

is frequency, and

\phi

is the work function of the metal. Since both experiments use the same frequency and same metal surface, the maximum kinetic energy is identical in both cases. The photoelectric current, however, measures the number of photoelectrons ejected per unit time. Higher power means more photons per second hitting the surface (since power = energy per unit time). With twice the power, Experiment A delivers twice as many photons per second, ejecting twice as many electrons and producing twice the current. Looking at the wrong answers: Choice A incorrectly assumes power affects maximum kinetic energy—this is a common misconception. Choice C reverses the relationship, suggesting kinetic energy depends on power while current doesn't. Choice D ignores the effect of power entirely, missing that more photons create more photoelectrons. The correct answer is B: same maximum kinetic energy (frequency determines this) but twice the photoelectric current (power determines photon rate). Remember this key distinction: in photoelectric effect problems, frequency controls the energy per photoelectron, while intensity (power) controls how many photoelectrons are produced. Don't confuse these two independent effects.

Question 16

Two photoelectric experiments are performed using the same metal but different light sources. Experiment 1 uses red light ( $\lambda = 650$ nm) and Experiment 2 uses blue light ( $\lambda = 450$ nm). If both experiments produce photoelectrons, how do the stopping potentials compare?

The stopping potential for red light is higher because red photons have longer wavelengths
The stopping potential for blue light is higher because blue photons have higher energy (correct answer)
Both stopping potentials are equal because they use the same metal surface
The stopping potential for red light is higher because red light typically has higher intensity
No comparison can be made without knowing the work function of the metal

Explanation: When you encounter photoelectric effect questions, focus on the relationship between photon energy and the kinetic energy of emitted electrons. The key equation is Einstein's photoelectric equation:

KE_{max} = hf - \phi

, where

h

is Planck's constant,

f

is frequency, and

\phi

is the work function. Since photon energy equals

hf = \frac{hc}{\lambda}

, shorter wavelengths mean higher photon energies. Blue light (450 nm) has higher energy photons than red light (650 nm). When a higher-energy photon strikes an electron, it transfers more energy beyond what's needed to overcome the work function, giving the electron greater kinetic energy. The stopping potential is directly proportional to this maximum kinetic energy, so blue light produces a higher stopping potential. Option A incorrectly suggests longer wavelengths create higher stopping potentials—this reverses the actual relationship. Longer wavelengths mean lower photon energies and lower stopping potentials. Option C fails because while the work function (determined by the metal) is constant, the photon energies differ, creating different stopping potentials. Option D confuses intensity with photon energy. Intensity affects the number of photoelectrons produced, not their individual energies or the stopping potential. Remember this pattern: in photoelectric effect problems, shorter wavelength always means higher photon energy, which translates to higher stopping potential. Don't confuse intensity (number of photons) with energy per photon—only the latter determines stopping potential.

Question 17

In a photoelectric experiment, the photocurrent is measured as a function of the applied voltage. When a small forward voltage is applied (helping electrons reach the collector), what happens to the photocurrent compared to zero applied voltage?

The photocurrent decreases because the forward voltage opposes electron emission
The photocurrent increases significantly because the voltage accelerates all photoelectrons
The photocurrent increases slightly because low-energy photoelectrons are now collected (correct answer)
The photocurrent remains exactly the same because voltage doesn't affect photoelectron production
The photocurrent becomes zero because the forward voltage prevents electron collection

Explanation: When you encounter photoelectric effect questions involving voltage and current, focus on understanding what limits the photocurrent at zero voltage versus what happens when you assist electron collection. At zero applied voltage, some photoelectrons emitted from the photocathode don't reach the collector because they have insufficient kinetic energy to overcome small contact potential differences or work function variations between the electrodes. These low-energy electrons are essentially "lost" and don't contribute to the measured current. When you apply a small forward voltage (positive potential on the collector), you're helping these marginal electrons reach the detector. The voltage doesn't create new photoelectrons—that depends only on photon energy and intensity—but it does ensure that essentially all emitted photoelectrons are collected. This results in a slight increase in photocurrent as you capture those low-energy electrons that were previously lost. Looking at the incorrect options: (A) misunderstands the voltage direction—forward voltage assists, not opposes, electron collection. (B) overestimates the effect; while current increases, it's not significant because most photoelectrons already had enough energy to reach the collector at zero voltage. (D) ignores the reality that some electrons need assistance to be detected, even though voltage doesn't affect the emission process itself. Remember this pattern: in photoelectric experiments, small forward voltages increase collection efficiency slightly, while you need reverse voltage (creating a stopping potential) to prevent current flow entirely. The current change with small forward voltage is modest because it only affects the lowest-energy photoelectrons.

Question 18

A photoelectric experiment is performed with caesium metal (work function 2.1 eV). Light with wavelength 400 nm is incident on the surface. What percentage of the incident photon energy is converted to kinetic energy of the most energetic photoelectrons?

32% (correct answer)
48%
68%
84%
100%

Explanation: When you encounter photoelectric effect problems, you're dealing with Einstein's equation: the energy of an incident photon either overcomes the work function (minimum energy to remove an electron) or becomes kinetic energy of the emitted photoelectron. Start by finding the photon energy. Using

E = \frac{hc}{\lambda}

with

h = 4.14 \times 10^{-15}

eV·s and

c = 3.00 \times 10^8

m/s:

E_{photon} = \frac{(4.14 \times 10^{-15})(3.00 \times 10^8)}{400 \times 10^{-9}} = 3.11 \text{ eV}

The photoelectric equation is

E_{photon} = \phi + KE_{max}

, where

\phi

is the work function (2.1 eV). Therefore:

KE_{max} = 3.11 - 2.1 = 1.01 \text{ eV}

The percentage of photon energy converted to kinetic energy is:

\frac{1.01}{3.11} \times 100\% = 32.5\% \approx 32\%

This confirms answer A (32%) is correct. Answer B (48%) likely comes from incorrectly using a rounded photon energy of 3.0 eV, giving

\frac{0.9}{3.0} \times 100\% = 30\%

, but then miscalculating the percentage. Answer C (68%) represents the work function percentage:

\frac{2.1}{3.11} \times 100\% = 67.5\%

. This is a common trap—confusing what fraction goes to overcoming the work function versus becoming kinetic energy. Answer D (84%) has no clear physical basis and likely represents a calculation error. Remember: in photoelectric problems, always check whether you're asked for the kinetic energy fraction or the work function fraction—they're complementary percentages that add to 100%.

Question 19

In a photoelectric experiment, light with wavelength 300 nm produces photoelectrons from a metal surface. When the wavelength is changed to 600 nm, no photoelectrons are observed. Which conclusion about the threshold wavelength $\lambda_{th}$ is correct?

$\lambda_{th} = 450$ nm (average of the two wavelengths)
$\lambda_{th} < 300$ nm (shorter than both wavelengths)
$\lambda_{th} > 600$ nm (longer than both wavelengths)
$300 \text{ nm} < \lambda_{th} < 600$ nm (between the two wavelengths) (correct answer)
$\lambda_{th} = 600$ nm (equal to the longer wavelength)

Explanation: When you encounter photoelectric effect problems, focus on the relationship between photon energy and the threshold energy needed to eject electrons. Photons must have enough energy to overcome the work function of the metal. The key insight is that photon energy is inversely related to wavelength:

E = \frac{hc}{\lambda}

. Shorter wavelengths mean higher energy photons, while longer wavelengths mean lower energy photons. Since 300 nm light produces photoelectrons, its photons have enough energy to overcome the work function. Since 600 nm light produces no photoelectrons, its photons lack sufficient energy. The threshold wavelength

\lambda_{th}

represents the exact boundary where photon energy equals the work function. Answer D is correct because

\lambda_{th}

must fall between 300 nm and 600 nm. Any wavelength shorter than

\lambda_{th}

(including 300 nm) will produce photoelectrons, while any wavelength longer than

\lambda_{th}

(including 600 nm) will not. Answer A incorrectly assumes you can simply average the wavelengths - there's no physical basis for this approach. Answer B suggests

\lambda_{th} < 300

nm, but if this were true, then 300 nm light wouldn't have enough energy to produce photoelectrons, contradicting the given information. Answer C suggests

\lambda_{th} > 600

nm, but then 600 nm light would produce photoelectrons, again contradicting the data. Remember: in photoelectric problems, always think about energy thresholds. The threshold wavelength is the longest wavelength (lowest energy) that can still eject electrons.

Question 20

A photoelectric effect experiment uses a variable-wavelength light source. When the wavelength is gradually increased from 200 nm, photoelectrons are observed until the wavelength reaches 350 nm, at which point the photoelectric current drops to zero. If the wavelength is then set to 300 nm and the light intensity is increased by a factor of 10, what happens to the stopping potential and the photoelectric current?

Stopping potential increases by factor of 10, current increases by factor of 10
Stopping potential increases by factor of 10, current remains constant
Stopping potential remains constant, current increases by factor of 10 (correct answer)
Both stopping potential and current remain constant

Explanation: When you encounter photoelectric effect problems, focus on Einstein's equation: the energy of incident photons determines the maximum kinetic energy of ejected electrons, while intensity only affects the number of photons (and thus current). The threshold wavelength of 350 nm tells us the work function of the material. At 300 nm, the photon energy is

E = \frac{hc}{\lambda} = \frac{hc}{300\text{ nm}}

, which is greater than the threshold energy

\frac{hc}{350\text{ nm}}

. The stopping potential depends only on the maximum kinetic energy of photoelectrons:

eV_s = E_{photon} - \phi

, where

\phi

is the work function. Since neither the wavelength (300 nm) nor the work function changes when intensity increases, the stopping potential remains constant. However, increasing intensity by a factor of 10 means 10 times more photons hit the surface per second. Since each photon above the threshold energy can eject one electron, the photoelectric current (number of electrons per second) increases proportionally by a factor of 10. Answer A incorrectly suggests stopping potential depends on intensity—it doesn't. The stopping potential is determined solely by photon energy (wavelength), not the number of photons. Answer B makes the same error about stopping potential while incorrectly claiming current stays constant. Answer D ignores the fundamental relationship between intensity and photocurrent. Remember: In photoelectric effect problems, wavelength determines stopping potential (via photon energy), while intensity determines current magnitude. These are independent effects—changing one doesn't affect the other.