CUET UG Physics Booster Test 3- Experimental Study of Photoelectric Effect -

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CUET UG Physics Booster Test 3- Experimental Study of Photoelectric Effect

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3- Experimental Study of Photoelectric Effect

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QUESTION 1 OF 20

Statements concerning the transmission of radiation through the experimental window:
1. If ordinary glass were used instead of quartz, zinc and magnesium might not exhibit photoelectric emission.
2. Alkali metals like lithium and sodium can exhibit emission even if the window only transmits visible light.
3. The quartz window strictly permits only visible light and blocks ultraviolet rays.
4. Transmitting UV radiation is crucial for materials with higher work functions.

Answer Review

1 Statements concerning the transmission of radiation through the experimental window:
1. If ordinary glass were used instead of quartz, zinc and magnesium might not exhibit photoelectric emission.
2. Alkali metals like lithium and sodium can exhibit emission even if the window only transmits visible light.
3. The quartz window strictly permits only visible light and blocks ultraviolet rays.
4. Transmitting UV radiation is crucial for materials with higher work functions.

A (1), (2), (4)

B (2), (3), (4)

C (1), (3)

D (1), (2), (3)

💡 Short Explanation

�� Quartz transmits ultraviolet radiation. �� High work-function metals need UV light. �� Ordinary glass absorbs most UV radiation.

✅ Correct Answer Explanation

Zinc and magnesium require ultraviolet radiation because of their relatively high work functions. Ordinary glass absorbs UV rays, so photoelectric emission may not occur if glass is used. → Statement 1 is correct. → Statement 2 is correct because alkali metals have lower work functions and can respond to visible light. → Statement 4 is correct because UV transmission is essential for metals requiring higher photon energies. → Statement 3 is incorrect because quartz transmits UV radiation rather than blocking it.

❓ Why Other Options Are Incorrect

�� Statement 3 → Opposite of the actual function of quartz.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement contradicting the known UV-transmitting property of quartz.

Final Logic:

�� Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Quartz = UV Gateway

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Quartz Window in Photoelectric Experiment

2 By reversing the commutator polarity and precisely locating the cut-off potential, a physicist effectively measures

A the exact number of electrons traversing the tube

B the kinetic energy of the least tightly bound electrons immediately after they absorb a photon

C the thermal emission rate of the cathode

D the continuous wave amplitude of the incident light

💡 Short Explanation

�� Stopping potential measures photoelectron energy. �� Retarding field opposes electron motion. �� Maximum kinetic energy is determined experimentally.

✅ Correct Answer Explanation

By applying a retarding potential and finding the stopping potential V₀, the maximum kinetic energy of emitted photoelectrons is determined using: Kmax = eV₀ → Although the wording is slightly unconventional, the question intends measurement of the kinetic energy of emitted photoelectrons through the stopping potential method. → Therefore option B is the best available answer.

❓ Why Other Options Are Incorrect

�� Option A → Number of electrons is measured through photocurrent.
�� Option C → Thermal emission is unrelated to photoelectric stopping potential.
�� Option D → Wave amplitude is not measured using a commutator.

🎯 Strategy

Used

�� Elimination

Application:

�� Select the option most closely related to stopping potential measurements.

Final Logic:

�� Stopping potential directly provides photoelectron kinetic energy.

🧠 Memory Trick

Stop Voltage → Electron Energy

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Stopping Potential

3 Incorrect statement about interpreting the voltmeter readings during the experiment:

A A positive reading of V implies the collector is at a higher potential than the emitter.

B Varying the potential difference alters the fraction of emitted electrons reaching the collector before saturation.

C A negative reading -V₀ corresponding to zero photocurrent gives the stopping potential.

D As long as the voltmeter shows a positive potential, the photocurrent is guaranteed to be strictly proportional to V.

💡 Short Explanation

�� Photocurrent is not proportional to voltage at all values. �� Saturation eventually occurs. �� Voltage affects collection efficiency.

✅ Correct Answer Explanation

Initially, increasing positive potential increases photocurrent because more photoelectrons reach the collector. However, after saturation current is reached, further increases in voltage do not increase the current. → Therefore the photocurrent is not strictly proportional to V. → Hence option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct interpretation of positive potential.
�� Option B → Correct because voltage influences collection efficiency.
�� Option C → Correct definition of stopping potential.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement contradicting the saturation current curve.

Final Logic:

�� Photocurrent eventually saturates and is not always proportional to voltage.

🧠 Memory Trick

Voltage ↑, Current ↑ ... Then Saturates

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photocurrent vs Applied Potential

4 For a constant frequency ν > ν₀, the saturation photocurrent I measured by the microammeter varies with the source distance r according to the relation:

A I ∝ r

B I ∝ r⁻¹

C I ∝ r⁻²

D I ∝ r²

💡 Short Explanation

�� Intensity follows inverse square law. �� Saturation current is proportional to intensity. �� Therefore current follows inverse square dependence.

✅ Correct Answer Explanation

For a point source: Intensity ∝ 1/r² Since saturation current is proportional to intensity: I ∝ Intensity Therefore: I ∝ 1/r² → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Current does not increase with distance.
�� Option B → Inverse dependence is weaker than actual relation.
�� Option D → Opposite trend.

🎯 Strategy

Used

�� Substitution

Application:

�� Combine inverse-square law with saturation current proportionality.

Final Logic:

�� I ∝ r⁻².

🧠 Memory Trick

Double Distance → Quarter Current

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Saturation Current and Intensity

5 When the source is moved closer to the emitter, the amplitude of electric and magnetic fields in the classical wave picture is ______, but in the photon picture, the energy of each individual photon remains ______.

A Increased, Increased

B Increased, Constant

C Constant, Increased

D Decreased, Constant

💡 Short Explanation

�� Moving closer increases intensity. �� Field amplitudes increase. �� Photon energy depends only on frequency.

✅ Correct Answer Explanation

In classical wave theory, greater intensity corresponds to larger electric and magnetic field amplitudes. → In the photon picture, the energy of an individual photon is: E = hν → Since frequency is unchanged, photon energy remains constant even though intensity increases. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Photon energy does not increase without frequency change.
�� Option C → Field amplitudes do increase.
�� Option D → Moving closer increases, not decreases, intensity.

🎯 Strategy

Used

�� Elimination

Application:

�� Distinguish intensity effects from frequency effects.

Final Logic:

�� Intensity changes amplitude, not photon energy.

🧠 Memory Trick

Intensity Changes Number, Frequency Changes Energy

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Intensity and Photon Energy

6 The strict linearity between photocurrent and intensity fundamentally rules out

A the conservation of energy in photoelectric interactions

B the particle nature of electromagnetic radiation

C a multi-photon absorption requirement for the ejection of a single electron under normal experimental intensities

D the existence of a threshold frequency

💡 Short Explanation

�� Photocurrent is directly proportional to intensity. �� One photon ejects one electron. �� Multi-photon processes would not generally produce strict linearity.

✅ Correct Answer Explanation

The experimentally observed linear relationship between photocurrent and intensity indicates that the number of emitted electrons is directly proportional to the number of incident photons. → This supports the one-photon–one-electron interaction model proposed by Einstein. → If emission required simultaneous absorption of multiple photons under ordinary experimental conditions, the dependence would generally be nonlinear. → Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Conservation of energy remains valid in photoelectric interactions.
�� Option B → Linearity actually supports the particle nature of radiation.
�� Option D → Threshold frequency is an experimentally verified property.

🎯 Strategy

Used

�� Elimination

Application:

�� Reject options that are supported rather than contradicted by photoelectric observations.

Final Logic:

�� Strict linearity supports single-photon absorption and rules out a multi-photon requirement.

🧠 Memory Trick

One Photon → One Electron

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photocurrent and Intensity Relationship

7 Correct statements analyzing emission rates:
1. Emission per second is independent of the incident photon's momentum.
2. Greater number of available energy quanta means more electrons absorb them.
3. Low intensity causes a delay in emission because fewer photons arrive per second.
4. Emission rate is zero if the incident photons have energy less than the work function, regardless of intensity.

A (1), (3)

B (2), (4)

C (2), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Emission rate depends on photon availability. �� Photon energy must exceed work function. �� No measurable delay occurs even at low intensity.

✅ Correct Answer Explanation

Statement 1 is correct because photoelectric emission depends primarily on photon energy rather than photon momentum. → Statement 2 is correct because increasing the number of available photons increases the probability of electron emission. → Statement 4 is correct because if hν < φ, photoelectric emission cannot occur regardless of intensity. → Statement 3 is incorrect because experiments show photoelectric emission occurs essentially instantaneously, even at very low intensities. → Therefore the correct combination is (1), (2), (4).

❓ Why Other Options Are Incorrect

�� Statement 3 → No measurable time lag is observed in photoelectric emission.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement contradicting the experimentally observed instantaneous emission.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Below Work Function → No Emission

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photoelectron Emission Rate

8 Match List I (Change in experimental parameter) with List II (Effect on proportionality):

List I	List II
1. Double the intensity (same frequency)	a. Number of emitted photons per second decreases (since each photon has higher energy)
2. Increase the frequency (same intensity in W/m²)	b. Saturation current doubles
3. Change to a metal with lower work function	c. Photocurrent reduces to zero
4. Apply stopping potential	d. Threshold frequency decreases

A (1)-b, (2)-c, (3)-d, (4)-a

B (1)-a, (2)-b, (3)-c, (4)-d

C (1)-b, (2)-a, (3)-d, (4)-c

D (1)-d, (2)-a, (3)-b, (4)-c

💡 Short Explanation

�� Intensity affects saturation current. �� Higher frequency means higher energy per photon. �� Lower work function lowers threshold frequency.

✅ Correct Answer Explanation

Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Increasing frequency does not directly reduce photocurrent to zero.
�� Option B → Intensity does not decrease photon count.
�� Option D → Lower work function does not directly double saturation current.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each physical change with its direct experimental consequence.

Final Logic:

�� Only Option A provides the correct set of pairings.

🧠 Memory Trick

Lower φ → Lower ν₀

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Intensity, Frequency and Work Function Effects

9 The reason the photocurrent eventually stops increasing and saturates at higher accelerating potentials is because

A The electric field inside the tube completely neutralizes the space charge

B the collector plate physically cannot hold any more electrons

C all the photoelectrons being emitted per second are already being successfully swept to the collector

D the work function of the emitter increases due to the depletion of electrons

💡 Short Explanation

�� Saturation means complete collection. �� No additional emitted electrons remain. �� Further voltage increase has no effect.

✅ Correct Answer Explanation

As accelerating potential increases, a greater fraction of emitted photoelectrons reaches the collector. → Eventually every emitted photoelectron is collected. → At this stage the photocurrent reaches saturation current and cannot increase further. → Therefore option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Space-charge effects are not the fundamental reason for saturation.
�� Option B → Collector capacity is not the limiting factor.
�� Option D → Work function remains unchanged.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate saturation current to complete electron collection.

Final Logic:

�� Saturation occurs when all emitted electrons are already collected.

🧠 Memory Trick

All Collected = Saturation

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Saturation Current

10 Incorrect statement regarding the mechanics of retarding potential:

A It applies an electric force on the emitted electrons in the direction opposite to their velocity.

B Only electrons with initial kinetic energy K > eV can reach the collector.

C The retarding potential alters the kinetic energy with which electrons are originally emitted from the surface.

D When V = -V₀, even the most energetic electrons are turned back just before reaching the collector.

💡 Short Explanation

�� Retarding potential acts after emission. �� Initial kinetic energy is fixed at emission. �� Frequency determines emitted kinetic energy.

✅ Correct Answer Explanation

The kinetic energy with which electrons are emitted is determined by Einstein's equation: Kmax = hν − φ → Retarding potential acts only after the electrons have been emitted and influences whether they can reach the collector. → Therefore it does not alter the original kinetic energy of emission. → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of retarding force.
�� Option B → Correct condition for reaching the collector.
�� Option D → Correct interpretation of stopping potential.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate emission processes from post-emission motion.

Final Logic:

�� Retarding potential affects electron motion, not emission energy.

🧠 Memory Trick

Frequency Sets Energy, Voltage Tests It

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Retarding Potential and Stopping Potential

11 Statements about the saturation state:
1. Different intensities I₃ > I₂ > I₁ yield different saturation currents.
2. For a given intensity, varying the frequency slightly changes the intensity definition.
3. Saturation current is independent of the accelerating potential once the threshold voltage is exceeded.
4. The stopping potential remains identical for I₃, I₂, I₁ assuming the same frequency.

A (1), (3), (4)

B (1), (2), (4)

C (2), (3)

D (1), (4)

💡 Short Explanation

�� Saturation current depends on intensity. �� Stopping potential depends on frequency. �� Saturation current becomes independent of voltage after saturation.

✅ Correct Answer Explanation

Statement 1 is correct because greater intensity produces more photoelectrons and hence larger saturation current. → Statement 3 is correct because once saturation is reached, increasing accelerating potential further does not increase photocurrent. → Statement 4 is correct because stopping potential depends on frequency and work function, not intensity. → Statement 2 is incorrect because frequency and intensity are independent physical quantities.

❓ Why Other Options Are Incorrect

�� Statement 2 → Changing frequency does not redefine intensity.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate intensity-dependent quantities from frequency-dependent quantities.

Final Logic:

�� Statements 1, 3 and 4 are correct.

🧠 Memory Trick

Intensity → Current, Frequency → Voltage

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Saturation Current and Stopping Potential

12 In the total electron collection regime (saturation), the role of space charge

A is maximized, severely restricting electron flow

B is effectively overcome by the strong positive potential of the collector

C forces the electrons to emit photons as they travel

D causes the electrons to return to the emitter

💡 Short Explanation

�� Strong positive potential attracts electrons. �� All emitted electrons are collected. �� Saturation current is achieved.

✅ Correct Answer Explanation

Space charge refers to the cloud of emitted electrons that can oppose further electron motion. In the saturation region, the collector is sufficiently positive to overcome this effect and collect all emitted photoelectrons. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Space-charge limitation is not dominant in the saturation region.
�� Option C → Electrons do not emit photons merely due to travel through the tube.
�� Option D → Electrons are attracted toward the collector, not forced back.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate saturation current to complete electron collection.

Final Logic:

�� Strong collector potential overcomes space-charge effects.

🧠 Memory Trick

Positive Collector Wins

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Saturation Current

13 Using Einstein's photoelectric equation, the cut-off point V₀ when a retarding potential is applied is precisely:

A (h/e)(ν - ν₀)

B (e/h)(ν - ν₀)

C (hν - φ₀)

D (h/e)(ν + ν₀)

💡 Short Explanation

�� eV₀ = hν − φ₀ �� φ₀ = hν₀ �� Substitute and simplify.

✅ Correct Answer Explanation

Einstein's equation: eV₀ = hν − φ₀ Since: φ₀ = hν₀ Therefore: eV₀ = h(ν − ν₀) Hence: V₀ = (h/e)(ν − ν₀) → Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal relationship is incorrect.
�� Option C → Represents energy, not potential.
�� Option D → Incorrect sign.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute φ₀ = hν₀ into Einstein's equation.

Final Logic:

�� V₀ = (h/e)(ν − ν₀).

🧠 Memory Trick

Stop Voltage = h/e × Excess Frequency

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Einstein's Photoelectric Equation

14 A sharply defined, critical value of negative potential V₀ exists because

A electrons are emitted with a continuous spectrum of kinetic energies up to a strictly defined maximum K_max

B all electrons are emitted with the exact same kinetic energy

C the potential field inside the tube is highly non-uniform

D the wave nature of light creates a sharp boundary of energy transfer

💡 Short Explanation

�� Electrons emerge with varying kinetic energies. �� Maximum kinetic energy is well defined. �� Stopping potential corresponds to K_max.

✅ Correct Answer Explanation

Photoelectrons are emitted with a range of kinetic energies from zero up to a maximum value K_max. → The stopping potential is the minimum retarding potential required to stop even the most energetic photoelectrons. → This leads to a sharply defined value of V₀. → Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Electrons do not all have identical kinetic energy.
�� Option C → Non-uniform fields are not the reason for stopping potential.
�� Option D → Wave theory does not explain the sharp stopping potential.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Connect stopping potential directly to maximum kinetic energy.

Final Logic:

�� A well-defined Kmax produces a well-defined stopping potential.

🧠 Memory Trick

Max Energy → Stop Voltage

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Stopping Potential

15 If the incident wavelength is 454 nm and the work function is 2.14 eV, what is the maximum kinetic energy in eV?
(hc/λ ≈ 19.89 × 10⁻¹⁹ J = 2.74 eV)

A 4.88 eV

B 2.74 eV

C 0.60 eV

D 1.60 eV

💡 Short Explanation

�� Kmax = Photon Energy − Work Function. �� Photon energy = 2.74 eV. �� Work function = 2.14 eV.

✅ Correct Answer Explanation

Using Einstein's equation: Kmax = hν − φ Given: Photon Energy = 2.74 eV Work Function = 2.14 eV Therefore: Kmax = 2.74 − 2.14 = 0.60 eV → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Sum rather than difference.
�� Option B → Represents photon energy only.
�� Option D → Numerical error.

🎯 Strategy

Used

�� Substitution

Application:

�� Directly apply Einstein's photoelectric equation.

Final Logic:

�� Kmax = 2.74 − 2.14 = 0.60 eV.

🧠 Memory Trick

Energy In − Energy Out

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Maximum Kinetic Energy

16 Correct statements analyzing the failure of wave theory regarding maximum kinetic energy:
1. Wave theory predicted Kmax should increase with intensity.
2. Wave theory predicted Kmax is independent of frequency.
3. Experiments showed Kmax is determined solely by photon energy and work function.
4. Experiments showed tighter bound electrons emerge with Kmax.

A (1), (3)

B (2), (3), (4)

C (1), (3), (4)

D (1), (2)

💡 Short Explanation

�� Classical theory linked energy to intensity. �� Experiments linked energy to frequency. �� Work function affects emitted energy.

✅ Correct Answer Explanation

Statement 1 is correct because classical wave theory predicted greater intensity should transfer more energy and produce larger kinetic energies. → Statement 3 is correct because experiments established: Kmax = hν − φ → Statement 2 is incorrect because wave theory did not specifically predict frequency independence in this form. → Statement 4 is incorrect because Kmax depends on photon energy and work function, not on tighter bound electrons emerging preferentially.

❓ Why Other Options Are Incorrect

�� Statement 2 → Not the experimentally established conclusion.
�� Statement 4 → Incorrect interpretation of photoelectron emission.

🎯 Strategy

Used

�� Elimination

Application:

�� Compare wave-theory predictions with Einstein's explanation.

Final Logic:

�� Statements 1 and 3 correctly describe the contrast.

🧠 Memory Trick

Wave → Intensity, Einstein → Frequency

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Failure of Classical Wave Theory

17 In Millikan's experimental verification, observing the shifts in stopping potential for different frequencies allowed the determination of ______, proving it was ______ of the nature of the material.

A Work function, Dependent

B Planck's constant, Independent

C Electron charge, Independent

D Threshold frequency, Dependent

💡 Short Explanation

�� Millikan verified Einstein's equation. �� Slope of V₀–ν graph equals h/e. �� Planck's constant is universal.

✅ Correct Answer Explanation

From: V₀ = (h/e)ν − φ/e the slope of the V₀ versus ν graph is h/e. → By measuring this slope, Millikan determined Planck's constant and confirmed that it is independent of the material used. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Work function depends on material.
�� Option C → Electron charge was already known and not the objective.
�� Option D → Threshold frequency varies with material.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Connect slope measurement with determination of Planck's constant.

Final Logic:

�� Millikan's experiment confirmed the universal value of h.

🧠 Memory Trick

Slope → h/e

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Millikan's Verification

18 Match List I (Graph features of V₀ vs ν) with List II (Physical constants):

List I	List II
1. Slope of the graph	a. -φ₀/e
2. x-intercept (frequency axis)	b. φ₀/e
3. y-intercept (potential axis)	c. ν₀
4. Value of V₀ at ν = 2ν₀	d. h/e

A (1)-d, (2)-c, (3)-a, (4)-b

B (1)-d, (2)-c, (3)-b, (4)-a

C (1)-a, (2)-b, (3)-c, (4)-d

D (1)-c, (2)-d, (3)-a, (4)-b

💡 Short Explanation

�� Slope = h/e. �� x-intercept = threshold frequency. �� y-intercept = −φ₀/e.

✅ Correct Answer Explanation

From: V₀ = (h/e)ν − φ₀/e → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong sign for y-intercept.
�� Option C → Multiple mismatches.
�� Option D → Slope and intercepts incorrectly assigned.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Compare each graph feature with the straight-line equation.

Final Logic:

�� Only Option A satisfies all graph relationships.

🧠 Memory Trick

Slope = h/e, Cutoff = ν₀

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: V₀ versus ν Graph

19 Incorrect statement about the time factor in photoelectric emission:

A Low intensity of incident light directly causes a significant delay in emission.

B The elementary process is the instantaneous absorption of a light quantum.

C Emission starts in a time of the order of 10⁻⁹ s or less.

D Wave theory incorrectly calculated that a single electron would take hours to absorb enough energy.

💡 Short Explanation

�� Photoemission is instantaneous. �� No measurable delay occurs. �� Even dim light can cause immediate emission.

✅ Correct Answer Explanation

Experiments show that photoelectric emission begins almost immediately after illumination, with a time lag less than about 10⁻⁹ s. → This remains true even for very low intensities. → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Correct photon explanation.
�� Option C → Agrees with experiments.
�� Option D → Correct criticism of classical wave theory.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement contradicting experimental observations.

Final Logic:

�� Low intensity affects rate, not emission delay.

🧠 Memory Trick

Dim Light ≠ Delay

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Instantaneous Emission

20 Statements concerning the nanosecond scale emission:
1. It proves energy is absorbed continuously over the wavefront.
2. It holds true even when the incident radiation is made exceedingly dim.
3. It demonstrates that intensity only determines how many electrons participate, not the time taken per electron.
4. It contradicts the classical wave picture of electromagnetism.

A (1), (2), (3)

B (2), (3), (4)

C (1), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Emission remains instantaneous. �� Intensity affects number of electrons. �� Classical wave theory fails to explain this behavior.

✅ Correct Answer Explanation

Statement 2 is correct because instantaneous emission occurs even at extremely low intensities. → Statement 3 is correct because intensity determines the number of emitted electrons, not the emission time per electron. → Statement 4 is correct because classical wave theory predicted measurable delays. → Statement 1 is incorrect because instantaneous emission supports photon absorption, not continuous energy accumulation.

❓ Why Other Options Are Incorrect

�� Statement 1 → Contradicts Einstein's photon explanation.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement inconsistent with experimental evidence.

Final Logic:

�� Statements 2, 3 and 4 correctly describe nanosecond-scale emission.

🧠 Memory Trick

Instant Photon → Instant Electron

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Instantaneous Nature of Photoelectric Effect

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