CUET UG Physics Booster Test 2- Experimental Study of Photoelectric Effect -

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CUET UG Physics Booster Test 2- 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 2- Experimental Study of Photoelectric Effect

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QUESTION 1 OF 20

Statements regarding the glass/quartz tube setup:
1. Quartz is specifically chosen because it permits UV rays to irradiate the plate.
2. It allows the study of UV-sensitive metals like zinc and magnesium.
3. Alkali metals would not show photoelectric effect if visible light is used.
4. The window ensures light reaches the emitter without being fully absorbed by standard glass.

Answer Review

1 Statements regarding the glass/quartz tube setup:
1. Quartz is specifically chosen because it permits UV rays to irradiate the plate.
2. It allows the study of UV-sensitive metals like zinc and magnesium.
3. Alkali metals would not show photoelectric effect if visible light is used.
4. The window ensures light reaches the emitter without being fully absorbed by standard glass.

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

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

C (1), (3)

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

💡 Short Explanation

�� Quartz transmits ultraviolet radiation. �� Zinc and magnesium require UV radiation. �� Ordinary glass absorbs UV radiation.

✅ Correct Answer Explanation

Quartz is used because ultraviolet radiation can pass through it, whereas ordinary glass absorbs most UV radiation. This enables experiments on metals such as zinc and magnesium whose threshold frequencies lie in the ultraviolet region. → Statement 1 is correct. → Statement 2 is correct. → Statement 4 is correct because quartz prevents UV absorption that would occur in ordinary glass. → Statement 3 is incorrect because alkali metals can exhibit photoelectric effect with visible light due to their low work functions.

❓ Why Other Options Are Incorrect

�� Statement 3 → Alkali metals such as cesium and potassium can emit photoelectrons under visible light.

🎯 Strategy

Used

�� Elimination

Application:

�� Reject the statement that contradicts the known behavior of alkali metals.

Final Logic:

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

🧠 Memory Trick

Quartz = UV Pass

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

2 Incorrect statement about the commutator's role in analyzing stopping potential:

A It switches the collector plate to a negative potential relative to the emitter.

B Reversing polarity prevents all electrons from reaching plate A immediately.

C It allows the measurement of the maximum kinetic energy by applying a retarding field.

D Reversing the polarity directly changes the work function of the emitter plate C.

💡 Short Explanation

�� Commutator reverses polarity. �� Retarding field helps determine stopping potential. �� Work function is a material property.

✅ Correct Answer Explanation

The commutator changes the polarity between the emitter and collector plates. This enables the application of a retarding potential required to determine the stopping potential and maximum kinetic energy of photoelectrons. → The work function depends only on the nature of the material and is not altered by polarity reversal. → Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct function of the commutator.
�� Option B → Correct because retarding potential prevents many electrons from reaching the collector.
�� Option C → Correct because stopping potential measurement gives maximum kinetic energy.

🎯 Strategy

Used

�� Odd One Out

Application:

�� Identify the statement unrelated to voltage reversal effects.

Final Logic:

�� Work function remains unchanged by circuit polarity.

🧠 Memory Trick

Work Function = Material Property

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

3 Match List I (Potential setup) with List II (Voltmeter indication and observation):

List I	List II
1. V = 0	a. Saturation current is reached
2. V > 0 (High)	b. Only sufficiently energetic electrons reach collector
3. V = -V₀	c. Photocurrent is exactly zero
4. V < 0 but > -V₀	d. Small photocurrent exists due to initial kinetic energy

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

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

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

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

💡 Short Explanation

�� Some electrons reach collector even at zero voltage. �� Large positive voltage gives saturation current. �� Stopping potential gives zero current.

✅ Correct Answer Explanation

Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching for saturation and stopping potential conditions.
�� Option C → Incorrect assignment of zero current and saturation current.
�� Option D → High positive voltage corresponds to saturation current, not selective collection.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each voltage condition with the corresponding photocurrent behavior.

Final Logic:

�� Only Option A correctly matches all four conditions.

🧠 Memory Trick

+V → Saturation, −V₀ → Stop

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

4 When a microammeter tracks the photocurrent while the accelerating potential is increased from zero, the readings

A decrease exponentially until reaching a null value

B increase continuously without any upper limit

C increase initially and then become constant despite further voltage increases

D fluctuate randomly due to the quantum nature of light

💡 Short Explanation

�� Accelerating voltage improves collection. �� Current rises initially. �� Saturation current is eventually reached.

✅ Correct Answer Explanation

As accelerating potential increases, more emitted photoelectrons are collected by the collector plate. The photocurrent therefore increases. Once all emitted photoelectrons are collected, saturation current is reached and the current remains constant even if voltage is increased further. → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite trend is observed.
�� Option B → Current cannot exceed saturation current.
�� Option D → Average photocurrent follows a predictable pattern.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate accelerating voltage with saturation current behavior.

Final Logic:

�� Current rises and then saturates.

🧠 Memory Trick

Rise → Saturate

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

5 If the distance between a point light source and the emitter is doubled (r' = 2r), the new saturation current I_sat' relates to the original I_sat as:

A I_sat' = I_sat

B I_sat' = 4 I_sat

C I_sat' = 0.25 I_sat

D I_sat' = 0.5 I_sat

💡 Short Explanation

�� Intensity follows inverse square law. �� Saturation current is proportional to intensity. �� Doubling distance reduces intensity fourfold.

✅ Correct Answer Explanation

For a point source: Intensity ∝ 1/r² When distance doubles: I' = I/(2²) = I/4 Since saturation current is directly proportional to intensity: Isat' = Isat/4 = 0.25 Isat → Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Intensity changes with distance.
�� Option B → Current decreases rather than increases.
�� Option D → Reduction is by a factor of four, not two.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply inverse square law directly.

Final Logic:

�� Doubling distance reduces saturation current to one-fourth.

🧠 Memory Trick

Double Distance → Quarter Intensity

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

6 The linear relationship between photocurrent and intensity implies that

A Each incident photon creates a cascading effect of multiple emitted electrons

B The elementary process is the absorption of a single quantum by a single electron

C The wave nature of light causes continuous energy accumulation

D The threshold frequency decreases linearly with increasing intensity

💡 Short Explanation

�� Photocurrent is proportional to photon number. �� One photon interacts with one electron. �� Supports Einstein's photon theory.

✅ Correct Answer Explanation

The linear dependence of photocurrent on intensity shows that increasing intensity increases the number of incident photons. Since each photon transfers its energy to a single electron, the number of emitted electrons increases proportionally. → This supports the concept that photoelectric emission is a one-photon-one-electron process. → Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → One photon does not normally eject multiple electrons in the photoelectric effect.
�� Option C → Continuous energy accumulation is a prediction of classical wave theory and cannot explain photoelectric observations.
�� Option D → Threshold frequency is a property of the material and is independent of intensity.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate linear photocurrent behavior to Einstein's photon model.

Final Logic:

�� Linear photocurrent confirms one photon transfers energy to one electron.

🧠 Memory Trick

One Photon → One Electron

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

7 A laser emits a power of 2.0 × 10⁻³ W with photon energy 3.98 × 10⁻¹⁹ J. What is the total number of photons emitted per second?

A 5.0 × 10¹⁵ photons/s

B 3.9 × 10¹⁶ photons/s

C 2.0 × 10⁻³ photons/s

D 8.0 × 10¹⁴ photons/s

💡 Short Explanation

�� Power = Energy per second. �� Number of photons = Total energy per second ÷ Energy per photon. �� Apply direct substitution.

✅ Correct Answer Explanation

Number of photons emitted per second: N = P/E = (2.0 × 10⁻³)/(3.98 × 10⁻¹⁹) ≈ 5.03 × 10¹⁵ photons/s ≈ 5.0 × 10¹⁵ photons/s → Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Overestimates the photon count.
�� Option C → Incorrect dimensional interpretation.
�� Option D → Much smaller than the calculated value.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute values into N = P/E.

Final Logic:

�� N = (2.0 × 10⁻³)/(3.98 × 10⁻¹⁹) ≈ 5.0 × 10¹⁵ photons/s.

🧠 Memory Trick

Photons = Power ÷ Photon Energy

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

8 If intensity of incident radiation is increased while keeping frequency constant, the number of energy quanta per unit area per unit time is ______ and the maximum kinetic energy of emitted electrons is ______.

A Constant, Increased

B Increased, Constant

C Increased, Increased

D Constant, Constant

💡 Short Explanation

�� Intensity determines photon flux. �� Frequency determines photon energy. �� Maximum kinetic energy depends on frequency.

✅ Correct Answer Explanation

Increasing intensity while keeping frequency constant increases the number of photons incident per unit area per unit time. → However, each photon still possesses the same energy because frequency remains unchanged. → According to Einstein's equation: Kmax = hν − ϕ Since ν is constant, Kmax remains constant. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Photon number does not remain constant when intensity increases.
�� Option C → Maximum kinetic energy does not increase without increasing frequency.
�� Option D → Photon flux definitely increases with intensity.

🎯 Strategy

Used

�� Elimination

Application:

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

Final Logic:

�� Intensity changes photon number, not photon energy.

🧠 Memory Trick

Intensity → Number, Frequency → Energy

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

9 Correct statements regarding the effect of accelerating potential:
1. Increasing it helps sweep all emitted electrons to the collector.
2. It increases the kinetic energy of electrons upon emission from the surface.
3. Beyond a certain voltage, it cannot extract any more electrons than what the intensity dictates.
4. It alters the work function of the material.

A (1), (3)

B (2), (4)

C (1), (2), (3)

D (3), (4)

💡 Short Explanation

�� Accelerating potential improves collection efficiency. �� Saturation current eventually occurs. �� Work function remains unchanged.

✅ Correct Answer Explanation

Increasing accelerating potential helps collect a larger fraction of emitted electrons. → Statement 1 is correct because more photoelectrons are swept toward the collector. → Statement 3 is correct because once saturation current is reached, increasing voltage further cannot collect more electrons than are being emitted. → Statement 2 is incorrect because electron kinetic energy at emission is determined by photon energy, not by the accelerating voltage. → Statement 4 is incorrect because work function is an intrinsic property of the material.

❓ Why Other Options Are Incorrect

�� Statement 2 → Initial kinetic energy depends on frequency of incident radiation.
�� Statement 4 → Work function is unaffected by accelerating potential.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove statements that contradict Einstein's photoelectric equation.

Final Logic:

�� Only statements 1 and 3 correctly describe accelerating potential effects.

🧠 Memory Trick

Accelerate Collects, Not Creates

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

10 Applying a retarding potential targets the kinetic energy of electrons; the current drops to zero precisely when

A The work function is perfectly balanced by the retarding field

B the retarding potential energy equals the maximum kinetic energy of the emitted photoelectrons

C The frequency of light is reduced below the threshold value

D The collector plate becomes completely saturated with negative charge

💡 Short Explanation

�� Retarding field opposes electron motion. �� Stopping potential stops even the fastest electrons. �� Kmax = eV₀.

✅ Correct Answer Explanation

Photocurrent becomes zero when the retarding potential is large enough to stop even the most energetic photoelectrons. → At stopping potential: eV₀ = Kmax → Thus the retarding potential energy equals the maximum kinetic energy of emitted photoelectrons. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Work function determines emission, not stopping condition.
�� Option C → Current can become zero even with frequency above threshold if sufficient retarding potential is applied.
�� Option D → Collector charge saturation is not the reason for stopping current.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate stopping potential directly to maximum kinetic energy.

Final Logic:

�� Current becomes zero when eV₀ equals Kmax.

🧠 Memory Trick

Stop Current → eV = Kmax

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

11 The maximum current value (saturation current) is fundamentally constrained by

A the number of photons incident per second (intensity)

B the maximum retarding potential the battery can apply

C the frequency of the incident radiation as long as ν > ν₀

D the speed of light in the evacuated tube

💡 Short Explanation

�� Saturation current depends on emitted electrons. �� Emitted electrons depend on photon flux. �� Photon flux is determined by intensity.

✅ Correct Answer Explanation

Saturation current is reached when all emitted photoelectrons are collected. Therefore, its value depends on the total number of photoelectrons emitted per second. → The number of emitted photoelectrons is directly proportional to the number of incident photons per second, which is determined by the intensity of the radiation. → Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Retarding potential affects stopping potential measurements, not the saturation current value.
�� Option C → Frequency affects kinetic energy, not the number of emitted electrons once ν > ν₀.
�� Option D → Speed of light has no direct role in determining saturation current.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the factor controlling the number of emitted photoelectrons.

Final Logic:

�� More incident photons produce more emitted electrons and larger saturation current.

🧠 Memory Trick

Intensity ↑ → Saturation Current ↑

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

12 Incorrect statement about the total collection of electrons:

A It defines the plateau region of the photocurrent-voltage graph.

B At this point, increasing the positive potential brings no new electrons to the collector.

C It implies that the rate of electron collection equals the rate of photoelectron emission.

D This state is only achievable if the incident light frequency is below the threshold frequency.

💡 Short Explanation

�� Saturation requires photoemission. �� Frequency must exceed threshold frequency. �� Plateau region corresponds to a complete collection.

✅ Correct Answer Explanation

Total collection of electrons corresponds to saturation current. In this region all emitted photoelectrons are collected by the collector plate. → Statements A, B and C correctly describe the saturation region. → Statement D is incorrect because photoelectric emission itself does not occur when the frequency is below the threshold frequency. Therefore saturation current cannot exist below threshold frequency.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of the saturation region.
�� Option B → Correct because all available electrons are already collected.
�� Option C → Correct since collection rate equals emission rate at saturation.

🎯 Strategy

Used

�� Elimination

Application:

�� Reject the statement that contradicts the threshold frequency condition.

Final Logic:

�� Saturation current requires ν > ν₀, not ν < ν₀.

🧠 Memory Trick

Below ν₀ → No Emission

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

13 Observations at the cut-off point:
1. The microammeter reads zero current.
2. Even the most energetic photoelectrons are repelled before reaching plate A.
3. The cut-off point shifts if the intensity of the light is doubled.
4. The cut-off point is heavily dependent on the frequency of incident radiation.

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

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

C (1), (3)

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

💡 Short Explanation

�� Stopping potential gives zero current. �� Fastest electrons are stopped. �� Frequency affects stopping potential.

✅ Correct Answer Explanation

At the stopping potential, photocurrent becomes exactly zero because even the most energetic photoelectrons fail to reach the collector. → Statement 1 is correct. → Statement 2 is correct. → Statement 4 is correct because stopping potential depends on frequency through Einstein's equation. → Statement 3 is incorrect because stopping potential is independent of intensity.

❓ Why Other Options Are Incorrect

�� Statement 3 → Doubling intensity changes photocurrent, not stopping potential.

🎯 Strategy

Used

�� Elimination

Application:

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

Final Logic:

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

🧠 Memory Trick

Stop Voltage → Zero Current

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

14 The stopping potential V₀ is expressed in terms of frequency ν and work function φ₀ as:

A V₀ = (hν / e) + (φ₀ / e)

B V₀ = (hν / e) - (φ₀ / e)

C V₀ = hν - eφ₀

D V₀ = (e / hν) - φ₀

💡 Short Explanation

�� Einstein's equation: Kmax = hν − φ₀. �� Kmax = eV₀. �� Combine both relations.

✅ Correct Answer Explanation

Einstein's photoelectric equation is: Kmax = hν − φ₀ Also, Kmax = eV₀ Therefore, eV₀ = hν − φ₀ or V₀ = (hν/e) − (φ₀/e) → Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect sign.
�� Option C → Dimensionally incorrect.
�� Option D → Not derived from Einstein's equation.

🎯 Strategy

Used

�� Substitution

Application:

�� Combine the two standard photoelectric equations.

Final Logic:

�� V₀ = (hν/e) − (φ₀/e).

🧠 Memory Trick

eV = hν − φ

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

15 Match List I (Parameters in Einstein's equation) with List II (Physical interpretation):

List I	List II
1. hν	a. Minimum energy required to escape metal surface
2. φ₀	b. Energy of a single quantum of incident radiation
3. Kmax	c. Maximum kinetic energy of emitted electron
4. ν₀	d. Threshold frequency for the material

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

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

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

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

💡 Short Explanation

�� hν represents photon energy. �� φ₀ is the work function. �� ν₀ is threshold frequency.

✅ Correct Answer Explanation

Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Work function and photon energy are interchanged.
�� Option C → Multiple physical meanings are mismatched.
�� Option D → Incorrect interpretation of all major parameters.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each symbol with its standard NCERT definition.

Final Logic:

�� Only Option A correctly pairs all quantities.

🧠 Memory Trick

hν = Photon, φ = Work Function

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

16 According to Einstein's picture, intensity alters the ______ of emitted electrons, while the energy of individual photons solely dictates the ______ of the electrons.

A Work function, Speed

B Number, Maximum kinetic energy

C Mass, Threshold frequency

D Number, Retarding potential polarity

💡 Short Explanation

�� Intensity changes electron count. �� Frequency changes electron energy. �� Supports photon theory.

✅ Correct Answer Explanation

Increasing intensity increases the number of incident photons and therefore the number of emitted electrons. → The maximum kinetic energy is determined by the energy of individual photons according to: Kmax = hν − φ₀ → Hence intensity affects number, while photon energy determines maximum kinetic energy. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Work function is a material property.
�� Option C → Mass and threshold frequency are not altered this way.
�� Option D → Retarding potential polarity is externally applied.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate intensity effects from frequency effects.

Final Logic:

�� Intensity controls quantity; frequency controls energy.

🧠 Memory Trick

Intensity → Number, Frequency → Energy

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

17 If the work function of caesium is 2.14 eV, what is its threshold frequency?
(h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C)

A 5.16 × 10¹⁴ Hz

B 3.22 × 10¹⁴ Hz

C 5.16 × 10⁻¹⁹ Hz

D 2.14 × 10¹⁴ Hz

💡 Short Explanation

�� Threshold frequency: ν₀ = φ₀/h. �� Convert eV into joules. �� Substitute values.

✅ Correct Answer Explanation

Work function: φ₀ = 2.14 × 1.6 × 10⁻¹⁹ = 3.424 × 10⁻¹⁹ J Threshold frequency: ν₀ = φ₀/h = (3.424 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) ≈ 5.16 × 10¹⁴ Hz → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Calculation error.
�� Option C → Wrong power of ten.
�� Option D → Incorrect numerical value.

🎯 Strategy

Used

�� Substitution

Application:

�� Use ν₀ = φ₀/h directly.

Final Logic:

�� ν₀ ≈ 5.16 × 10¹⁴ Hz.

🧠 Memory Trick

Threshold: ν₀ = φ/h

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

18 Correct statements regarding the V₀ versus ν graph:
1. It is a straight line for a given photosensitive material.
2. The slope of the line is given by h/e.
3. The slope depends on the nature of the material.
4. The intercept on the frequency axis is the threshold frequency.

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

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

C (1), (2), (4)

D (1), (3), (4)

💡 Short Explanation

�� V₀–ν graph is linear. �� Slope equals h/e. �� Frequency-axis intercept gives ν₀.

✅ Correct Answer Explanation

From: V₀ = (h/e)ν − (φ₀/e) → Statement 1 is correct because the graph is a straight line. → Statement 2 is correct because slope = h/e. → Statement 4 is correct because V₀ = 0 gives ν = ν₀. → Statement 3 is incorrect because slope depends only on h and e, not on the material.

❓ Why Other Options Are Incorrect

�� Statement 3 → Slope is universal and independent of the material.

🎯 Strategy

Used

�� Elimination

Application:

�� Use the equation of a straight line and identify slope and intercept.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Slope = h/e

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

19 The classical wave theory failed to explain the absence of time lag because it predicted that

A Light energy is concentrated in discrete quanta that are instantly absorbed

B Energy absorption takes place continuously over the entire wavefront, requiring hours for a single electron to gain enough energy

C electrons require a specific threshold frequency to vibrate

D The electric field is completely independent of the magnetic field in a wave

💡 Short Explanation

�� Wave theory assumed continuous energy transfer. �� Energy accumulation should take time. �� Experiments showed instantaneous emission.

✅ Correct Answer Explanation

Classical wave theory proposed that light energy is spread continuously across the wavefront. According to this idea, an electron would need a long time to accumulate enough energy for escape, especially at low intensities. → Experiments showed photoemission occurs almost instantaneously, contradicting this prediction. → Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → This is Einstein's photon explanation.
�� Option C → Threshold frequency is not a wave-theory prediction.
�� Option D → Unrelated to the time lag problem.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Compare wave-theory prediction with experimental observations.

Final Logic:

�� Continuous energy accumulation predicts delay, but none is observed.

🧠 Memory Trick

Wave Theory → Wait Time

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

20 The fact that photoelectric emission occurs in about 10⁻⁹ seconds or less

A proves that the work function is practically zero for all metals

B implies that the basic elementary process is the instantaneous absorption of a single light quantum by a single electron

C supports the continuous distribution of energy in an electromagnetic wave

D indicates that saturation current is independent of intensity

💡 Short Explanation

�� Emission occurs almost instantly. �� Photon transfers energy in one interaction. �� Supports Einstein's photon theory.

✅ Correct Answer Explanation

The extremely small time lag observed in photoelectric emission indicates that an electron receives sufficient energy in a single interaction with a photon and escapes immediately. → This observation strongly supports Einstein's concept of light quanta (photons). → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Work functions are finite and material-dependent.
�� Option C → Continuous energy distribution cannot explain instantaneous emission.
�� Option D → Saturation current remains proportional to intensity.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement consistent with the observed instantaneous emission.

Final Logic:

�� Instantaneous emission supports one-photon-one-electron interaction.

🧠 Memory Trick

Instant Photon → Instant Electron

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

CUET UG Physics Booster Test 2- Experimental Study of Photoelectric Effect

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