CUET UG Physics Booster Test 2- Einstein’s Quantum Theory and Photons -

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CUET UG Physics Booster Test 2- Einstein’s Quantum Theory and Photons

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2- Einstein’s Quantum Theory and Photons

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QUESTION 1 OF 20

In the classical wave theory of light, the energy distribution

Answer Review

1 In the classical wave theory of light, the energy distribution

A is concentrated in discrete isolated packets.

B depends solely on the frequency of the incident radiation.

C is continuous over the region of space where the wave is extended.

D successfully proves the existence of a threshold frequency.

💡 Short Explanation

�� Classical theory treats light as a wave. �� Energy is spread continuously. �� No discrete packets are assumed.

✅ Correct Answer Explanation

According to classical wave theory, energy is distributed continuously throughout the wavefront and over the region occupied by the wave. → This continuous distribution was one of the reasons the theory failed to explain the photoelectric effect. → Therefore option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes Einstein's photon theory.
�� Option B → Energy distribution is not solely determined by frequency.
�� Option D → Classical theory failed to explain threshold frequency.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement consistent with classical wave assumptions.

Final Logic:

�� Classical theory assumes continuous energy distribution.

🧠 Memory Trick

Wave = Continuous Energy

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

2 In a classical theoretical model, if the energy absorbed by an electron is continuously 1.0 × 10⁻²⁰ W, what will be the time required for a single electron to absorb 2.0 × 10⁻¹⁹ J?

A 20.0 s

B 2.0 × 10⁻³⁹ s

C 0.05 s

D 200.0 s

💡 Short Explanation

�� Power = Energy/Time. �� Time = Energy/Power. �� Direct substitution.

✅ Correct Answer Explanation

Given: Power = 1.0 × 10⁻²⁰ W Energy = 2.0 × 10⁻¹⁹ J Using: t = E/P = (2.0 × 10⁻¹⁹)/(1.0 × 10⁻²⁰) = 20 s → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect power-of-ten calculation.
�� Option C → Incorrect division.
�� Option D → Ten times larger than the correct value.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply t = E/P directly.

Final Logic:

�� t = 20 s.

🧠 Memory Trick

Time = Energy ÷ Power

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

3 Correct statements about the quantum hypothesis
1. Radiation energy is built up of discrete units called quanta.
2. Each quantum has an energy proportional to its wavelength.
3. Each quantum has energy hν.
4. The intensity determines the exact energy of a single quantum.

A (1), (3)

B (2), (3)

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

D (1), (4)

💡 Short Explanation

�� Energy exists as quanta. �� Quantum energy equals hν. �� Energy depends on frequency.

✅ Correct Answer Explanation

Statement 1 is correct because Planck proposed that radiation energy consists of discrete quanta. → Statement 3 is correct because each quantum has energy: E = hν → Statement 2 is incorrect because energy is proportional to frequency, not wavelength. → Statement 4 is incorrect because intensity changes the number of photons, not the energy of each photon.

❓ Why Other Options Are Incorrect

�� Statement 2 → E ∝ ν, not λ.
�� Statement 4 → Photon energy depends only on frequency.

🎯 Strategy

Used

�� Elimination

Application:

�� Retain statements directly derived from Planck's hypothesis.

Final Logic:

�� Statements 1 and 3 are correct.

🧠 Memory Trick

Quantum Energy = hν

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

4 Match List I with List II for Planck's constant and energy relationships

List I	List II
1. Energy E	a. 6.626 × 10⁻³⁴ J s
2. Momentum p	b. hν/c
3. Planck's constant h	c. hν
4. Wavelength λ	d. h/p

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

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

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

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

💡 Short Explanation

�� E = hν. �� p = hν/c. �� λ = h/p.

✅ Correct Answer Explanation

Correct matching: List I — List II 1. Energy E — c. hν 2. Momentum p — b. hν/c 3. Planck's constant h — a. 6.626 × 10⁻³⁴ J s 4. Wavelength λ — d. h/p → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Energy and momentum relations are interchanged.
�� Option C → Planck's constant and energy relation mismatched.
�� Option D → Incorrect wavelength relation.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each physical quantity with its standard formula.

Final Logic:

�� Only Option A gives all correct pairings.

🧠 Memory Trick

E = hν, p = h/λ

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

5 Incorrect statement about Einstein's 1905 proposal

A He introduced the revolutionary notion of light quanta.

B It completely supported continuous energy absorption from a wave front.

C It elegantly explained all primary features of the photoelectric effect.

D It posited that the photoelectric emission process is instantaneous.

💡 Short Explanation

�� Einstein proposed photons. �� Energy transfer is discrete. �� Continuous absorption belongs to wave theory.

✅ Correct Answer Explanation

Einstein's 1905 proposal introduced light quanta (photons) and discrete energy transfer. → It rejected the idea that energy is absorbed continuously from an extended wavefront. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of Einstein's proposal.
�� Option C → Correctly explains major photoelectric observations.
�� Option D → Instantaneous emission follows naturally from photon absorption.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate photon theory from classical wave theory.

Final Logic:

�� Continuous absorption is not part of Einstein's proposal.

🧠 Memory Trick

Einstein = Photons, Not Continuous Waves

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

6 For his explanation of the photoelectric effect and contribution to theoretical physics

A Einstein was awarded the Nobel Prize in Physics in 1905.

B Einstein was awarded the Nobel Prize in Physics in 1921.

C Millikan was awarded the Nobel Prize in Physics in 1921.

D Hertz was awarded the Nobel Prize for discovering photons.

💡 Short Explanation

�� Einstein explained photoelectric effect. �� Nobel Prize awarded in 1921. �� Recognition was for theoretical physics contributions.

✅ Correct Answer Explanation

Albert Einstein received the 1921 Nobel Prize in Physics primarily for his explanation of the photoelectric effect. → This work provided strong support for the quantum nature of light. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Nobel Prize was not awarded in 1905.
�� Option C → Millikan did not receive the 1921 Nobel Prize for this work.
�� Option D → Hertz did not discover photons.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Associate Nobel recognition with Einstein's explanation.

Final Logic:

�� Einstein received the 1921 Nobel Prize.

🧠 Memory Trick

Einstein → Nobel 1921

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

7 Energy conservation application statements
1. The kinetic energy of the electron equals photon energy minus work function.
2. Energy is entirely dissipated into the medium without allowing the electron to escape.
3. Maximum kinetic energy depends linearly on the incident frequency.
4. Kmax = hν − φ₀ represents conservation of energy for the absorption process.

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

B (1), (2), (3)

C (2), (4)

D (1), (2), (3), (4)

💡 Short Explanation

�� Einstein's equation conserves energy. �� Kmax depends on frequency. �� Work function must first be overcome.

✅ Correct Answer Explanation

Statement 1 is correct from Einstein's equation: Kmax = hν − φ₀ → Statement 3 is correct because Kmax varies linearly with ν. → Statement 4 is correct because the equation directly represents conservation of energy. → Statement 2 is incorrect because photoelectric emission occurs when sufficient energy is available.

❓ Why Other Options Are Incorrect

�� Statement 2 → Contradicts photoelectric emission.

🎯 Strategy

Used

�� Elimination

Application:

�� Retain statements directly derived from Einstein's equation.

Final Logic:

�� Statements 1, 3 and 4 are correct.

🧠 Memory Trick

hν = φ + K

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

8 If the stopping potential is V₀, and the elementary charge is e, the expression for the maximum kinetic energy Kmax will be

A e/V₀

B eV₀

C V₀/e

D e + V₀

💡 Short Explanation

�� Stopping potential measures maximum kinetic energy. �� Charge × potential gives energy. �� Standard photoelectric relation.

✅ Correct Answer Explanation

The maximum kinetic energy of emitted photoelectrons is: Kmax = eV₀ where e is the electronic charge and V₀ is the stopping potential. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect relation.
�� Option C → Dimensionally incorrect.
�� Option D → Physically meaningless.

🎯 Strategy

Used

�� Dimensional/Unit Analysis

Application:

�� Use energy = charge × potential difference.

Final Logic:

�� Kmax = eV₀.

🧠 Memory Trick

Stop Voltage → eV Energy

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

9 The basic elementary process in Einstein's picture involves the absorption of a _______ light quantum by a _______ electron.

A Single, Multiple

B Multiple, Single

C Single, Single

D Continuous, Discrete

💡 Short Explanation

�� One photon interacts with one electron. �� Energy transfer is discrete. �� Basis of photoelectric theory.

✅ Correct Answer Explanation

Einstein proposed that a single photon transfers its entire energy to a single electron. → If the transferred energy exceeds the work function, photoemission occurs. → Therefore option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → One photon is not shared among many electrons.
�� Option B → Multiple photons are not required in ordinary photoelectric emission.
�� Option D → Continuous absorption belongs to classical theory.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Recall the one-photon–one-electron interaction model.

Final Logic:

�� Single photon → Single electron.

🧠 Memory Trick

1 Photon → 1 Electron

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photon-Electron Interaction

10 Incorrect statement regarding the kinetic energy of emitted photoelectrons

A The intensity of incident light dictates the maximum kinetic energy.

B More tightly bound electrons emerge with less than maximum kinetic energy.

C Kmax is independent of the intensity of incident radiation.

D The least tightly bound electrons have kinetic energy equal to hν − φ₀.

💡 Short Explanation

�� Frequency determines Kmax. �� Intensity affects number of electrons. �� Einstein's equation contains frequency.

✅ Correct Answer Explanation

Maximum kinetic energy is given by: Kmax = hν − φ₀ → Thus Kmax depends on frequency and work function, not intensity. → Therefore option A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct statement.
�� Option C → Correct experimental result.
�� Option D → Correct expression for least bound electrons.

🎯 Strategy

Used

�� Elimination

Application:

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

Final Logic:

�� Intensity changes electron count, not Kmax.

🧠 Memory Trick

Frequency → Energy, Intensity → Number

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

11 If a metal has a work function of 3.315 × 10⁻¹⁹ J and h = 6.63 × 10⁻³⁴ J s, what will be the threshold frequency?

A 5.0 × 10¹⁴ Hz

B 2.0 × 10¹⁴ Hz

C 5.0 × 10¹⁵ Hz

D 2.0 × 10¹⁵ Hz

💡 Short Explanation

�� Threshold frequency: ν₀ = φ₀/h. �� Use direct substitution. �� Work function equals threshold photon energy.

✅ Correct Answer Explanation

Using: ν₀ = φ₀/h = (3.315 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 0.5 × 10¹⁵ = 5.0 × 10¹⁴ Hz → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect numerical calculation.
�� Option C → Ten times larger than the correct value.
�� Option D → Four times larger than the correct value.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply ν₀ = φ₀/h directly.

Final Logic:

�� ν₀ = 5.0 × 10¹⁴ Hz.

🧠 Memory Trick

Threshold Frequency = Work Function ÷ h

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

12 The minimum emission condition states that

A photoemission occurs only if hν ≤ φ₀.

B photoemission occurs only if ν > ν₀.

C high intensity light always causes emission regardless of frequency.

D minimum emission strictly requires high potential difference.

💡 Short Explanation

�� Frequency must exceed threshold frequency. �� Intensity cannot replace frequency. �� Work function must be overcome.

✅ Correct Answer Explanation

Photoelectric emission occurs only when photon energy exceeds the work function: hν > φ₀ Since: φ₀ = hν₀ Therefore: ν > ν₀ → Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reversed inequality.
�� Option C → Intensity cannot cause emission when ν < ν₀.
�� Option D → Potential difference is not the minimum emission condition.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the fundamental threshold condition.

Final Logic:

�� Emission requires ν > ν₀.

🧠 Memory Trick

Above Threshold → Emission

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

13 From Einstein's photoelectric equation, the slope of the cut-off potential V₀ versus frequency ν graph will be

A h/e

B e/h

C he

D h − e

💡 Short Explanation

�� V₀ and ν are linearly related. �� Compare with y = mx + c. �� Slope equals h/e.

✅ Correct Answer Explanation

Einstein's equation: eV₀ = hν − φ₀ or V₀ = (h/e)ν − φ₀/e Comparing with: y = mx + c Slope = h/e → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal relation.
�� Option C → Incorrect dimensions.
�� Option D → Not derived from the equation.

🎯 Strategy

Used

�� Substitution

Application:

�� Rewrite Einstein's equation in straight-line form.

Final Logic:

�� Slope = h/e.

🧠 Memory Trick

V₀–ν Slope = h/e

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

14 Correct statements about Millikan's precision testing
1. He performed experiments during 1906–1916.
2. He originally aimed to disprove Einstein's equation.
3. He ended up proving the validity of Einstein's equation.
4. He found a value of h inconsistent with Planck's original value.

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

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

C (1), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� Millikan tested Einstein's theory rigorously. �� Initial skepticism existed. �� Experimental results supported Einstein.

✅ Correct Answer Explanation

Statement 1 is correct because Millikan's investigations extended over roughly a decade. → Statement 2 is correct because he initially approached the theory critically. → Statement 3 is correct because his experimental data strongly verified Einstein's photoelectric equation. → Statement 4 is incorrect because his measured value of h agreed closely with Planck's value.

❓ Why Other Options Are Incorrect

�� Statement 4 → Experimental value of h was consistent with Planck's constant.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement contradicting Millikan's findings.

Final Logic:

�� Statements 1, 2 and 3 are correct.

🧠 Memory Trick

Tried to Disprove → Ended Up Proving

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

15 Photons travel at the speed of _______ and are electrically _______.

A Sound, Positive

B Light, Neutral

C Light, Negative

D Sound, Neutral

💡 Short Explanation

�� Photons are massless particles. �� Travel at speed c in vacuum. �� Carry no electric charge.

✅ Correct Answer Explanation

Photons travel at the speed of light in vacuum. → Photons are electrically neutral particles and therefore are not deflected by electric or magnetic fields. → Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Photons do not travel at the speed of sound and are not positive.
�� Option C → Photons are neutral, not negative.
�� Option D → Speed of sound is incorrect.

🎯 Strategy

Used

�� Elimination

Application:

�� Use fundamental photon properties.

Final Logic:

�� Photon = Speed of Light + Neutral Charge.

🧠 Memory Trick

Photon = c + Neutral

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

16 Intensity as photon flux statements
1. Increasing intensity increases the number of photons per second.
2. Photon energy is independent of intensity.
3. Increased intensity raises the maximum kinetic energy.
4. Intensity directly dictates the threshold frequency.

A (1), (2)

B (1), (3)

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

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

💡 Short Explanation

�� Intensity controls photon flux. �� Frequency controls photon energy. �� Threshold frequency is a material property.

✅ Correct Answer Explanation

Statement 1 is correct because increasing intensity increases the number of photons incident per second. → Statement 2 is correct because photon energy depends on frequency: E = hν → Statement 3 is incorrect because Kmax depends on frequency, not intensity. → Statement 4 is incorrect because threshold frequency depends on work function.

❓ Why Other Options Are Incorrect

�� Statement 3 → Kmax remains unchanged when only intensity changes.
�� Statement 4 → ν₀ is determined by the material.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate intensity effects from frequency effects.

Final Logic:

�� Statements 1 and 2 are correct.

🧠 Memory Trick

Intensity → Number, Frequency → Energy

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

17 Match List I with List II for photon formulas

List I	List II
1. Energy E	a. νλ
2. Momentum p	b. hν
3. Speed c	c. h/λ
4. Wavelength λ	d. h/p

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

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

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

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

💡 Short Explanation

�� E = hν. �� p = h/λ. �� c = νλ.

✅ Correct Answer Explanation

Correct matching: List I — List II 1. Energy E — b. hν 2. Momentum p — c. h/λ 3. Speed c — a. νλ 4. Wavelength λ — d. h/p → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Momentum and wavelength relations interchanged.
�� Option C → Multiple mismatches.
�� Option D → Energy relation incorrectly assigned.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each physical quantity with its standard formula.

Final Logic:

�� Only Option A gives all correct matches.

🧠 Memory Trick

E = hν, p = h/λ

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

18 In phenomena such as the photoelectric effect and Compton effect

A radiation behaves entirely as a continuous wave without momentum transfer.

B radiation behaves as if it is made up of a bunch of particles transferring energy and momentum.

C electrons solely transfer momentum to the incoming electromagnetic wave.

D energy is transferred without any corresponding transfer of momentum.

💡 Short Explanation

�� Photons carry energy and momentum. �� Photoelectric effect demonstrates energy transfer. �� Compton effect demonstrates momentum transfer.

✅ Correct Answer Explanation

Both the photoelectric effect and Compton scattering support the particle nature of light. → Photons transfer both energy and momentum during interactions with electrons. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Contradicted by experimental observations.
�� Option C → Momentum transfer is not one-sided.
�� Option D → Momentum transfer accompanies energy transfer.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Connect experimental evidence with photon behavior.

Final Logic:

�� Light behaves like particles carrying energy and momentum.

🧠 Memory Trick

Photon = Energy + Momentum

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

19 Incorrect statement about conservation laws in photon-particle collisions

A Total energy before and after the collision is conserved.

B Total momentum of the system remains constant.

C The number of photons is perfectly conserved during every collision.

D A photon can transfer its entire energy to an electron and cease to exist.

💡 Short Explanation

�� Energy is conserved. �� Momentum is conserved. �� Photon number need not be conserved.

✅ Correct Answer Explanation

In photon-particle interactions, total energy and momentum are conserved. → However, photons may be absorbed or created. → Therefore photon number is not always conserved. → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct conservation law.
�� Option B → Correct conservation law.
�� Option D → Occurs in photoelectric absorption.

🎯 Strategy

Used

�� Elimination

Application:

�� Distinguish conserved quantities from non-conserved quantities.

Final Logic:

�� Photon number is not a conserved quantity.

🧠 Memory Trick

Energy Conserved, Photon Count Not Always

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Conservation Laws in Photon Interactions

20 In a photon-electron collision, the number of photons

A is always strictly conserved without exception.

B must continuously increase due to radiation pressure.

C may not be conserved because the photon may be absorbed.

D dictates the total mass change of the combined system.

💡 Short Explanation

�� Photon absorption is possible. �� Photon creation is also possible. �� Photon number is not a conserved quantity.

✅ Correct Answer Explanation

During interactions such as the photoelectric effect, a photon may transfer all its energy to an electron and disappear. → Therefore the total number of photons need not remain constant. → Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Photon number is not universally conserved.
�� Option B → No such conservation rule exists.
�� Option D → Photon count does not directly determine system mass change.

🎯 Strategy

Used

�� Elimination

Application:

�� Recall photon absorption processes.

Final Logic:

�� Photon number may change during interactions.

🧠 Memory Trick

Photon Absorbed → Photon Gone

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Photon-Electron Collisions

CUET UG Physics Booster Test 2- Einstein’s Quantum Theory and Photons

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