CUET UG Physics Booster Test 3- Einstein’s Quantum Theory and Photons -

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CUET UG Physics Booster Test 3- 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 3- Einstein’s Quantum Theory and Photons

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QUESTION 1 OF 20

Wave theory failure statements
1. Greater intensity implies greater energy continuously absorbed per electron.
2. Sufficiently intense radiation should overcome the work function eventually, regardless of frequency.
3. A threshold frequency arises naturally from continuous wave absorption.
4. Absorption occurs continuously over the entire wavefront.

Answer Review

1 Wave theory failure statements
1. Greater intensity implies greater energy continuously absorbed per electron.
2. Sufficiently intense radiation should overcome the work function eventually, regardless of frequency.
3. A threshold frequency arises naturally from continuous wave absorption.
4. Absorption occurs continuously over the entire wavefront.

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

B (1), (3), (4)

C (2), (3)

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

💡 Short Explanation

�� Classical theory assumes continuous energy absorption. �� Higher intensity means greater energy transfer. �� Threshold frequency cannot be explained classically.

✅ Correct Answer Explanation

Statement 1 is correct according to classical wave theory because energy absorption is continuous and increases with intensity. → Statement 2 is correct because classical theory predicts that sufficiently intense radiation should eventually provide enough energy for emission, irrespective of frequency. → Statement 4 is correct because wave theory assumes energy is distributed continuously over the wavefront. → Statement 3 is incorrect because classical theory cannot naturally explain the existence of a threshold frequency.

❓ Why Other Options Are Incorrect

�� Statement 3 → Threshold frequency is unexplained by classical wave theory.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement inconsistent with classical wave predictions.

Final Logic:

�� Statements 1, 2 and 4 correctly describe the classical picture.

🧠 Memory Trick

Wave Theory = Continuous + No Threshold

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

2 Incorrect statement about emission delay in classical theory versus experiment

A Classical theory predicted it could take hours for an electron to absorb enough energy.

B Wave theory accurately predicted the instantaneous emission of photoelectrons.

C Experimental emission starts in 10⁻⁹ s or less.

D Low intensity in wave theory means a significantly longer delay in emission.

💡 Short Explanation

�� Classical theory predicted delay. �� Experiments showed immediate emission. �� This was a major failure of wave theory.

✅ Correct Answer Explanation

Classical wave theory predicted that electrons would require time to accumulate sufficient energy before emission. → Experiments showed photoemission begins almost instantaneously (≈10⁻⁹ s or less). → Therefore wave theory did not accurately predict instantaneous emission. → Hence option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct classical prediction.
�� Option C → Correct experimental observation.
�� Option D → Correct consequence of classical theory.

🎯 Strategy

Used

�� Elimination

Application:

�� Compare theoretical prediction with experimental results.

Final Logic:

�� Instantaneous emission contradicted classical wave theory.

🧠 Memory Trick

Wave Theory → Delay, Experiment → Instant

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

3 Match List I with List II concerning quantum versus classical properties

List I	List II
1. Continuous energy distribution	a. Quantum hypothesis
2. Discrete energy units	b. Wave theory
3. Group velocity of matter wave	c. Single photon absorption
4. Instantaneous emission	d. Velocity of the particle

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

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

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

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

💡 Short Explanation

�� Continuous distribution belongs to wave theory. �� Quanta belong to quantum theory. �� Group velocity equals particle velocity.

✅ Correct Answer Explanation

Correct matching: List I — List II 1. Continuous energy distribution — b. Wave theory 2. Discrete energy units — a. Quantum hypothesis 3. Group velocity of matter wave — d. Velocity of the particle 4. Instantaneous emission — c. Single photon absorption → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect pairings between wave and quantum concepts.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each concept with its defining principle.

Final Logic:

�� Only Option A provides all correct associations.

🧠 Memory Trick

Wave → Continuous, Quantum → Discrete

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

4 The power emitted by a source is 2.0 × 10⁻³ W with photon energy 3.98 × 10⁻¹⁹ J. What will be the total number of photons emitted per second?

A 7.9 × 10⁻²² s⁻¹

B 2.0 × 10¹⁵ s⁻¹

C 5.0 × 10¹⁵ s⁻¹

D 3.98 × 10¹⁹ s⁻¹

💡 Short Explanation

�� Power = Energy per second. �� Number of photons = Power ÷ Photon energy. �� 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 s⁻¹ ≈ 5.0 × 10¹⁵ s⁻¹ → Therefore option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Wrong order of magnitude.
�� Option B → Incorrect division.
�� Option D → Excessively large value.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply N = P/E.

Final Logic:

�� N ≈ 5.0 × 10¹⁵ photons/s.

🧠 Memory Trick

Photon Count = Power ÷ Photon Energy

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

5 Einstein's radical assumption fundamentally differed from classical physics because

A it assumed radiation energy is completely continuous over space.

B it asserted that radiant energy is built up of discrete units.

C it denied the overarching conservation of total energy.

D it stated that intensity determines the energy of single individual quanta.

💡 Short Explanation

�� Einstein introduced photons. �� Energy exchange is discrete. �� Conservation laws remain valid.

✅ Correct Answer Explanation

Einstein proposed that radiation consists of discrete packets of energy called quanta or photons. → This was fundamentally different from the classical continuous-wave picture. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Classical, not Einsteinian, view.
�� Option C → Einstein's theory obeys conservation laws.
�� Option D → Photon energy depends on frequency.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the key distinction between quantum and classical theories.

Final Logic:

�� Einstein replaced continuity with discreteness.

🧠 Memory Trick

Einstein = Photons

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

6 In Einstein's photoelectric equation, if the maximum kinetic energy is set to zero, the expression for the threshold frequency ν₀ will be

A φ₀/h

B h/φ₀

C hφ₀

D φ₀ − h

💡 Short Explanation

�� At threshold, Kmax = 0. �� Photon energy equals work function. �� Use Einstein's equation.

✅ Correct Answer Explanation

Einstein's equation: Kmax = hν − φ₀ At threshold: Kmax = 0 Therefore: hν₀ = φ₀ ν₀ = φ₀/h → Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal form.
�� Option C → Incorrect dimensions.
�� Option D → Not physically meaningful.

🎯 Strategy

Used

�� Substitution

Application:

�� Set Kmax = 0 in Einstein's equation.

Final Logic:

�� ν₀ = φ₀/h.

🧠 Memory Trick

Threshold = φ/h

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

7 The photoelectric equation shows that an increase in incident frequency leads to a _______ in maximum kinetic energy, while an increase in intensity leads to _______ in maximum kinetic energy.

A Linear increase, No change

B Decrease, Linear increase

C No change, Decrease

D Linear increase, Linear increase

💡 Short Explanation

�� Kmax depends on frequency. �� Kmax is independent of intensity. �� Einstein's equation gives a linear relation.

✅ Correct Answer Explanation

From: Kmax = hν − φ₀ → Increasing frequency increases Kmax linearly. → Intensity changes the number of emitted electrons, not their maximum kinetic energy. → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite frequency dependence.
�� Option C → Incorrect for both quantities.
�� Option D → Intensity does not affect Kmax.

🎯 Strategy

Used

�� Elimination

Application:

�� Separate frequency effects from intensity effects.

Final Logic:

�� Frequency affects energy; intensity affects count.

🧠 Memory Trick

Frequency → Energy, Intensity → Number

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

8 In evaluating the stopping potential V₀ corresponding to maximum kinetic energy,

A it is found that V₀ decreases as the incident frequency increases.

B V₀ multiplied by the elementary charge e equals Kmax.

C V₀ is directly proportional to the incident light's intensity.

D it represents the accelerating positive potential needed to emit electrons.

💡 Short Explanation

�� Stopping potential measures Kmax. �� eV₀ equals maximum kinetic energy. �� Frequency increases V₀.

✅ Correct Answer Explanation

At stopping potential: eV₀ = Kmax → Thus the product of electronic charge and stopping potential gives the maximum kinetic energy. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → V₀ increases with frequency.
�� Option C → V₀ is independent of intensity.
�� Option D → V₀ is a retarding, not accelerating, potential.

🎯 Strategy

Used

�� Substitution

Application:

�� Use the stopping potential relation directly.

Final Logic:

�� Kmax = eV₀.

🧠 Memory Trick

Stop Voltage = Electron Energy

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

9 Correct statements about single quantum absorption
1. Low intensity does not imply a delay in emission.
2. A single electron absorbs only a single quantum of radiation.
3. The basic elementary process depends directly on the total intensity.
4. High intensity ensures multiple quanta are absorbed by one electron.

A (1), (2)

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

C (1), (4)

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

💡 Short Explanation

�� Emission is instantaneous. �� One photon interacts with one electron. �� Intensity affects photon number only.

✅ Correct Answer Explanation

Statement 1 is correct because photoelectric emission occurs almost instantaneously even at low intensity. → Statement 2 is correct because the elementary process involves absorption of a single photon by a single electron. → Statements 3 and 4 are incorrect.

❓ Why Other Options Are Incorrect

�� Statement 3 → Intensity affects emission rate, not the elementary process.
�� Statement 4 → High intensity does not guarantee multi-photon absorption.

🎯 Strategy

Used

�� Elimination

Application:

�� Retain only statements consistent with Einstein's photon model.

Final Logic:

�� Statements 1 and 2 are correct.

🧠 Memory Trick

One Photon → One Electron

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

10 Match List I with List II for electron emission energies

List I	List II
1. Maximum kinetic energy	a. Energy absorbed < φ₀
2. Lower kinetic energy	b. Least tightly bound electrons
3. Zero kinetic energy	c. Tightly bound electrons
4. Cannot escape	d. Incident frequency equals threshold

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

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

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

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

💡 Short Explanation

�� Least bound electrons have Kmax. �� Tightly bound electrons emerge with less energy. �� Threshold frequency gives zero kinetic energy.

✅ Correct Answer Explanation

Correct matching: List I — List II 1. Maximum kinetic energy — b. Least tightly bound electrons 2. Lower kinetic energy — c. Tightly bound electrons 3. Zero kinetic energy — d. Incident frequency equals threshold 4. Cannot escape — a. Energy absorbed < φ₀ → Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Options A, B and C contain incorrect energy interpretations.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each emission condition with the corresponding physical situation.

Final Logic:

�� Only Option D provides all correct pairings.

🧠 Memory Trick

Threshold → Zero K.E.

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

11 If the threshold frequency for caesium is 5.16 × 10¹⁴ Hz, what will be its work function?
(h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)

A 2.14 eV

B 3.42 eV

C 4.20 eV

D 1.76 eV

💡 Short Explanation

�� Work function: φ₀ = hν₀. �� Convert joules into eV. �� Direct substitution.

✅ Correct Answer Explanation

Using: φ₀ = hν₀ = (6.63 × 10⁻³⁴)(5.16 × 10¹⁴) = 3.42 × 10⁻¹⁹ J Converting to eV: φ₀ = (3.42 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2.14 eV → Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Value in joules incorrectly treated as eV.
�� Option C → Calculation error.
�� Option D → Incorrect conversion.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply φ₀ = hν₀ and convert units.

Final Logic:

�� φ₀ = 2.14 eV.

🧠 Memory Trick

Work Function = hν₀

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

12 Minimum emission condition statements
1. Requires ν > ν₀.
2. Dictates that photoelectric emission cannot occur if hν < φ₀.
3. Shows that threshold frequency is exactly the same for all metals.
4. Ensures Kmax is positive for emission.

A (3), (4)

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

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

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

💡 Short Explanation

�� Frequency must exceed threshold. �� Photon energy must exceed work function. �� Different metals have different threshold frequencies.

✅ Correct Answer Explanation

Statement 1 is correct because emission requires ν > ν₀. → Statement 2 is correct because if hν < φ₀, electrons cannot escape. → Statement 4 is correct because for ν > ν₀: Kmax = hν − φ₀ > 0 → Statement 3 is incorrect since threshold frequency depends on the metal.

❓ Why Other Options Are Incorrect

�� Statement 3 → Different metals possess different work functions and threshold frequencies.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement contradicting material dependence.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Above ν₀ → Positive K.E.

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

13 Incorrect statement about Millikan's slope measurements

A The slope of the V₀ versus ν curve is experimentally equal to h/e.

B The magnitude of the slope is fundamentally dependent on the nature of the specific metal used.

C Using the independently known value of e, Millikan determined h.

D Millikan's precise measurements verified the validity of Einstein's equation.

💡 Short Explanation

�� Slope equals h/e. �� h and e are universal constants. �� Slope is independent of metal.

✅ Correct Answer Explanation

Einstein's equation: V₀ = (h/e)ν − φ₀/e → The slope is h/e and therefore depends only on universal constants. → It is independent of the photosensitive material used. → Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct experimental result.
�� Option C → Correct method used by Millikan.
�� Option D → Correct conclusion from the experiment.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify which statement contradicts Einstein's equation.

Final Logic:

�� Slope depends on h/e, not on the metal.

🧠 Memory Trick

Slope = Universal Constant

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

14 The experimental verification of the photoelectric equation by Millikan

A was initially aimed at proving Einstein theoretically correct.

B utilized precise measurements over a wide range of radiation frequencies for alkali metals.

C resulted in a value of h drastically different from Planck's original value.

D proved that light has absolutely no particle nature.

💡 Short Explanation

�� Millikan performed precise experiments. �� Wide frequency range was used. �� Results supported Einstein's equation.

✅ Correct Answer Explanation

Millikan carried out highly accurate measurements using different frequencies and photosensitive metals. → The resulting V₀–ν graph verified Einstein's photoelectric equation and provided an accurate value of Planck's constant. → Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → He initially attempted to test and challenge Einstein's proposal.
�� Option C → Measured h agreed closely with Planck's value.
�� Option D → The experiment supported the particle nature of light.

🎯 Strategy

Used

�� Elimination

Application:

�� Select the experimentally verified statement.

Final Logic:

�� Millikan used precise measurements across frequencies.

🧠 Memory Trick

Millikan Measured, Einstein Verified

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

15 Let an electric field be E and the charge of a photon be q. Since photons are electrically neutral, their electrostatic force in the field will be

A qE

B cE

C hE

D 0E

💡 Short Explanation

�� Photon charge is zero. �� Electric force = qE. �� Neutral particles experience no electrostatic force.

✅ Correct Answer Explanation

Electrostatic force is: F = qE → For a photon: q = 0 Therefore: F = 0 × E = 0 → Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Applies only to charged particles.
�� Option B → Speed of light is unrelated to electrostatic force.
�� Option C → Planck's constant is unrelated to electric force.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute q = 0 into F = qE.

Final Logic:

�� Neutral photons experience zero electrostatic force.

🧠 Memory Trick

Photon Charge = 0

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

16 For a fixed frequency, an increase in intensity results in a _______ in the number of photons per second, but _______ in the energy of each photon.

A Decrease, Increase

B Increase, No change

C Increase, Increase

D No change, Decrease

💡 Short Explanation

�� Intensity controls photon flux. �� Frequency controls photon energy. �� Photon energy remains constant.

✅ Correct Answer Explanation

Photon energy is: E = hν → Since frequency is fixed, energy per photon remains unchanged. → Increasing intensity only increases the number of photons arriving per second. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite trend.
�� Option C → Photon energy does not increase.
�� Option D → Photon number increases with intensity.

🎯 Strategy

Used

�� Elimination

Application:

�� Distinguish intensity effects from frequency effects.

Final Logic:

�� More photons, same photon energy.

🧠 Memory Trick

Intensity → Count, Frequency → Energy

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

17 Correct statements about the momentum of light
1. The momentum of a photon is given by p = h/λ.
2. All photons of a particular wavelength have the exact same momentum.
3. Photon momentum is directly proportional to intensity.
4. Momentum confirms a particle-like nature of light.

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

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

C (1), (3)

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

💡 Short Explanation

�� Photon momentum depends on wavelength. �� Same wavelength implies same momentum. �� Momentum supports particle behavior.

✅ Correct Answer Explanation

Statement 1 is correct: p = h/λ → Statement 2 is correct because photons of identical wavelength possess identical momentum. → Statement 4 is correct because momentum transfer demonstrates particle nature. → Statement 3 is incorrect because photon momentum depends on wavelength, not intensity.

❓ Why Other Options Are Incorrect

�� Statement 3 → Intensity changes photon number, not momentum per photon.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement confusing intensity with momentum.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Momentum = h/λ

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

18 Match List I with List II regarding radiation interaction

List I	List II
1. Wavelength λ	a. Wave picture
2. Group velocity of matter wave	b. Velocity of the particle
3. Compton effect	c. Scattering of X-rays from electrons
4. Eye-lens gathering light	d. Momentum p = h/λ

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

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

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

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

💡 Short Explanation

�� λ relates to momentum. �� Group velocity equals particle velocity. �� Compton effect involves X-ray scattering.

✅ Correct Answer Explanation

Correct matching: List I — List II 1. Wavelength λ — d. Momentum p = h/λ 2. Group velocity of matter wave — b. Velocity of the particle 3. Compton effect — c. Scattering of X-rays from electrons 4. Eye-lens gathering light — a. Wave picture → Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Options A, B and D contain incorrect conceptual pairings.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each phenomenon with its defining concept.

Final Logic:

�� Only Option A gives all correct matches.

🧠 Memory Trick

Compton → X-ray Scattering

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

19 In analyzing a photon-electron collision process, it is essential to note that

A the initial and final numbers of photons must remain mathematically identical.

B total energy and total momentum of the isolated system are conserved.

C photons always gain potential energy at the direct expense of electrons.

D momentum conservation applies strictly only to matter, not to radiation.

💡 Short Explanation

�� Energy is conserved. �� Momentum is conserved. �� Radiation also carries momentum.

✅ Correct Answer Explanation

In photon-electron interactions such as Compton scattering, both total energy and total momentum remain conserved. → These conservation laws are fundamental to all isolated systems. → Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Photon number need not be conserved.
�� Option C → Not a general rule.
�� Option D → Radiation possesses momentum.

🎯 Strategy

Used

�� Elimination

Application:

�� Apply universal conservation laws.

Final Logic:

�� Both energy and momentum are conserved.

🧠 Memory Trick

Collision = Energy + Momentum Conserved

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

20 Non-conservation of photon number statements
1. A photon may be entirely absorbed by an electron.
2. The total number of photons before and after a collision may completely differ.
3. A new photon may be created during emission or collision.
4. Photons possess rest mass which must be rigorously conserved.

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

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

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

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

💡 Short Explanation

�� Photon number is not conserved. �� Photons may be absorbed or created. �� Photons have zero rest mass.

✅ Correct Answer Explanation

Statement 1 is correct because photons can be completely absorbed in processes such as the photoelectric effect. → Statement 2 is correct because photon number may change during interactions. → Statement 3 is correct because photons can be emitted or created in physical processes. → Statement 4 is incorrect because photons possess zero rest mass.

❓ Why Other Options Are Incorrect

�� Statement 4 → Photons do not have rest mass.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the statement contradicting photon properties.

Final Logic:

�� Statements 1, 2 and 3 are correct.

🧠 Memory Trick

Photon Count Changes, Rest Mass Zero

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

CUET UG Physics Booster Test 3- Einstein’s Quantum Theory and Photons

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