CUET UG Physics Booster Test 2- Historical Discoveries and Electron Properties -

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CUET UG Physics Booster Test 2- Historical Discoveries and Electron Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2- Historical Discoveries and Electron Properties

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QUESTION 1 OF 20

Maxwell's equations and Hertz's discoveries

Answer Review

1 Maxwell's equations and Hertz's discoveries

A proved that light requires a physical material medium

B contradicted the wave nature of light

C strongly established the wave nature of light

D directly measured the specific charge of an electron

💡 Short Explanation

�� Maxwell predicted electromagnetic waves. �� Hertz experimentally verified their existence. �� Together they established the wave nature of light.

✅ Correct Answer Explanation

Maxwell's equations predicted the existence of electromagnetic waves and showed that light itself is an electromagnetic wave. Hertz later generated and detected electromagnetic waves experimentally in 1887. These discoveries strongly established the wave nature of light. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Maxwell's theory did not prove the need for a material medium for light.
�� Option B → Maxwell's and Hertz's work supported, not contradicted, the wave nature of light.
�� Option D → Measurement of specific charge (e/m) was done by J.J. Thomson.

🎯 Strategy

Used

Elimination

Application:

Remove options unrelated to Maxwell's theory and Hertz's experiments.

Final Logic:

Only Option C correctly describes the significance of their discoveries.

🧠 Memory Trick

Maxwell Predicts → Hertz Proves

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Maxwell's Theory and Hertz's Experiments

2 Match List I with List II (Hertz's framework)

List I	List II (Correct Match)
(1) Electromagnetic waves	(a) Year Hertz performed defining experiments
(2) 1887	(b) Generated and detected by Hertz
(3) Wave nature	(c) Theory foundational to Hertz's work
(4) Maxwell's equations	(d) Strongly established by late 19th century

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

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

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

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

💡 Short Explanation

�� Hertz generated and detected electromagnetic waves. �� The experiments were performed in 1887. �� Maxwell's equations formed the theoretical foundation.

✅ Correct Answer Explanation

The correct matching is: → (1) Electromagnetic waves → (b) Generated and detected by Hertz → (2) 1887 → (a) Year Hertz performed defining experiments → (3) Wave nature → (d) Strongly established by late 19th century → (4) Maxwell's equations → (c) Theory foundational to Hertz's work Hertz's experiments verified Maxwell's predictions and strongly established the wave nature of light.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of (1), (2), and (3).
�� Option C → Incorrectly associates 1887 and electromagnetic waves.
�� Option D → Incorrectly matches Maxwell's equations and wave nature.

🎯 Strategy

Used

Option Grouping

Application:

Match each term with its correct scientific significance.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

Waves → Hertz → 1887 → Maxwell

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Hertz's Verification of Maxwell's Theory

3 Low pressure conduction statements
1. Discharge occurs at about 0.001 mm Hg.
2. An electric field must be applied to the gas.
3. It happens without electrodes.
4. A fluorescent glow appears on the glass.

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

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

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

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

💡 Short Explanation

�� Gas discharge requires very low pressure. �� An electric field is necessary. �� Fluorescence is observed on the glass wall.

✅ Correct Answer Explanation

Statement 1 is correct because discharge occurs at approximately 0.001 mm Hg. → Statement 2 is correct because an electric field must be applied across the electrodes. → Statement 3 is incorrect because electrodes are required in the discharge tube. → Statement 4 is correct because a fluorescent glow appears on the glass opposite the cathode. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Check each numbered statement against discharge tube observations.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Low Pressure + Electric Field + Glow

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Discharge Tube Experiment

4 Incorrect statement about cathode fluorescence

A It appears on the glass opposite to the cathode.

B The colour of the glow depends on the type of glass.

C It is yellowish-green for soda glass.

D It is caused by positive radiation originating from the anode.

💡 Short Explanation

�� Cathode rays strike the glass. �� Fluorescence depends on glass type. �� Soda glass produces yellowish-green fluorescence.

✅ Correct Answer Explanation

Cathode fluorescence is produced when cathode rays strike the glass wall opposite the cathode. The colour depends on the nature of the glass and appears yellowish-green in soda glass. It is not produced by positive radiation from the anode. Therefore, Option D is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct observation of cathode ray experiments.
�� Option B → Correct because fluorescence colour depends on glass composition.
�� Option C → Correct for soda glass.

🎯 Strategy

Used

Elimination

Application:

Identify the statement inconsistent with cathode ray observations.

Final Logic:

Fluorescence is produced by cathode rays, not positive radiation.

🧠 Memory Trick

Cathode Rays → Green Glow

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Cathode Ray Observations

5 The entity discovered in 1895, and its discoverer:

A Cathode rays, Crookes

B X-rays, Roentgen

C Electrons, Thomson

D Specific charge, Millikan

💡 Short Explanation

�� Roentgen discovered X-rays in 1895. �� This was a landmark discovery. �� It contributed significantly to atomic physics.

✅ Correct Answer Explanation

Wilhelm Roentgen discovered X-rays in 1895 while studying cathode ray tubes. This discovery became one of the most important milestones in the development of atomic structure. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Cathode rays were known before 1895.
�� Option C → Electron discovery is associated with Thomson in 1897.
�� Option D → Millikan measured electronic charge, not specific charge.

🎯 Strategy

Used

Direct Recall

Application:

Recall the scientist and year associated with X-rays.

Final Logic:

1895 corresponds to X-rays discovered by Roentgen.

🧠 Memory Trick

1895 = Roentgen = X-rays

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Discovery of X-rays

6 When assessing the history of atomic physics, the discoveries of X-rays and electrons

A proved that electric charge behaves as a continuous fluid

B occurred primarily due to high-pressure gas observations

C were critical milestones in the understanding of atomic structure

D were entirely reliant on mathematical theory without experiments

💡 Short Explanation

�� X-rays and electrons transformed atomic theory. �� They revealed subatomic structure. �� Both discoveries were experimentally based.

✅ Correct Answer Explanation

The discoveries of X-rays and electrons demonstrated that atoms possessed internal structure and were not indivisible particles. These discoveries became major milestones in the understanding of atomic structure and initiated modern atomic physics. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Charge is quantised, not a continuous fluid.
�� Option B → These discoveries did not arise from high-pressure gas observations.
�� Option D → Both discoveries relied heavily on experiments.

🎯 Strategy

Used

Elimination

Application:

Reject options that contradict historical facts.

Final Logic:

The discoveries were key milestones in understanding atomic structure.

🧠 Memory Trick

X-rays + Electrons = Modern Atom

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Historical Development of Atomic Structure

7 Correct statements about William Crookes' hypothesis
1. He discovered cathode rays in 1870.
2. He suggested they were streams of positively charged particles.
3. He proposed in 1879 that they were fast moving negative particles.
4. His hypothesis was later confirmed by J.J. Thomson.

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

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

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

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

💡 Short Explanation

�� Cathode rays were discovered before Thomson's work. �� Crookes proposed they were negatively charged particles. �� Thomson later confirmed this idea.

✅ Correct Answer Explanation

Statement 1 is correct because cathode rays were discovered around 1870. → Statement 2 is incorrect because Crookes proposed that cathode rays were negatively charged particles, not positively charged particles. → Statement 3 is correct because Crookes suggested in 1879 that cathode rays were streams of fast moving negative particles. → Statement 4 is correct because Thomson later confirmed the nature of cathode rays through his experiments. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 2.
�� Option C → Includes incorrect statement 2.
�� Option D → Includes incorrect statement 2.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each numbered statement separately.

Final Logic:

Only statements 1, 3 and 4 are correct.

🧠 Memory Trick

Crookes → Negative Particles → Thomson Confirms

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Cathode Rays and Crookes' Hypothesis

8 If an electron has a charge of -x(e) and another identical electron in the stream has a charge of -y(e), the total charge of these two specific particles will be

A (x - y) e

B (x + y) e

C -(x + y) e

D (y - x) e

💡 Short Explanation

�� Charges add algebraically. �� Both charges are negative. �� Total charge remains negative.

✅ Correct Answer Explanation

Total charge = -x(e) + -y(e) = -(x + y)e Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect sign.
�� Option A → Charges are added, not subtracted.
�� Option D → Incorrect algebraic expression.

🎯 Strategy

Used

Substitution

Application:

Add the charges directly.

Final Logic:

Sum of two negative charges is -(x + y)e.

🧠 Memory Trick

Negative + Negative = More Negative

📚 Class XI – Physics Part I – Chapter: Electric Charges and Fields – Topic: Addition of Charges

9 If a particle identical to an electron has a mass 2 times larger but the same charge, its specific charge relative to 1.76 × 10¹¹ C kg⁻¹ would be:

A 3.52 × 10¹¹ C kg⁻¹

B 0.88 × 10¹¹ C kg⁻¹

C 1.76 × 10¹¹ C kg⁻¹

D 1.60 × 10⁻¹⁹ C kg⁻¹

💡 Short Explanation

�� Specific charge = e/m. �� Mass doubles while charge remains constant. �� Therefore e/m becomes half.

✅ Correct Answer Explanation

Given: e/m = 1.76 × 10¹¹ C kg⁻¹ If mass becomes 2m and charge remains e, New specific charge = e/2m = (1/2)(1.76 × 10¹¹) = 0.88 × 10¹¹ C kg⁻¹ Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents double the specific charge.
�� Option C → Represents the original value.
�� Option D → Represents elementary charge, not specific charge.

🎯 Strategy

Used

Substitution

Application:

Apply e/m directly.

Final Logic:

Doubling mass halves the specific charge.

🧠 Memory Trick

Double Mass → Half e/m

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Specific Charge of Electron

10 J.J. Thomson's 1906 Nobel Prize recognition

A was for the discovery of electromagnetic waves

B rewarded his theoretical and experimental investigations on conduction of electricity by gases

C was given directly for the oil-drop experiment

D proved the existence of the work function

💡 Short Explanation

�� Thomson received the Nobel Prize in Physics in 1906. �� The award recognized work on gas conduction. �� It was unrelated to Millikan's experiment.

✅ Correct Answer Explanation

J.J. Thomson received the 1906 Nobel Prize in Physics for his theoretical and experimental investigations on the conduction of electricity by gases. These investigations led to important discoveries regarding electrons and atomic structure. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electromagnetic waves were experimentally demonstrated by Hertz.
�� Option C → Oil-drop experiment was performed by Millikan.
�� Option D → Work function was not the basis of the Nobel Prize.

🎯 Strategy

Used

Direct Recall

Application:

Recall the official reason for Thomson's Nobel Prize.

Final Logic:

The Nobel Prize recognized his investigations on electricity conduction by gases.

🧠 Memory Trick

Thomson Nobel = Gas Conduction

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: J.J. Thomson and Electron Discovery

11 During the determination of specific charge, the value obtained

A depends highly on the gas introduced in the discharge tube

B is independent of the nature of the material used as the cathode

C proves that cathode rays are uniquely tied to soda glass

D changes radically if different mutually perpendicular fields are used

💡 Short Explanation

�� Thomson found the same e/m value under different conditions. �� The cathode material did not affect the result. �� This supported the universality of electrons.

✅ Correct Answer Explanation

J.J. Thomson observed that the value of e/m remained the same irrespective of the cathode material used in the discharge tube. This indicated that cathode ray particles were identical and universal in nature. The result was one of the strongest pieces of evidence for the existence of electrons as fundamental constituents of matter. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The value of e/m was found to be independent of the gas used.
�� Option C → Cathode rays are not uniquely associated with soda glass.
�� Option D → Mutually perpendicular fields are used in measurement, but they do not cause radical changes in e/m.

🎯 Strategy

Used

Elimination

Application:

Remove statements that contradict Thomson's observations.

Final Logic:

The measured e/m value was independent of cathode material.

🧠 Memory Trick

Same Cathode or Different Cathode → Same e/m

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Universality of Electron

12 Statements about electron universality
1. Particles emitted by UV light irradiation have identical e/m.
2. Heated metals emit particles with the same e/m.
3. The specific charge alters dynamically based on the source.
4. Thomson suggested these identical particles are fundamental constituents of matter.

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

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

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

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

💡 Short Explanation

�� Electrons from different sources have identical e/m. �� Thomson proposed electrons are universal. �� e/m does not depend on the source.

✅ Correct Answer Explanation

Statement 1 is correct because particles emitted through UV light irradiation possess the same e/m as electrons. → Statement 2 is correct because electrons emitted from heated metals also have the same e/m value. → Statement 3 is incorrect because the specific charge does not change based on the source. → Statement 4 is correct because Thomson suggested that these identical particles are fundamental constituents of matter. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 3.
�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Check each numbered statement using the concept of electron universality.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Different Sources → Same Electron

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Universality of Electron

13 The speed range of cathode rays compared to light, and the fields used to measure it:

A 0.1c to 0.2c, mutually perpendicular electric and magnetic

B 0.5c to 0.9c, Gravitational only

C 0.1c to 0.2c, Electric only

D 1.0c, Magnetic only

💡 Short Explanation

�� Cathode rays move at high speeds. �� Their speeds are lower than the speed of light. �� Electric and magnetic fields were used in Thomson's experiment.

✅ Correct Answer Explanation

Cathode ray particles travel with speeds approximately between 0.1c and 0.2c. J.J. Thomson used mutually perpendicular electric and magnetic fields to determine the specific charge and velocity of these particles. These measurements helped establish the properties of electrons. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect speed range and incorrect field description.
�� Option C → Both electric and magnetic fields were used.
�� Option D → Electrons do not travel at the speed of light.

🎯 Strategy

Used

Direct Recall

Application:

Recall the speed range and experimental arrangement used by Thomson.

Final Logic:

Cathode rays travel at 0.1c to 0.2c and are studied using mutually perpendicular electric and magnetic fields.

🧠 Memory Trick

0.1c–0.2c + E ⟂ B

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Determination of e/m of Electron

14 Correct statements about the accepted e/m value
1. The presently accepted value is 1.76 × 10¹¹ C/kg.
2. It proved electrons carry positive charge.
3. It confirmed particles from varied emission sources are identical.
4. It depends on the target gas used.

A (1), (3)

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

C (3), (4)

D (1), (4)

💡 Short Explanation

�� Accepted e/m value is 1.76 × 10¹¹ C/kg. �� Electrons carry negative charge. �� e/m is independent of emission source.

✅ Correct Answer Explanation

Statement 1 is correct because the accepted value of e/m is 1.76 × 10¹¹ C/kg. → Statement 2 is incorrect because electrons carry negative charge, not positive charge. → Statement 3 is correct because identical e/m values from different emission sources established the universality of electrons. → Statement 4 is incorrect because e/m does not depend on the target gas. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 2.
�� Option C → Includes incorrect statement 4.
�� Option D → Includes incorrect statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Verify each numbered statement individually.

Final Logic:

Only statements 1 and 3 are correct.

🧠 Memory Trick

1.76 × 10¹¹ → Universal Electron

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Specific Charge of Electron

15 If an oil drop holds a total measured charge of 4.806 × 10⁻¹⁹ C, and the elementary charge is 1.602 × 10⁻¹⁹ C, how many elementary charges are present?

A 2

B 3

C 4

D 5

💡 Short Explanation

�� Charge on a droplet is an integral multiple of e. �� Divide total charge by elementary charge. �� Result gives the number of elementary charges.

✅ Correct Answer Explanation

Number of elementary charges = (4.806 × 10⁻¹⁹ C)/(1.602 × 10⁻¹⁹ C) = 3 Therefore, the oil drop contains 3 elementary charges. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Gives a smaller value than calculated.
�� Option C → Greater than the actual value.
�� Option D → Greater than the actual value.

🎯 Strategy

Used

Substitution

Application:

Substitute the given values into q/e.

Final Logic:

4.806 ÷ 1.602 = 3.

🧠 Memory Trick

Charge ÷ e = Number of Electrons

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Millikan's Oil-Drop Experiment

16 Incorrect statement about charge quantization

A The charge on an oil droplet is always an integral multiple of an elementary charge.

B The elementary charge value is approximately 1.602 × 10⁻¹⁹ C.

C Millikan's experiment established that electric charge is continuous.

D From the values of e and e/m, the electron's mass can be determined.

💡 Short Explanation

�� Charge is quantized. �� Millikan established discrete charge values. �� Electron mass can be calculated using e and e/m.

✅ Correct Answer Explanation

Millikan's oil-drop experiment showed that electric charge exists in discrete packets and is not continuous. The charge on any droplet is always an integral multiple of the elementary charge 1.602 × 10⁻¹⁹ C. Combining e with e/m allows determination of electron mass. Therefore, Option B is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement regarding quantization of charge.
�� Option B → Correct value of elementary charge.
�� Option D → Correct because electron mass can be calculated from e and e/m.

🎯 Strategy

Used

Elimination

Application:

Identify the statement that contradicts charge quantization.

Final Logic:

Millikan proved quantized charge, not continuous charge.

🧠 Memory Trick

Millikan → Quantized Charge

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Quantization of Charge

17 Match List I with List II (Metal properties)

List I	List II (Correct Match)
(1) Free electrons	(a) Responsible for carrying current
(2) Inside metal	(b) Held inside by attractive forces
(3) Positive ions	(c) Exert attractive forces to pull electrons back
(4) Conductivity	(d) Free to move but generally unable to escape

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

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

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

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

💡 Short Explanation

�� Free electrons move inside metals. �� Positive ions attract electrons. �� Conductivity results from electron motion.

✅ Correct Answer Explanation

The correct matching is: → (1) Free electrons → (d) Free to move but generally unable to escape → (2) Inside metal → (b) Held inside by attractive forces → (3) Positive ions → (c) Exert attractive forces to pull electrons back → (4) Conductivity → (a) Responsible for carrying current Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of (1), (3) and (4).
�� Option C → Multiple pairings are incorrect.
�� Option D → Multiple pairings are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Match each metal property with its physical meaning.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

Free → Held → Pull → Current

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Free Electrons in Metals

18 The attractive forces in a metal

A repel free electrons actively out of the surface

B are entirely caused by neighbouring negative ions

C hold the free electrons inside the metal surface boundary

D diminish to zero when an electron attempts to escape

💡 Short Explanation

�� Positive ions attract electrons. �� Electrons remain bound to the metal. �� Escape requires additional energy.

✅ Correct Answer Explanation

The attractive forces due to positive ions inside a metal hold free electrons within the metal surface boundary. These forces prevent electrons from escaping freely and are responsible for the existence of the work function. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Attractive forces do not push electrons outward.
�� Option B → Metals contain positive ions, not neighbouring negative ions.
�� Option D → Attractive forces do not vanish during escape attempts.

🎯 Strategy

Used

Elimination

Application:

Apply the concept of electron binding in metals.

Final Logic:

Attractive forces keep electrons confined within the metal.

🧠 Memory Trick

Positive Ions Hold Electrons

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Surface Barrier in Metals

19 When a free electron attempts to leave the metal surface without sufficient energy

A It generates a secondary photon

B The metal surface acquires a positive charge and pulls the electron back

C The electron accelerates outward instantly

D The internal ions repel the electron away

💡 Short Explanation

�� Electron loss leaves positive charge behind. �� Positive charge attracts the electron. �� The electron is pulled back.

✅ Correct Answer Explanation

When an electron attempts to leave the metal surface without sufficient energy, the metal becomes positively charged due to loss of negative charge. This positive charge attracts the electron and pulls it back into the metal. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Secondary photon generation is not the expected outcome.
�� Option C → The electron cannot escape without sufficient energy.
�� Option D → Positive ions attract rather than repel electrons.

🎯 Strategy

Used

Elimination

Application:

Use electrostatic attraction between opposite charges.

Final Logic:

Positive surface charge pulls the electron back.

🧠 Memory Trick

Electron Leaves → Surface Positive

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

20 Correct statements about escape energy requirements
1. A certain minimum amount of energy is required to escape.
2. This threshold energy is known as the work function.
3. It is typically denoted by φ₀ and measured in electron volts (eV).
4. It proves that electrons have zero mass.

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

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

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

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

💡 Short Explanation

�� Electron escape requires minimum energy. �� This energy is called work function. �� Work function is measured in eV.

✅ Correct Answer Explanation

Statement 1 is correct because electrons require a minimum amount of energy to escape from a metal surface. → Statement 2 is correct because this minimum energy is called the work function. → Statement 3 is correct because the work function is commonly denoted by φ₀ and measured in electron volts (eV). → Statement 4 is incorrect because the work function has no relation to electrons having zero mass. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 4.
�� Option C → Includes incorrect statement 4.
�� Option D → Includes incorrect statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Check each numbered statement using the definition of work function.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Work Function = Minimum Escape Energy

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Work Function (φ₀)

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