CUET UG Physics Booster Test 3- Historical Discoveries and Electron Properties -

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UET UG Physics Booster Test 3- Historical Discoveries and Electron Properties

⏱️ 30 Minutes

❓ 20 Questions

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

UET UG Physics Booster Test 3- Historical Discoveries and Electron Properties

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QUESTION 1 OF 20

Statements concerning Early Electromagnetic Foundations
1. Maxwell's equations and Hertz's experiments established the wave nature of light.
2. Hertz's experiments involved the generation of electromagnetic waves.
3. These foundations perfectly predicted the instantaneous nature of the photoelectric effect.
4. The wave nature of light was strongly established by 1887.

Answer Review

1 Statements concerning Early Electromagnetic Foundations
1. Maxwell's equations and Hertz's experiments established the wave nature of light.
2. Hertz's experiments involved the generation of electromagnetic waves.
3. These foundations perfectly predicted the instantaneous nature of the photoelectric effect.
4. The wave nature of light was strongly established by 1887.

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

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

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

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

💡 Short Explanation

�� Maxwell predicted electromagnetic waves. �� Hertz generated and detected electromagnetic waves. �� The photoelectric effect was not explained by classical wave theory.

✅ Correct Answer Explanation

Statement 1 is correct because Maxwell's equations and Hertz's experiments strongly established the wave nature of light. → Statement 2 is correct because Hertz successfully generated and detected electromagnetic waves. → Statement 3 is incorrect because classical electromagnetic theory could not explain the instantaneous nature of the photoelectric effect. → Statement 4 is correct because Hertz's experiments in 1887 provided strong experimental support for the wave nature of light. → 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 using Maxwell's theory and Hertz's experiments.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Maxwell Predicts → Hertz Proves

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

2 In the context of late 19th-century physics, Hertz's 1887 experiments

A provided the very first direct evidence of the electron's mass

B successfully demonstrated generation and detection of electromagnetic waves, affirming Maxwell

C mathematically calculated the specific charge of cathode rays

D proved that cathode rays are merely electromagnetic waves

💡 Short Explanation

�� Hertz tested Maxwell's predictions. �� Electromagnetic waves were generated and detected. �� The experiments supported electromagnetic wave theory.

✅ Correct Answer Explanation

Hertz successfully generated and detected electromagnetic waves in 1887. His experiments provided experimental verification of Maxwell's electromagnetic theory and strongly supported the wave nature of light. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Hertz did not determine electron mass.
�� Option C → Specific charge (e/m) was determined by J.J. Thomson.
�� Option D → Cathode rays are streams of electrons, not electromagnetic waves.

🎯 Strategy

Used

Direct Recall

Application:

Recall Hertz's major experimental achievement.

Final Logic:

Hertz experimentally confirmed Maxwell's electromagnetic wave theory.

🧠 Memory Trick

Hertz = EM Waves

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

3 If a discharge tube is operating optimally at 0.001 mm Hg, and standard atmospheric pressure is 760 mm Hg, what approximate fraction of atmospheric pressure is this?

A ~1.3 × 10⁻⁶

B ~1.3 × 10⁻³

C ~1.0 × 10⁻⁴

D ~7.6 × 10⁻²

💡 Short Explanation

�� Fraction = operating pressure ÷ atmospheric pressure. �� Atmospheric pressure is 760 mm Hg. �� The result is extremely small.

✅ Correct Answer Explanation

Fraction of atmospheric pressure = 0.001 / 760 ≈ 1.315 × 10⁻⁶ ≈ 1.3 × 10⁻⁶ Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Much larger than the calculated value.
�� Option C → Incorrect order of magnitude.
�� Option D → Far larger than the actual fraction.

🎯 Strategy

Used

Substitution

Application:

Substitute the given values into pressure ratio.

Final Logic:

0.001 ÷ 760 ≈ 1.3 × 10⁻⁶.

🧠 Memory Trick

0.001 ÷ 760 → 10⁻⁶

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

4 The fluorescence observed in historical gas discharge tubes

A occurs on the glass adjacent immediately to the anode

B is independent of the composition of the glass used

C is attributed to invisible radiation originating from the cathode

D only happens effectively at extremely high pressures

💡 Short Explanation

�� Cathode rays travel from the cathode. �� They produce fluorescence on striking glass. �� The glow depends on glass composition.

✅ Correct Answer Explanation

The fluorescence observed in discharge tubes is caused by invisible cathode rays originating from the cathode. When these rays strike the glass wall, fluorescence is produced. The colour depends on the type of glass used. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The glow appears on the glass opposite the cathode.
�� Option B → The colour depends on the glass composition.
�� Option D → The effect is observed at very low pressures, not extremely high pressures.

🎯 Strategy

Used

Elimination

Application:

Compare each option with cathode ray observations.

Final Logic:

Cathode rays originating from the cathode cause fluorescence.

🧠 Memory Trick

Cathode Rays → Fluorescence

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

5 Match List I with List II (Late 19th-century milestones)

List I (Scientist & Year)	List II (Correct Milestone)
(1) Roentgen (1895)	(a) Generation of EM waves
(2) Thomson (1897)	(b) Hypothesized streams of fast moving negative particles
(3) Crookes (1879)	(c) Historic discovery of X-rays
(4) Hertz (1887)	(d) Historic discovery of the electron

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

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

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

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

💡 Short Explanation

�� Roentgen discovered X-rays. �� Thomson discovered the electron. �� Crookes proposed the negative-particle hypothesis.

✅ Correct Answer Explanation

The correct matching is: → (1) Roentgen (1895) → (c) Historic discovery of X-rays → (2) Thomson (1897) → (d) Historic discovery of the electron → (3) Crookes (1879) → (b) Hypothesized streams of fast moving negative particles → (4) Hertz (1887) → (a) Generation of EM waves These milestones played major roles in the development of modern atomic physics.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly exchanges the contributions of Roentgen and Thomson.
�� Option C → Multiple scientist-discovery pairings are incorrect.
�� Option D → Incorrectly matches Thomson and Crookes.

🎯 Strategy

Used

Chronological Matching

Application:

Match each scientist with the historically correct discovery.

Final Logic:

Only Option A correctly matches all scientists and discoveries.

🧠 Memory Trick

Crookes → Hertz → Roentgen → Thomson

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

6 Incorrect statement about atomic structure implications

A The discovery of X-rays was a major milestone for atomic structure.

B Thomson's discovery of the electron occurred in 1897.

C Experimental investigations on conduction of electricity at low pressure led to these historic discoveries.

D The established wave nature of light directly predicted the quantized mass of the electron.

💡 Short Explanation

�� X-rays and electrons were major discoveries. �� Low-pressure discharge experiments were important. �� Wave theory did not predict electron mass quantization.

✅ Correct Answer Explanation

The wave nature of light explained electromagnetic radiation but did not directly predict the quantized mass of the electron. The discovery of X-rays and electrons arose from experimental investigations and became major milestones in atomic structure research. Therefore, Option D is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct historical fact.
�� Option C → Correct statement regarding discharge tube experiments.

🎯 Strategy

Used

Elimination

Application:

Identify the statement unsupported by electromagnetic wave theory.

Final Logic:

Wave theory does not predict quantized electron mass.

🧠 Memory Trick

Wave Nature ≠ Electron Mass

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

7 The specific nature hypothesized by Crookes and the scientist who confirmed it:

A Fast-moving negative particles, Thomson

B Continuous wave medium, Roentgen

C X-ray radiation, Millikan

D Positive ions, Hertz

💡 Short Explanation

�� Crookes proposed cathode rays were negative particles. �� Thomson experimentally confirmed this idea. �� The particles were later identified as electrons.

✅ Correct Answer Explanation

Crookes proposed that cathode rays were streams of fast moving negative particles. J.J. Thomson later confirmed this through his experiments and identified these particles as electrons. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Roentgen discovered X-rays, not a continuous wave medium.
�� Option C → Millikan measured electronic charge.
�� Option D → Hertz worked on electromagnetic waves.

🎯 Strategy

Used

Direct Recall

Application:

Recall Crookes' hypothesis and Thomson's confirmation.

Final Logic:

Crookes proposed negative particles; Thomson confirmed them.

🧠 Memory Trick

Crookes Proposes → Thomson Confirms

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Cathode Rays and Electron Discovery

8 During a theoretical calculation, if one oil drop acquires a discrete charge of -x(e) and another drop acquires -y(e), the combined net charge of both drops mathematically evaluates to

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

🎯 Strategy

Used

Substitution

Application:

Add the given charges directly.

Final Logic:

Two negative charges add to give -(x + y)e.

🧠 Memory Trick

Negative + Negative = More Negative

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

9 By applying mutually perpendicular electric and magnetic fields across the discharge tube, J.J. Thomson

A found the specific charge of cathode rays depends intimately on the specific gas

B calculated the precise exact mass of the electron without using its charge

C determined the specific charge (e/m) of the cathode ray particles

D observed that cathode rays are totally unaffected by magnetic fields

💡 Short Explanation

�� Thomson used electric and magnetic fields. �� He measured e/m. �� The value was independent of gas and cathode material.

✅ Correct Answer Explanation

J.J. Thomson applied mutually perpendicular electric and magnetic fields to cathode rays and determined the specific charge (e/m) of the particles. This experiment played a crucial role in identifying the electron. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → e/m was independent of the gas used.
�� Option B → Thomson did not directly determine electron mass.
�� Option D → Cathode rays are affected by magnetic fields.

🎯 Strategy

Used

Direct Recall

Application:

Recall the purpose of Thomson's experiment.

Final Logic:

The experiment determined the specific charge e/m.

🧠 Memory Trick

Thomson → e/m

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

10 Correct statements about J.J. Thomson's 1906 Nobel Prize
1. It was awarded for contributions to Physics.
2. It recognized his investigations on the conduction of electricity by gases.
3. It was awarded after Millikan's pioneering 1913 oil-drop experiment.
4. It established electrons as fundamental, universal constituents of matter.

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

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

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

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

💡 Short Explanation

�� Thomson received the Nobel Prize in Physics in 1906. �� The award recognized work on gas conduction. �� His work established the universal nature of electrons.

✅ Correct Answer Explanation

Statement 1 is correct because Thomson received the Nobel Prize in Physics. → Statement 2 is correct because the award recognized his investigations on conduction of electricity by gases. → Statement 3 is incorrect because the Nobel Prize was awarded in 1906, before Millikan's 1913 oil-drop experiment. → Statement 4 is correct because Thomson's work established electrons as universal constituents of matter. → 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 using historical facts.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

1906 Nobel → Gas Conduction → Electron

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

11 The empirical significance of the e/m value remaining independent of the cathode material

A implies that electrons are uniquely constrained to alkali metals

B proved that cathode rays are actually electromagnetic waves

C demonstrated that ambient gases alter the physical mass of the electron

D heavily suggested the universality of cathode ray particles

💡 Short Explanation

�� e/m remained constant for different cathode materials. �� This indicated identical particles in all cases. �� The observation supported electron universality.

✅ Correct Answer Explanation

J.J. Thomson found that the value of e/m remained independent of the cathode material used in the discharge tube. This showed that cathode ray particles were identical regardless of their source. Such observations strongly suggested that electrons are universal constituents of matter. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electrons are present in all matter, not only alkali metals.
�� Option B → Cathode rays are streams of electrons, not electromagnetic waves.
�� Option C → Ambient gases do not alter the mass of electrons.

🎯 Strategy

Used

Elimination

Application:

Reject statements that contradict Thomson's experimental findings.

Final Logic:

Constant e/m implies universal cathode ray particles.

🧠 Memory Trick

Same e/m → Same Electron

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

12 Match List I with List II (Universality of particles)

List I	List II (Correct Match)
(1) UV-irradiated metals	(a) Discovered to emit negatively charged particles with small speeds
(2) Heated metals	(b) Suggested these are fundamental matter constituents
(3) Cathode rays	(c) Their e/m value matches that of particles from thermal and UV sources
(4) J.J. Thomson	(d) Discovered to emit identical particles at high temperatures

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

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

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

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

💡 Short Explanation

�� UV irradiated and heated metals emit electrons. �� Cathode ray particles have the same e/m. �� Thomson proposed electrons as fundamental constituents of matter.

✅ Correct Answer Explanation

The correct matching is: → (1) UV irradiated metals → (a) Discovered to emit negatively charged particles with small speeds → (2) Heated metals → (d) Discovered to emit identical particles at high temperatures → (3) Cathode rays → (c) Their e/m value matches that of particles from thermal and UV sources → (4) J.J. Thomson → (b) Suggested these are fundamental matter constituents These observations collectively established the universality of electrons.

❓ Why Other Options Are Incorrect

�� Option A → Multiple pairings are incorrect.
�� Option B → Incorrectly matches (3) and (4).
�� Option C → Multiple pairings are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Match each source of electron emission with its correct description.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

UV → Heated → Cathode Rays → Thomson

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

13 The functional speed range of electrons relative to light, and the principal force constraining their escape from metal:

A 0.1c to 0.2c, Attractive ion forces

B 0.5c to 0.9c, Strong nuclear forces

C 0.1c to 0.2c, Repulsive internal forces

D 1.0c, Magnetic forces

💡 Short Explanation

�� Electrons move at about 0.1c to 0.2c in cathode rays. �� Positive ions attract electrons. �� This attraction opposes electron escape.

✅ Correct Answer Explanation

Cathode ray electrons typically travel at speeds ranging from about 0.1c to 0.2c. Inside a metal, positive ions exert attractive forces on electrons and help keep them confined within the metal. These attractive ion forces contribute to the surface barrier and work function. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect speed range and unrelated force.
�� Option C → Attractive, not repulsive, forces constrain electrons.
�� Option D → Electrons do not travel at the speed of light and magnetic forces are not the principal restraining force.

🎯 Strategy

Used

Elimination

Application:

Compare known electron speeds and binding forces.

Final Logic:

Electrons move at 0.1c–0.2c and are held by attractive ion forces.

🧠 Memory Trick

0.1c–0.2c + Positive Ions Hold

📚 Class XI – Chemistry Part I – Chapter: Structure of Atom – Topic: Properties of Cathode Rays

14 If the accepted e/m value is 1.76 × 10¹¹ C/kg, what is the calculated value of m/e?

A 1.60 × 10⁻¹⁹ kg/C

B 1.76 × 10¹¹ kg/C

C 5.68 × 10⁻¹² kg/C

D 9.11 × 10⁻³¹ kg/C

💡 Short Explanation

�� m/e is the reciprocal of e/m. �� Use inversion of the given value. �� Reciprocal calculation gives the answer.

✅ Correct Answer Explanation

Given: e/m = 1.76 × 10¹¹ C/kg Therefore, m/e = 1 / (1.76 × 10¹¹) = 5.68 × 10⁻¹² kg/C Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents elementary charge magnitude, not m/e.
�� Option B → Same numerical value as e/m, not its reciprocal.
�� Option D → Represents electron mass, not m/e.

🎯 Strategy

Used

Substitution

Application:

Take the reciprocal of the given e/m value.

Final Logic:

m/e = 1 ÷ (1.76 × 10¹¹) = 5.68 × 10⁻¹² kg/C.

🧠 Memory Trick

m/e = 1 ÷ (e/m)

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

15 Statements about Millikan's elementary charge measurement
1. It involved the pioneering oil-drop experiment in 1913.
2. It proved that charge inherently behaves in a quantized manner.
3. It found the fundamental charge to be 1.602 × 10⁻¹⁹ C.
4. It was primarily used to find the work function of alkali metals.

A (2), (3), (4)

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

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

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

💡 Short Explanation

�� Millikan measured the elementary charge. �� The experiment established charge quantization. �� It was not designed to determine work function.

✅ Correct Answer Explanation

Statement 1 is correct because Millikan's oil-drop experiment is associated with 1913. → Statement 2 is correct because the experiment established the quantized nature of electric charge. → Statement 3 is correct because the measured elementary charge was approximately 1.602 × 10⁻¹⁹ C. → Statement 4 is incorrect because the experiment was not primarily used to determine the work function of alkali metals. → Therefore Option C is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each numbered statement independently.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Millikan → Quantized Charge → 1.602 × 10⁻¹⁹ C

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

16 Incorrect statement regarding charge quantization and derived properties

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

B The mass of the electron cannot be determined even when given e and e/m.

C Millikan performed the defining experiment in 1913.

D The precise elementary charge is 1.602 × 10⁻¹⁹ C.

💡 Short Explanation

�� Charge is quantized. �� e and e/m together can give electron mass. �� Millikan measured the elementary charge.

✅ Correct Answer Explanation

If the values of e and e/m are known, the electron mass can be calculated using: m = e ÷ (e/m) Therefore, the statement that electron mass cannot be determined is incorrect. The remaining statements are consistent with Millikan's experiment and charge quantization. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement regarding quantized charge.
�� Option C → Correct historical statement.
�� Option D → Correct value of elementary charge.

🎯 Strategy

Used

Elimination

Application:

Check whether electron mass can be derived from e and e/m.

Final Logic:

Knowing e and e/m allows calculation of electron mass.

🧠 Memory Trick

e + e/m → Electron Mass

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

17 While free electrons are responsible for establishing metal conductivity, they paradoxically

A can normally escape out of the metal surface effortlessly

B are held securely inside by the attractive forces of the fixed ions

C completely ignore the surface energy barrier

D are rigidly bonded continuously to specific lattice nuclei

💡 Short Explanation

�� Free electrons conduct electricity. �� Positive ions attract and retain electrons. �� Escape requires additional energy.

✅ Correct Answer Explanation

Although free electrons are responsible for conductivity, they are still held within the metal by the attractive forces of fixed positive ions. Because of these forces, electrons cannot normally escape without sufficient external energy. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electrons cannot escape effortlessly.
�� Option C → Surface energy barrier affects electron escape.
�� Option D → Free electrons are not permanently bound to specific nuclei.

🎯 Strategy

Used

Elimination

Application:

Apply the free electron model of metals.

Final Logic:

Free electrons move within the metal but remain confined by attractive ion forces.

🧠 Memory Trick

Free to Move, Not Free to Escape

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

18 Correct statements concerning ion attractive forces
1. They pull the escaping electron back into the metal surface.
2. They cause the surface to acquire an effective positive charge when an electron attempts to escape.
3. They are easily bypassed at room temperature without required external energy.
4. They create the conceptual surface barrier that defines the work function.

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

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

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

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

💡 Short Explanation

�� Positive ions attract escaping electrons. �� Escape attempts leave the surface effectively positive. �� These effects contribute to the work function.

✅ Correct Answer Explanation

Statement 1 is correct because attractive forces tend to pull escaping electrons back into the metal. → Statement 2 is correct because the loss of an electron leaves the surface effectively positively charged. → Statement 3 is incorrect because room-temperature energy is generally insufficient to overcome the surface barrier. → Statement 4 is correct because these attractive forces are responsible for the surface barrier associated with the work function. → 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 work function.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Pull Back → Positive Surface → Work Function

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

19 The positive charge pullback mechanism mathematically and conceptually

A accelerates electrons radically out of the metal boundary

B requires the electron to supply energy spontaneously to the metal

C dictates that a strict minimum energy threshold must be met to escape

D acts independently of the nature of the specific metal surface

💡 Short Explanation

�� Escaping electrons face a surface barrier. �� Minimum energy is required for escape. �� This energy is called the work function.

✅ Correct Answer Explanation

The attractive forces acting on electrons inside a metal create a surface barrier. To overcome this barrier, an electron must possess a minimum threshold energy. This minimum energy requirement is the basis of the work function. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The mechanism opposes escape rather than accelerating electrons outward.
�� Option B → No spontaneous energy supply occurs.
�� Option D → Work function depends on the nature of the metal surface.

🎯 Strategy

Used

Elimination

Application:

Apply the concept of work function and surface barrier.

Final Logic:

A minimum threshold energy must be supplied for escape.

🧠 Memory Trick

Barrier → Threshold Energy

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

20 If the minimum escape energy (work function) of a metal is exactly 2.0 eV, what is this specific barrier energy evaluated in Joules? (1 eV = 1.602 × 10⁻¹⁹ J)

A 1.602 × 10⁻¹⁹ J

B 3.204 × 10⁻¹⁹ J

C 4.806 × 10⁻¹⁹ J

D 8.010 × 10⁻¹⁹ J

💡 Short Explanation

�� Convert eV to Joules. �� Multiply by 1.602 × 10⁻¹⁹ J. �� Use the given work function value.

✅ Correct Answer Explanation

Given: Work function = 2.0 eV 1 eV = 1.602 × 10⁻¹⁹ J Therefore, 2.0 × 1.602 × 10⁻¹⁹ = 3.204 × 10⁻¹⁹ J Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to 1.0 eV.
�� Option C → Corresponds to 3.0 eV.
�� Option D → Corresponds to 5.0 eV.

🎯 Strategy

Used

Substitution

Application:

Substitute the given conversion factor directly.

Final Logic:

2.0 eV × 1.602 × 10⁻¹⁹ J/eV = 3.204 × 10⁻¹⁹ J.

🧠 Memory Trick

2 eV = 2 × 1.602 × 10⁻¹⁹ J

📚 Class XII – Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Work Function and Electron Volt Conversion

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