CUET UG Physics Booster Test 2-Alpha-Particle Scattering -

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CUET UG Physics Booster Test 2-Alpha-Particle Scattering

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-Alpha-Particle Scattering

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QUESTION 1 OF 20

Statements about the proposed strategy of the classic scattering experiment:
1. Ernst Rutherford originally proposed the scattering of alpha particles by atoms in 1906.
2. It aimed to investigate how positive charge and electrons are arranged inside the atom.
3. It leveraged alpha-particles emitted by radioactive elements as a probing tool.
4. The experiment was physically performed from start to finish solely by J.J. Thomson.

Answer Review

1 Statements about the proposed strategy of the classic scattering experiment:
1. Ernst Rutherford originally proposed the scattering of alpha particles by atoms in 1906.
2. It aimed to investigate how positive charge and electrons are arranged inside the atom.
3. It leveraged alpha-particles emitted by radioactive elements as a probing tool.
4. The experiment was physically performed from start to finish solely by J.J. Thomson.

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

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

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

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

💡 Short Explanation

�� Rutherford proposed the experiment. �� Atomic structure was the target of investigation. �� Alpha particles acted as probes.

✅ Correct Answer Explanation

Statement 1 is correct because Rutherford proposed alpha-particle scattering in 1906. Statement 2 is correct because the objective was to determine the arrangement of positive charge and electrons. Statement 3 is correct because naturally emitted alpha particles from radioactive substances were used as probes. Statement 4 is incorrect because the experiment was performed by Geiger and Marsden under Rutherford's guidance. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the historical facts of the Rutherford experiment.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Rutherford Proposed, Geiger Performed

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Experiment

2 Rutherford primarily acted as the ________ of the experiment, while Geiger and Marsden were the ones who ultimately ________ it.

A Proposer, Performed

B Opponent, Ignored

C Student, Proposed

D Observer, Theorized

💡 Short Explanation

�� Rutherford designed the strategy. �� Geiger and Marsden carried it out. �� Experimental work validated the idea.

✅ Correct Answer Explanation

Rutherford conceived and proposed the alpha-particle scattering experiment. The actual experimental work was conducted by Hans Geiger and Ernest Marsden. Therefore, the correct combination is Proposer, Performed.

❓ Why Other Options Are Incorrect

�� Option B → Rutherford supported and proposed the experiment.
�� Option C → Rutherford was not a student in this context.
�� Option D → Geiger and Marsden performed experiments rather than merely theorizing.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate scientists with their roles.

Final Logic:

Rutherford proposed; Geiger and Marsden performed.

🧠 Memory Trick

Think → Rutherford, Do → Geiger-Marsden

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Experiment

3 In the experimental setup designed by Geiger and Marsden,

A The entire apparatus is placed in a vacuum chamber, though it is not explicitly shown in the schematic diagram

B The vacuum chamber physically houses only the radioactive source and nothing else

C the vacuum chamber is pumped full of dense hydrogen gas

D the apparatus must be placed in a lead chamber filled with water

💡 Short Explanation

�� Vacuum minimizes unwanted collisions. �� Entire setup is enclosed. �� Alpha particles travel freely.

✅ Correct Answer Explanation

The entire scattering apparatus was placed inside a vacuum chamber to prevent alpha particles from interacting with air molecules before reaching the foil and detector. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because the entire setup was housed inside the vacuum chamber.
�� Option C → Incorrect because dense gas would interfere with alpha particles.
�� Option D → Incorrect because no water-filled lead chamber was used.

🎯 Strategy

Used

Elimination

Application:

Identify the purpose and use of the vacuum chamber.

Final Logic:

Vacuum ensures unobstructed alpha-particle motion.

🧠 Memory Trick

Vacuum = Clear Path

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Experimental Arrangement

4 Match List I with List II regarding the experimental components of the setup.

List I	List II
1. Radioactive source	a. 214/83 Bi emitting 5.5 MeV particles
2. Lead bricks	b. Material used to collimate alpha-particles
3. Narrow beam	c. Directed onto the gold foil
4. Metal foil target	d. Has a thickness of 2.1 × 10⁻⁷ m

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

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

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

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

💡 Short Explanation

�� Bi source emits alpha particles. �� Lead bricks collimate the beam. �� Gold foil acts as target.

✅ Correct Answer Explanation

1 → a : Radioactive bismuth source emits alpha particles. 2 → b : Lead bricks are used for collimation. 3 → c : The narrow beam is directed toward the gold foil. 4 → d : Gold foil thickness is approximately 2.1 × 10⁻⁷ m. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Source and lead bricks are mismatched.
�� Option C → Multiple incorrect pairings.
�� Option D → Incorrect assignment of foil and source.

🎯 Strategy

Used

Option Grouping

Application:

Match each experimental component with its function.

Final Logic:

Only option A correctly matches all components.

🧠 Memory Trick

Source-Lead-Beam-Foil

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Experimental Setup

5 If an alpha particle of natural origin has a maximum kinetic energy of 7.7 MeV, this exact energy converted into Joules (given e = 1.6 × 10⁻¹⁹ C) mathematically equals

A 1.2 × 10⁻¹² J

B 1.6 × 10⁻¹⁹ J

C 1.2 × 10⁻¹³ J

D 7.7 × 10⁻¹² J

💡 Short Explanation

�� 1 eV = 1.6 × 10⁻¹⁹ J. �� 7.7 MeV = 7.7 × 10⁶ eV. �� Convert by multiplication.

✅ Correct Answer Explanation

Energy in joules: E = 7.7 × 10⁶ × 1.6 × 10⁻¹⁹ = 12.32 × 10⁻¹³ = 1.23 × 10⁻¹² J ≈ 1.2 × 10⁻¹² J Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents only 1 eV.
�� Option C → One order of magnitude smaller.
�� Option D → Numerical conversion is incorrect.

🎯 Strategy

Used

Substitution

Application:

Convert MeV to joules using 1 eV = 1.6 × 10⁻¹⁹ J.

Final Logic:

7.7 MeV ≈ 1.2 × 10⁻¹² J.

🧠 Memory Trick

MeV × 10⁻¹³ ≈ Joules

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy Conversion in Rutherford Analysis

6 Incorrect statement about the gold foil used in the experiment:

A The atomic number (Z) of the gold atom is 79.

B The foil is extremely thin to allow the assumption that alpha-particles suffer not more than one scattering.

C Its measured thickness is exactly 2.1 × 10⁻¹⁰ m.

D The nucleus of gold is approximately 50 times heavier than an incident alpha-particle.

💡 Short Explanation

�� Gold foil thickness is 2.1 × 10⁻⁷ m. �� Gold has Z = 79. �� Thin foil minimizes multiple scattering.

✅ Correct Answer Explanation

The thickness of the gold foil used in the experiment is approximately 2.1 × 10⁻⁷ m, not 2.1 × 10⁻¹⁰ m. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct atomic number of gold.
�� Option B → Correct reason for using a thin foil.
�� Option D → Correct approximation regarding nuclear mass.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare the actual foil thickness with the given value.

Final Logic:

2.1 × 10⁻¹⁰ m is much smaller than the actual foil thickness.

🧠 Memory Trick

Gold Foil = 10⁻⁷ m

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Gold Foil Characteristics

7 Correct statements regarding the detector apparatus:
1. It consists of a zinc sulphide screen coupled with a microscope.
2. It can be rotated to study the particle distribution as a function of scattering angle.
3. Flashes on the screen are produced when alpha-particles strike it.
4. It measures the exact mass of individual alpha particles on impact.

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

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

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

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

💡 Short Explanation

�� ZnS screen detects particles. �� Microscope observes flashes. �� Detector rotates to measure angles.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct descriptions of the detector system used in the Geiger-Marsden experiment. Statement 4 is incorrect because the detector counted scintillations and did not measure particle mass directly. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify the functions of the detector apparatus.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

ZnS + Microscope = Detector

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Detection Apparatus

8 During the continuous detection process in the experiment,

A A steady, uninterrupted beam of continuous visible light is emitted by the

B screenBrief light flashes or scintillations are observed through the microscope when scattered particles hit the screen

C light flashes occur exclusively when alpha-particles bypass the foil completely

D scintillations are generated internally by the gold foil, independently of the screen

💡 Short Explanation

�� Alpha particles strike ZnS. �� Tiny flashes appear. �� Microscope is used for observation.

✅ Correct Answer Explanation

Scattered alpha particles striking the zinc sulphide screen produce small flashes of light known as scintillations. These flashes are observed through a microscope. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because flashes are discrete, not continuous.
�� Option C → Incorrect because scintillations occur whenever particles strike the screen.
�� Option D → Incorrect because flashes originate in the ZnS screen.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate scintillations to detector operation.

Final Logic:

ZnS screen produces visible flashes.

🧠 Memory Trick

Alpha + ZnS = Scintillation

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Scintillation Detection

9 Approximately what percentage of the incident alpha-particles in the experiment scatter by an angle greater than 1°?

A 14%

B 1.4%

C 0.014%

D 0.14%

💡 Short Explanation

�� Most particles pass straight through. �� Only a small fraction scatter noticeably. �� Large-angle scattering is very rare.

✅ Correct Answer Explanation

Experimental observations showed that only about 0.14% of the incident alpha particles were scattered by angles greater than 1°. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Far larger than experimental observations.
�� Option B → Ten times larger than the observed value.
�� Option C → Ten times smaller than the observed value.

🎯 Strategy

Used

Recall-Based Elimination

Application:

Recall the quantitative results of Rutherford scattering.

Final Logic:

Observed value ≈ 0.14%.

🧠 Memory Trick

More than 1° → 0.14%

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Data

10 The theoretical prediction curve plotted against the data points for particle distribution
Choose correct:

A Only (A)

B Only (B)

C Both (A) and (C)

D (B) and (D)

💡 Short Explanation

�� Rutherford theory assumes a nucleus. �� Particles are not absorbed. �� Thomson's model fails.

✅ Correct Answer Explanation

Statement A is correct because Rutherford's theoretical distribution assumes a small, dense, positively charged nucleus. Statements B, C and D are incorrect because the theory neither assumes absorption nor uniform charge distribution and does not support Thomson's model. Therefore, only statement A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Particle absorption is not assumed.
�� Option C → Rutherford rejected uniform charge distribution.
�� Option D → Thomson's model could not explain the observed data.

🎯 Strategy

Used

Elimination

Application:

Compare Rutherford's assumptions with alternative models.

Final Logic:

Only statement A matches Rutherford's theory.

🧠 Memory Trick

Curve = Nucleus Model

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Theory and Data Comparison

11 Match List I with List II for the observed scattering probabilities and mechanisms.

List I	List II
1. Deflection > 1°	a. Accounts for roughly 0.14% of particles
2. Deflection > 90°	b. Accounts for about 1 in 8000 particles
3. Pass straight through foil	c. Happens to the vast majority of incident particles
4. Rebounds backward	d. Occurs when the impact parameter is minimum

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

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

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

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

💡 Short Explanation

�� Most particles pass straight through. �� Large-angle scattering is rare. �� Backscattering occurs in head-on encounters.

✅ Correct Answer Explanation

1 → a : About 0.14% of alpha particles scatter by more than 1°. 2 → b : Approximately 1 in 8000 particles scatter by more than 90°. 3 → c : Most alpha particles pass through the foil undeflected. 4 → d : Backward scattering occurs when the impact parameter is minimum. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of scattering probabilities.
�� Option C → Multiple mismatches.
�� Option D → Incorrect association of rebound and straight-through events.

🎯 Strategy

Used

Option Grouping

Application:

Match each scattering event with its experimentally observed frequency.

Final Logic:

Only option A correctly matches all observations.

🧠 Memory Trick

0.14%, 1/8000, Straight, Rebound

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Observations

12 What specific fraction of incident alpha-particles experience a drastic deflection of more than 90°?

A 1 in 100000

B 1 in 1000

C 1 in 8000

D 1 in 800

💡 Short Explanation

�� Large-angle scattering is extremely rare. �� Most particles miss the nucleus. �� Only a few approach very closely.

✅ Correct Answer Explanation

Experimental observations showed that only about one alpha particle in every 8000 incident particles was scattered through an angle greater than 90°. This rare event provided strong evidence for a compact nucleus. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Much smaller than the observed frequency.
�� Option B → Much larger than the observed frequency.
�� Option D → Ten times larger than the experimental value.

🎯 Strategy

Used

Recall-Based Elimination

Application:

Recall the quantitative scattering results.

Final Logic:

Backward scattering occurs for approximately 1 in 8000 particles.

🧠 Memory Trick

90°+ → 1 in 8000

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Results

13 Incorrect statement regarding Rutherford's central concentration hypothesis:

A Greater part of the atom's mass must be tightly concentrated at its centre to provide adequate deflection force.

B It explains massive deflections as the result of close encounters with a tight positive charge.

C An incoming alpha-particle gets very close without penetrating the concentrated mass.

D This hypothesis disproved the nuclear model and favored the plum pudding model.

💡 Short Explanation

�� Rutherford proposed the nuclear model. �� Large deflections require concentrated charge. �� Experimental evidence supported the nucleus.

✅ Correct Answer Explanation

Rutherford's central concentration hypothesis led directly to the nuclear model. The observed scattering data strongly contradicted Thomson's plum pudding model and supported a concentrated positive nucleus. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct inference regarding mass concentration.
�� Option B → Correct explanation of large-angle scattering.
�� Option C → Correct description of closest approach.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement opposite to Rutherford's conclusions.

Final Logic:

The nuclear hypothesis supported, not disproved, the nuclear model.

🧠 Memory Trick

Concentrated Charge → Nuclear Model

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Nuclear Hypothesis

14 Statements regarding the electrostatic repulsive force on an alpha-particle:
1. It is governed purely by Coulomb's law for electrostatic force.
2. It is directed strictly along the line joining the alpha-particle and the target nucleus.
3. Its magnitude continuously changes as the particle approaches and recedes from the nucleus.
4. It functions as an attractive gravitational force.

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

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

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

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

💡 Short Explanation

�� Coulomb force governs scattering. �� Force acts along the line joining charges. �� Magnitude varies with distance.

✅ Correct Answer Explanation

Statement 1 is correct because scattering is explained by electrostatic Coulomb repulsion. Statement 2 is correct because Coulomb force acts along the line connecting the charges. Statement 3 is correct because force varies as 1/r². Statement 4 is incorrect because the interaction is electrostatic repulsion, not gravitational attraction. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Apply Coulomb's law to each statement.

Final Logic:

Only statements 1, 2 and 3 correctly describe the force.

🧠 Memory Trick

Coulomb → Line → Variable

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Coulomb Scattering Force

15 The experimentally derived upper limit to the size of the nucleus was determined to be ________, which is drastically smaller than the established atom size from ________ theory.

A 10⁻¹⁴ m, Kinetic

B 10⁻¹⁰ m, Electromagnetic

C 10⁻¹⁵ m, Quantum

D 10⁻⁷ m, Relativity

💡 Short Explanation

�� Nucleus is extremely small. �� Atomic size was known from kinetic theory. �� Rutherford estimated nuclear dimensions.

✅ Correct Answer Explanation

Rutherford's analysis gave an upper limit of approximately 10⁻¹⁴ m for nuclear size. Prior estimates from kinetic theory indicated atomic dimensions of about 10⁻¹⁰ m. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → 10⁻¹⁰ m corresponds to atomic size.
�� Option C → Not the basis of Rutherford's estimate.
�� Option D → Completely incorrect scale and theory.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare atomic and nuclear size scales.

Final Logic:

Nucleus ≈ 10⁻¹⁴ m, atom ≈ 10⁻¹⁰ m.

🧠 Memory Trick

Atom 10⁻¹⁰, Nucleus 10⁻¹⁴

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Nuclear Dimensions

16 If the calculated radius of an atom is represented as Rₐ and the size of the nucleus as Rₙ, the proportion Rₐ / Rₙ based on Rutherford's deductions is expressed algebraically as:

A 10² ≤ Rₐ / Rₙ ≤ 10³

B 10⁴ ≤ Rₐ / Rₙ ≤ 10⁵

C 10⁶ ≤ Rₐ / Rₙ ≤ 10⁷

D 10 ≤ Rₐ / Rₙ ≤ 100

💡 Short Explanation

�� Atom is much larger than nucleus. �� Ratio comes from size comparison. �� Most atomic volume is empty.

✅ Correct Answer Explanation

Atomic radius is approximately 10⁻¹⁰ m, while nuclear radius lies around 10⁻¹⁴–10⁻¹⁵ m. Thus, Rₐ/Rₙ ≈ 10⁴ to 10⁵ Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too small.
�� Option C → Too large.
�� Option D → Far below the observed ratio.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare typical atomic and nuclear dimensions.

Final Logic:

Atom is 10,000–100,000 times larger than nucleus.

🧠 Memory Trick

Atom/Nucleus = 10⁴–10⁵

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Atomic and Nuclear Dimensions

17 The intense electric field originating from the central nucleus
Choose correct:

A Both (A) and (C)

B Only (B)

C Only (A)

D (A) and (D)

💡 Short Explanation

�� Nucleus creates a strong electric field. �� Alpha particles are repelled. �� Scattering results from this repulsion.

✅ Correct Answer Explanation

The strong electric field near the positively charged nucleus produces the electrostatic repulsion responsible for large-angle scattering of alpha particles. Statements B, C and D are incorrect. Therefore, only statement C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Scattering is due to electric, not magnetic, effects.
�� Option B → Electrons do not completely neutralize the nuclear field.
�� Option D → Alpha particles do not orbit the nucleus.

🎯 Strategy

Used

Elimination

Application:

Identify the physical cause of scattering.

Final Logic:

Only nuclear electric repulsion explains the observation.

🧠 Memory Trick

Strong Field → Strong Scatter

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Nuclear Electric Field

18 Match the participant particle to its role/characteristic in the scattering dynamics.

List I	List II
1. Alpha-particle	a. Carries a positive charge of +2e
2. Gold nucleus	b. Modeled as remaining stationary due to heavy mass
3. Atomic electrons	c. So light they do not appreciably affect scattering paths
4. Incident beam	d. Composed of particles with 5.5 MeV kinetic energy

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

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

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

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

💡 Short Explanation

�� Alpha particles carry +2e. �� Gold nucleus is very heavy. �� Electrons have negligible effect.

✅ Correct Answer Explanation

1 → a : Alpha particle has charge +2e. 2 → b : Gold nucleus is treated as stationary. 3 → c : Electrons are too light to significantly influence scattering. 4 → d : Incident beam consists of 5.5 MeV alpha particles. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignments.
�� Option C → Multiple mismatches.
�� Option D → Incorrect charge and mass associations.

🎯 Strategy

Used

Option Grouping

Application:

Match each particle with its known property.

Final Logic:

Only option A correctly matches all entries.

🧠 Memory Trick

+2e, Heavy, Light, 5.5 MeV

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Scattering Dynamics

19 Correct statements concerning the computation of an alpha-particle's trajectory:
1. The computation leverages Newton's second law of motion.
2. It incorporates Coulomb's law for calculating the electrostatic force.
3. It practically assumes the much heavier gold nucleus remains stationary.
4. It requires solving for the exact real-time position of all atomic electrons simultaneously.

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

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

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

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

💡 Short Explanation

�� Newton's law determines motion. �� Coulomb force provides interaction. �� Heavy nucleus is treated as fixed.

✅ Correct Answer Explanation

Statement 1 is correct because trajectory calculation uses Newton's second law. Statement 2 is correct because the force is obtained from Coulomb's law. Statement 3 is correct because the gold nucleus is much heavier than the alpha particle. Statement 4 is incorrect because electron positions are not required for Rutherford's scattering analysis. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Identify the assumptions used in Rutherford's theoretical treatment.

Final Logic:

Only statements 1, 2 and 3 are required.

🧠 Memory Trick

Newton + Coulomb + Fixed Nucleus

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Theory

20 Statements framing Coulomb repulsion in Rutherford's model:
1. The electrostatic force is inversely proportional to the square of the distance between them.
2. The alpha-particle acts as a moving point charge of +2e.
3. The target gold nucleus acts as a stationary point charge of +Ze.
4. The severe repulsion causes the gold nucleus to violently recoil out of the foil.

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

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

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

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

💡 Short Explanation

�� Coulomb force follows inverse-square law. �� Alpha particle has charge +2e. �� Nucleus is treated as stationary.

✅ Correct Answer Explanation

Statement 1 is correct because Coulomb force varies as 1/r². Statement 2 is correct because an alpha particle carries charge +2e. Statement 3 is correct because the gold nucleus is modeled as a stationary charge +Ze. Statement 4 is incorrect because the massive gold nucleus does not violently recoil out of the foil. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Apply Coulomb's law and Rutherford's assumptions.

Final Logic:

Only statements 1, 2 and 3 accurately describe the model.

🧠 Memory Trick

1/r², +2e, +Ze

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Coulomb Repulsion in Rutherford Scattering

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