CUET UG Physics Booster Test 3-Alpha-Particle Scattering -

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CUET UG Physics Booster Test 3-Alpha-Particle Scattering

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Alpha-Particle Scattering

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QUESTION 1 OF 20

The classic experiment proposed by Rutherford in 1906
Choose correct:

Answer Review

1 The classic experiment proposed by Rutherford in 1906
Choose correct:

A Only (A)

B Only (C)

C Both (A) and (C)

D (B) and (D)

💡 Short Explanation

�� Rutherford wanted to probe atomic structure. �� Alpha particles were used as probes. �� The experiment was unrelated to Balmer's formula.

✅ Correct Answer Explanation

Statement A is correct because Rutherford proposed alpha-particle scattering to determine the internal arrangement of positive charge and electrons within atoms. Statements B, C and D are incorrect. The experiment eventually disproved Thomson's model, was not designed to test Balmer's formula, and was unrelated to the study of emission spectra. Therefore, only statement A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement C is incorrect.
�� Option C → Statement C is incorrect.
�� Option D → Both statements B and D are incorrect.

🎯 Strategy

Used

Elimination

Application:

Check the original objective of Rutherford's proposal.

Final Logic:

The experiment was designed to investigate atomic structure.

🧠 Memory Trick

Alpha Probe → Atom Structure

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

2 Incorrect statement regarding the historical timeline and roles in atomic models:

A Rutherford executed the complex alpha-scattering experiment completely alone in 1906.

B J.J. Thomson proposed the uniform plum pudding model of the atom in 1898.

C Hans Geiger and 20-year-old student Ernst Marsden actually performed the scattering experiment around 1911.

D Rutherford, who proposed the nuclear model, was once a former research student of J.J. Thomson.

💡 Short Explanation

�� Rutherford proposed the experiment. �� Geiger and Marsden performed it. �� Thomson supervised Rutherford earlier.

✅ Correct Answer Explanation

Rutherford did not perform the entire experiment alone. He proposed the experiment, while Hans Geiger and Ernest Marsden carried out the actual scattering measurements around 1911. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Correct historical statement.
�� Option C → Correct description of the experimenters.
�� Option D → Correct historical relationship.

🎯 Strategy

Used

Odd One Out

Application:

Identify the historically inaccurate statement.

Final Logic:

Rutherford proposed the experiment but did not perform it alone.

🧠 Memory Trick

Rutherford Planned, Geiger-Marsden Measured

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Historical Development of Atomic Models

3 Match List I (Apparatus part) with List II (Analytical Function/Condition)

List I	List II
1. Vacuum Chamber	a. Environment containing the entire scattering apparatus
2. Lead bricks	b. Shapes the emitted alpha-particles into a concentrated narrow beam
3. Radioactive Source	c. Decays to continuously emit 5.5 MeV alpha particles
4. Rotatable detector	d. Allows empirical measurement of particle distribution at varying angles θ

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

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

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

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

💡 Short Explanation

�� Vacuum prevents unwanted collisions. �� Lead bricks collimate the beam. �� Detector measures angular distribution.

✅ Correct Answer Explanation

1 → a : Vacuum chamber contains the entire apparatus. 2 → b : Lead bricks collimate alpha particles. 3 → c : Radioactive source emits 5.5 MeV alpha particles. 4 → d : Rotatable detector records scattering at different angles. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiple mismatches.
�� Option C → Incorrect matching of detector and source.
�� Option D → Incorrect assignment of apparatus functions.

🎯 Strategy

Used

Option Grouping

Application:

Match each apparatus component with its function.

Final Logic:

Only option B correctly matches all pairs.

🧠 Memory Trick

Vacuum–Lead–Source–Detector

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

4 The distance of closest approach d for an alpha particle in a head-on collision is mathematically established by equating its initial kinetic energy (K) to its final electrostatic potential energy. Which expression precisely defines d?

A d = (Ze)/(4πε₀K²)

B d = (Ze²)/(4πε₀K)

C d = (4Ze²)/(4πε₀K)

D d = (2Ze²)/(4πε₀K)

💡 Short Explanation

�� At closest approach, KE becomes PE. �� Coulomb repulsion stops the alpha particle. �� Energy conservation is applied.

✅ Correct Answer Explanation

For a head-on collision, K = (1/4πε₀)(2Ze²/d) Rearranging, d = (2Ze²)/(4πε₀K) Thus, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect dimensions and relation.
�� Option B → Missing factor 2.
�� Option C → Extra factor 2.

🎯 Strategy

Used

Substitution

Application:

Apply energy conservation and Coulomb potential energy.

Final Logic:

KE = PE gives d = (2Ze²)/(4πε₀K).

🧠 Memory Trick

Closest Approach → KE = PE

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Distance of Closest Approach

5 Correct analytical statements concerning the radioactive source:
1. It was a specific isotope of Bismuth, designated as ²¹⁴₈₃Bi.
2. It served as the origin point for alpha-particles carrying 5.5 MeV of kinetic energy.
3. The maximum theoretical kinetic energy found in alpha particles of natural origin is 7.7 MeV.
4. It emitted dense continuous X-ray spectra that interfered with the zinc sulphide screen.

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

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

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

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

💡 Short Explanation

�� Bismuth source emitted alpha particles. �� Alpha energy was 5.5 MeV. �� Natural alpha energies can reach 7.7 MeV.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct descriptions of the source and alpha-particle energies used in Rutherford's analysis. Statement 4 is incorrect because continuous X-ray interference was not a feature of the experiment. 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:

Evaluate each statement using experimental facts.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Bi Source → 5.5 MeV

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Alpha-Particle Source

6 The gold foil target must be exceptionally ________ so that physicists can reliably assume the traversing alpha-particles suffer not more than ________ scattering event during their passage.

A Dense, Zero

B Thick, Multiple

C Thin, One

D Charged, Elastic

💡 Short Explanation

�� Thin foil minimizes repeated collisions. �� Single-scattering assumption is important. �� Simplifies theoretical analysis.

✅ Correct Answer Explanation

The gold foil was made extremely thin so that most alpha particles would experience at most one significant scattering event. This assumption is essential for Rutherford's theoretical calculations. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Zero scattering is not assumed.
�� Option B → Opposite of the experimental requirement.
�� Option D → Charge and elasticity are irrelevant here.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate foil properties to scattering assumptions.

Final Logic:

Thin foil ensures predominantly single scattering.

🧠 Memory Trick

Thin Foil → One Scatter

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

7 Statements critically evaluating the detection methodology:
1. The detector fundamentally requires a zinc sulphide screen paired with a microscope.
2. Its rotatability is crucial for mapping scattering magnitude as a function of angle.
3. Individual impact flashes are brief but distinctly viewable through the microscope optics.
4. It electrically counts the exact number of atomic electrons displaced by the collision.

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

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

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

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

💡 Short Explanation

�� ZnS screen detects alpha particles. �� Microscope views scintillations. �� Rotation allows angular measurements.

✅ Correct Answer Explanation

Statements 1, 2 and 3 accurately describe the Geiger-Marsden detector. Statement 4 is incorrect because the apparatus observed scintillations and did not count displaced electrons. 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 actual detector functions.

Final Logic:

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

🧠 Memory Trick

ZnS + Microscope + Rotation

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

8 If statistically 1 in 8000 particles deflects by more than 90° causing a flash on the far sides of the detector, what is this specific probability expressed as a decimal fraction?

A 0.00125

B 0.000125

C 0.0125

D 0.125

💡 Short Explanation

�� Probability = Favorable/Total. �� Here it is 1/8000. �� Convert directly to decimal.

✅ Correct Answer Explanation

Probability = 1/8000 = 0.000125 Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equal to 1/800.
�� Option C → Equal to 1/80.
�� Option D → Equal to 1/8.

🎯 Strategy

Used

Substitution

Application:

Convert the given fraction into decimal form.

Final Logic:

1 ÷ 8000 = 0.000125.

🧠 Memory Trick

1/8000 → Five Zeros

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

9 In analyzing the graph detailing the angular distribution of scattered particles,

A the continuous solid curve visualizes the theoretical prediction based on a nuclear model, while the plotted dots represent empirical data

B the solid curve is merely random error bounds, while the dots represent plum pudding model expectations

C the number of particles is shown to increase exponentially as the scattering angle widens

D the angle of scattering reaches exactly 180 degrees for over 50% of the particles

💡 Short Explanation

�� Dots represent observations. �� Curve represents theory. �� Agreement supports the nuclear model.

✅ Correct Answer Explanation

In Rutherford scattering plots, experimental measurements are shown as data points, while the theoretical nuclear-model prediction appears as a smooth curve. Their agreement strongly supports the nuclear model. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect interpretation of graph elements.
�� Option C → Particle count decreases with increasing angle.
�� Option D → Backscattering is extremely rare.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the meaning of graph components.

Final Logic:

Dots = data, Curve = theory.

🧠 Memory Trick

Dots Data, Curve Theory

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

10 Match List I with List II representing Rutherford's statistical scattering data.

List I	List II
1. Scattered by more than 1°	a. Approximately 0.14% of incidence
2. Scattered by more than 90°	b. Statistically 1 in 8000 occurrences
3. Passes through with no collision	c. Behavior of the overwhelming majority
4. Distribution mapping	d. Studied intricately as a function of angle θ

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

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

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

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

💡 Short Explanation

�� Small-angle scattering is uncommon. �� Large-angle scattering is very rare. �� Most particles pass through unaffected.

✅ Correct Answer Explanation

1 → a : About 0.14% scatter by more than 1°. 2 → b : About 1 in 8000 scatter by more than 90°. 3 → c : Most particles pass straight through. 4 → d : Distribution was studied as a function of scattering angle θ. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect probability assignments.
�� Option C → Multiple mismatches.
�� Option D → Incorrect statistical associations.

🎯 Strategy

Used

Option Grouping

Application:

Match each scattering observation with its statistical result.

Final Logic:

Only option A correctly matches all entries.

🧠 Memory Trick

0.14%, 1/8000, Straight, θ

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

11 If N represents the total number of incident alpha-particles, the mathematical expectation for the number of particles passing forward without exceeding a 1° deflection is analytically approximated by:

A N × 0.0014

B N × (1 − 0.0014)

C N × (1/8000)

D N × 0.14

💡 Short Explanation

�� About 0.14% scatter by more than 1°. �� The remaining particles undergo less than 1° deflection. �� Use complement probability.

✅ Correct Answer Explanation

Experimental observations show that approximately 0.14% = 0.0014 of alpha particles scatter by more than 1°. Therefore, the fraction not exceeding 1° is: 1 − 0.0014 = 0.9986 Hence, the expected number is: N × (1 − 0.0014) Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents particles scattered by more than 1°.
�� Option C → Represents particles scattered by more than 90°.
�� Option D → Uses 14% instead of 0.14%.

🎯 Strategy

Used

Substitution

Application:

Use the complementary probability of scattering greater than 1°.

Final Logic:

Forward transmission ≈ Total particles − strongly scattered particles.

🧠 Memory Trick

Forward = 100% − 0.14%

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

12 The extreme rarity of alpha particles deflecting by massive angles mathematically implies that
Statements:
1. The atom's positive charge is tightly concentrated in a volume vastly smaller than the atom.
2. The atom behaves strictly as a solid sphere of uniform mass and charge density.
3. Alpha particles are heavily attracted and absorbed into the nucleus via strong force.
4. Planetary electrons play the dominant role in back-scattering the alpha particles.

A Only (A)

B Both (B) and (D)

C Only (C)

D All of the above

💡 Short Explanation

�� Large deflections are rare. �� Positive charge is concentrated. �� Thomson's model cannot explain observations.

✅ Correct Answer Explanation

Statement 1 is correct because rare large-angle scattering indicates a tiny, dense, positively charged nucleus. Statements 2, 3 and 4 are incorrect. Uniform charge distribution cannot produce the observed scattering, alpha particles are repelled rather than absorbed, and electrons contribute negligibly. Therefore, only statement 1 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Includes multiple incorrect statements.

🎯 Strategy

Used

Elimination

Application:

Relate scattering frequency to atomic structure.

Final Logic:

Rare backscattering implies a tiny concentrated nucleus.

🧠 Memory Trick

Rare Scatter → Tiny Nucleus

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

13 Given the orbital radius of the earth is ~ 1.5 × 10¹¹ m and the sun's radius is 7 × 10⁸ m. If the solar system's dimensions were proportionally scaled to match the empty space ratio of an atom (roughly 10⁵), what would the earth's new orbital radius algebraically become?

A 1.0 × 10¹³ m

B 1.5 × 10¹⁶ m

C 7 × 10³ m

D 7 × 10¹³ m

💡 Short Explanation

�� Atomic empty-space ratio ≈ 10⁵. �� Solar ratio ≈ 1.5 × 10¹¹ / 7 × 10⁸ ≈ 214. �� Scale accordingly.

✅ Correct Answer Explanation

To achieve an atom-like ratio: New orbital radius = (10⁵) × (Sun radius) = (10⁵)(7 × 10⁸) = 7 × 10¹³ m Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Does not satisfy the required ratio.
�� Option B → Incorrect scaling.
�� Option C → Far too small.

🎯 Strategy

Used

Substitution

Application:

Apply the required empty-space ratio directly.

Final Logic:

10⁵ × 7 × 10⁸ = 7 × 10¹³ m.

🧠 Memory Trick

Atomic Ratio = 10⁵

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Empty Space in Atom

14 The determining force drastically altering the path of an incoming alpha-particle is heavily ________ and acts in strict accordance with ________ law.

A Repulsive, Coulomb's

B Attractive, Newton's

C Neutral, Faraday's

D Magnetic, Ampere's

💡 Short Explanation

�� Both nucleus and alpha particle are positive. �� Like charges repel. �� Coulomb's law governs the interaction.

✅ Correct Answer Explanation

The gold nucleus carries charge +Ze and the alpha particle carries charge +2e. The interaction is therefore electrostatic repulsion described by Coulomb's law. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Interaction is not attractive.
�� Option C → No neutral force exists here.
�� Option D → Scattering is not governed by Ampere's law.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the physical force responsible for scattering.

Final Logic:

Positive charges repel according to Coulomb's law.

🧠 Memory Trick

and + → Repel

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

15 Identify the incorrect statement derived from Rutherford's atomic dimensions analysis:

A The actual calculated radius of a gold nucleus is established to be exactly 30 fm.

B The radius of the electron's orbit is about 10⁵ times larger than the radius of the nucleus itself.

C Experimental discrepancy arose because the closest approach distance is larger than the sum of nuclear radii.

D The size of an atom based on kinetic theory was accepted to be roughly 10⁻¹⁰ m.

💡 Short Explanation

�� Rutherford estimated only an upper limit. �� Nuclear radius was not measured exactly. �� Atomic size was already known.

✅ Correct Answer Explanation

Rutherford's analysis provided only an upper limit on nuclear dimensions. It did not establish an exact nuclear radius of 30 fm. Statements B, C and D are consistent with the conclusions of Rutherford's analysis. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Correct order of magnitude.
�� Option C → Correct interpretation of closest approach.
�� Option D → Correct atomic dimension estimate.

🎯 Strategy

Used

Extreme Word Filter

Application:

Watch for absolute terms such as "exactly established."

Final Logic:

Rutherford estimated limits, not exact nuclear radii.

🧠 Memory Trick

Upper Limit ≠ Exact Radius

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

16 Analytical statements regarding the conceptual "empty space" in an atom:
1. Most of an atom's volume is empty space, explaining the high forward transmission rate of alpha-particles.
2. The nuclear atom implies it contains a much greater fraction of empty space than our solar system does.
3. A given alpha-particle rarely comes near enough to a nucleus to encounter the intense electric field.
4. The empty space is permeated by a positively charged jelly-like substance.

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

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

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

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

💡 Short Explanation

�� Most atomic volume is empty. �� Nuclear model implies enormous emptiness. �� Close nuclear encounters are rare.

✅ Correct Answer Explanation

Statements 1, 2 and 3 follow directly from Rutherford's nuclear model. Statement 4 describes Thomson's plum pudding model, not Rutherford's model. 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:

Separate Rutherford's model from Thomson's model.

Final Logic:

Positive jelly belongs to the plum pudding model.

🧠 Memory Trick

Rutherford = Empty Space

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Empty Space in Atom

17 Match List I to List II concerning the mechanics of a single scattering interaction.

List I	List II
1. Intense localized electric field	a. Sufficiently strong to scatter passing particles through large angles
2. Extremely light atomic electrons	b. Do not possess the inertia to appreciably affect passing alpha-particles
3. Heavy gold nucleus (Z = 79)	c. Assumed stationary due to being ~50 times heavier than the projectile
4. Alpha-particle charge profile	d. Two units of positive charge (+2e)

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

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

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

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

💡 Short Explanation

�� Strong electric field causes scattering. �� Electrons are too light. �� Alpha particle carries +2e.

✅ Correct Answer Explanation

1 → a : Strong electric field causes large-angle scattering. 2 → b : Electrons negligibly affect trajectories. 3 → c : Gold nucleus is treated as stationary. 4 → d : Alpha particle has charge +2e. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

Match each physical quantity with its role.

Final Logic:

Only option A provides all correct matches.

🧠 Memory Trick

Field–Electron–Nucleus–Alpha

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

18 Statements fundamentally explaining why atomic electrons are neglected in the formal trajectory computation:
1. They possess an insignificantly small mass compared to the highly massive alpha particles.
2. They lack the momentum required to appreciably deflect the incoming projectile's path.
3. The dominant scattering force is overwhelmingly provided by the massive, concentrated positive nucleus.
4. Their negative charge perfectly neutralises the nuclear charge precisely at the minimum impact parameter.

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

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

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

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

💡 Short Explanation

�� Electrons are very light. �� Their effect is negligible. �� Nucleus dominates scattering.

✅ Correct Answer Explanation

Statements 1, 2 and 3 correctly explain why electrons are ignored in Rutherford's trajectory calculations. Statement 4 is incorrect because nuclear charge is not perfectly neutralized at minimum impact parameter. 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 dominant contributor to scattering.

Final Logic:

Nucleus dominates; electrons are negligible.

🧠 Memory Trick

Light Electrons, Heavy Nucleus

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

19 In establishing the strict mathematical computation of an alpha-particle's trajectory,

A the entire trajectory formulation relies solely on the transient magnetic field generated by the orbiting electrons

B the impact parameter, b, represents the perpendicular distance of the initial velocity vector from the centre of the target nucleus

C an infinitely large impact parameter inevitably corresponds to a complete head-on collision

D statistical mechanics dictates that all incident particles in the beam inherently share the exact same impact parameter

💡 Short Explanation

�� Impact parameter determines scattering. �� It is measured perpendicular to motion. �� Larger b gives smaller deflection.

✅ Correct Answer Explanation

The impact parameter b is defined as the perpendicular distance between the initial velocity direction of the alpha particle and the centre of the target nucleus. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Rutherford's analysis is based on Coulomb force, not electron-generated magnetic fields.
�� Option C → Head-on collision corresponds to minimum impact parameter.
�� Option D → Different particles have different impact parameters.

🎯 Strategy

Used

Definition Recall

Application:

Use the standard definition of impact parameter.

Final Logic:

Impact parameter is the perpendicular offset from the nuclear centre.

🧠 Memory Trick

b = Side Offset

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Impact Parameter

20 In evaluating the closest approach distance during a theoretical head-on collision,

A the initial kinetic energy of the alpha particle is considered to be entirely converted into electrostatic potential energy at the exact stopping point

B the alpha particle successfully penetrates the target nucleus due to its overwhelming 7.7 MeV kinetic energy

C the potential energy strictly becomes zero at the exact moment the particle momentarily stops

D the theoretical stopping distance is exclusively equal to the sum of the physical radii of the alpha-particle and gold nucleus

💡 Short Explanation

�� Energy conservation is used. �� Velocity becomes zero momentarily. �� KE converts into PE.

✅ Correct Answer Explanation

At the distance of closest approach, the alpha particle momentarily comes to rest. Therefore, all its initial kinetic energy is converted into electrostatic potential energy. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Alpha particles do not penetrate the nucleus in this analysis.
�� Option C → Potential energy is maximum, not zero.
�� Option D → Closest approach is determined by energy conservation, not physical contact.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply conservation of energy during head-on scattering.

Final Logic:

At closest approach, KE = PE.

🧠 Memory Trick

Closest Approach → KE = PE

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Distance of Closest Approach and Energy Conservation

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