CUET UG Physics Booster Test 2-Nuclear Dimensions and Density -

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CUET UG Physics Booster Test 2-Nuclear Dimensions and Density

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-Nuclear Dimensions and Density

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QUESTION 1 OF 20

Incorrect statement about the scattering of alpha particles:

Answer Review

1 Incorrect statement about the scattering of alpha particles:

A At the distance of closest approach, the scattering is governed by Coulomb repulsion.

B Deviations from Rutherford's calculations occur when short-range nuclear forces begin to affect the scattering.

C The actual size of the nucleus must be greater than the distance of closest approach.

D By performing experiments with fast electrons, the sizes of nuclei have been accurately measured.

💡 Short Explanation

�� Closest approach is larger than nuclear radius. �� Coulomb repulsion governs Rutherford scattering. �� Fast electrons measure nuclear dimensions.

✅ Correct Answer Explanation

The distance of closest approach represents the minimum separation reached by an alpha particle before it turns back due to Coulomb repulsion. The actual nuclear radius is much smaller than this distance. Statements A, B and D are correct. Statement C is incorrect because the nuclear size must be smaller than the distance of closest approach. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of Rutherford scattering.
�� Option B → Correct; nuclear forces become important at very small distances.
�� Option D → Fast electron scattering is used to determine nuclear size.

🎯 Strategy

Used

Elimination

Application:

Compare nuclear radius with distance of closest approach.

Final Logic:

Nuclear radius < Distance of closest approach.

🧠 Memory Trick

Closest Approach > Nuclear Radius

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

2 According to Rutherford's model, if an alpha particle of higher energy than 5.5 MeV is used:

A the distance of closest approach will be smaller, and scattering will eventually be affected by short-range nuclear forces.

B it will definitely be captured by the nucleus forming a new element immediately.

C the distance of closest approach will increase proportionally.

D the Coulomb repulsive force will become entirely attractive.

💡 Short Explanation

�� Higher kinetic energy allows closer approach. �� Coulomb barrier is penetrated further. �� Nuclear forces become significant at short distances.

✅ Correct Answer Explanation

The distance of closest approach is inversely proportional to the kinetic energy of the alpha particle. Increasing the energy decreases the closest approach distance. At sufficiently small separations, short-range nuclear forces begin to influence the scattering process, causing deviations from Rutherford's predictions. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Capture is not guaranteed.
�� Option C → Distance decreases, not increases.
�� Option D → Coulomb force remains repulsive.

🎯 Strategy

Used

Substitution

Application:

Use the inverse relation between closest approach and kinetic energy.

Final Logic:

Higher energy → Smaller closest approach.

🧠 Memory Trick

More Energy → More Penetration

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

3 If the volume of an atom is Vₐ and the volume of its nucleus is Vₙ, their ratio Vₐ/Vₙ based on the radius scaling factor of 10⁴ will be:

A 10⁴

B 10⁸

C 10¹²

D 10¹⁶

💡 Short Explanation

�� Volume depends on cube of radius. �� Radius ratio is 10⁴. �� Volume ratio becomes (10⁴)³.

✅ Correct Answer Explanation

Given: Rₐ/Rₙ = 10⁴ Volume ratio: Vₐ/Vₙ = (Rₐ/Rₙ)³ = (10⁴)³ = 10¹² Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Radius ratio, not volume ratio.
�� Option B → Incorrect power.
�� Option D → Larger than actual value.

🎯 Strategy

Used

Substitution

Application:

Use Volume ∝ Radius³.

Final Logic:

(10⁴)³ = 10¹².

🧠 Memory Trick

Radius³ → Volume

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Atomic and Nuclear Volumes

4 Statements about atomic and nuclear volumes:
1. If an atom's radius is 10⁻¹⁰ m, its nuclear radius is approximately 10⁻¹⁴ m.
2. The nucleus takes up approximately one-trillionth (10⁻¹²) of the atomic volume.
3. The volume of the nucleus is proportional to A.
4. Most of the atom's volume is occupied by densely packed protons.

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

B (2), (3)

C (1), (4)

D (1), (2)

💡 Short Explanation

�� Nuclear radius is about 10⁻¹⁴ m. �� Nuclear volume is extremely small. �� Volume varies directly with A.

✅ Correct Answer Explanation

Statement 1 is correct because atomic radius is about 10⁻¹⁰ m and nuclear radius is about 10⁻¹⁴ m. Statement 2 is correct because nuclear volume is approximately 10⁻¹² times atomic volume. Statement 3 is correct because V ∝ A. Statement 4 is incorrect because most atomic volume is empty space. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement 1.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Omits statement 3.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using nuclear size relations.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

10⁻¹⁰ vs 10⁻¹⁴

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

5 For an alpha particle with initial kinetic energy K, the distance of closest approach is r₀ based on Coulomb repulsion. If the kinetic energy is doubled to 2K, what is the new distance of closest approach?

A 0.5 r₀

B 0.25 r₀

C 2.0 r₀

D 4.0 r₀

💡 Short Explanation

�� Closest approach is inversely proportional to kinetic energy. �� Doubling energy halves the distance. �� Based on energy conservation.

✅ Correct Answer Explanation

At closest approach: K ∝ 1/r Thus: r ∝ 1/K If kinetic energy doubles: r' = r₀/2 = 0.5 r₀ Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Distance becomes one-fourth only if energy quadruples.
�� Option C → Opposite trend.
�� Option D → Incorrect proportionality.

🎯 Strategy

Used

Substitution

Application:

Apply r ∝ 1/K.

Final Logic:

Doubling K halves r.

🧠 Memory Trick

Double K → Half r

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

6 Statements about using fast electrons instead of alpha particles:
1. Fast electrons probe the charge distribution of the nucleus.
2. Fast electrons measure the exact same radius value as alpha particles.
3. Alpha particles sense the nuclear matter.
4. Electron scattering experiments are identical in results to alpha scattering experiments.

A (1), (3)

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

C (2), (4)

D (1), (4)

💡 Short Explanation

�� Electrons probe charge distribution. �� Alpha particles probe matter distribution. �� Results are not exactly identical.

✅ Correct Answer Explanation

Statement 1 is correct because electrons interact electromagnetically and reveal charge distribution. Statement 2 is incorrect because radii obtained may differ slightly. Statement 3 is correct because alpha particles probe nuclear matter. Statement 4 is incorrect because the two methods provide different information. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Differentiate charge-distribution measurements from matter-distribution measurements.

Final Logic:

Electrons → Charge; Alpha → Matter.

🧠 Memory Trick

Electron = Charge, Alpha = Matter

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Electron and Alpha Scattering

7 If a nucleus has a mass number A = 64, its nuclear radius R expressed in terms of the empirical constant R₀ will be:

A 4R₀

B 8R₀

C 16R₀

D 64R₀

💡 Short Explanation

�� Nuclear radius follows A¹ᐟ³. �� 64 = 4³. �� Radius = 4R₀.

✅ Correct Answer Explanation

Using: R = R₀A¹ᐟ³ For A = 64: R = R₀(64)¹ᐟ³ = R₀(4) = 4R₀ Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Uses incorrect cube root.
�� Option C → Too large.
�� Option D → Directly proportional to A, which is wrong.

🎯 Strategy

Used

Substitution

Application:

Take cube root of 64.

Final Logic:

64¹ᐟ³ = 4.

🧠 Memory Trick

64 = 4³

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Radius Formula

8 Statements comparing two nuclei with mass numbers A₁ = 8 and A₂ = 64:
1. The ratio of their radii is 1:2.
2. The ratio of their volumes is 1:8.
3. The ratio of their densities is 1:8.
4. The ratio of their radii is 1:8.

A (1), (2)

B (2), (3)

C (1), (4)

D (1), (3)

💡 Short Explanation

�� Radius ∝ A¹ᐟ³. �� Volume ∝ A. �� Density is constant.

✅ Correct Answer Explanation

Radius ratio: R₁/R₂ = (8/64)¹ᐟ³ = (1/8)¹ᐟ³ = 1/2 Statement 1 is correct. Volume ratio: V₁/V₂ = 8/64 = 1/8 Statement 2 is correct. Statement 3 is incorrect because density is constant. Statement 4 is incorrect because radius ratio is 1:2. Therefore, statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3.
�� Option C → Includes statement 4.
�� Option D → Includes statement 3.

🎯 Strategy

Used

Substitution

Application:

Use R ∝ A¹ᐟ³ and V ∝ A.

Final Logic:

Radius ratio 1:2 and volume ratio 1:8.

🧠 Memory Trick

Cube Root for Radius

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Radius and Volume Relations

9 The relationship between nuclear volume V and mass number A implies that nuclear density is ___, while the volume is ___ proportional to A.

A Constant, Directly

B Variable, Inversely

C Constant, Inversely

D Variable, Directly

💡 Short Explanation

�� Mass ∝ A. �� Volume ∝ A. �� Density remains constant.

✅ Correct Answer Explanation

Since: Mass ∝ A and Volume ∝ A Density = Mass/Volume remains constant. Therefore, volume is directly proportional to A and density is constant. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Both descriptions are incorrect.
�� Option C → Volume is not inversely proportional.
�� Option D → Density is not variable.

🎯 Strategy

Used

Definition Recall

Application:

Use mass-volume proportionality.

Final Logic:

Mass and volume increase together.

🧠 Memory Trick

A/A = Constant

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Density

10 Match List I (Nuclear Properties) with List II (Implications)

List I	List II
1. Constant density	a. Implies nuclei are like a drop of liquid
2. Volume proportional to A	b. Adding nucleons increases the total volume linearly
3. Mass mostly in nucleus	c. Over 99.9% of atomic mass is highly concentrated
4. Radius proportional to A¹ᐟ³	d. Derived from fast electron scattering measurements

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

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

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

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

💡 Short Explanation

�� Constant density supports liquid-drop model. �� Volume increases with nucleon number. �� Mass is concentrated in the nucleus.

✅ Correct Answer Explanation

1 → a : Constant density supports the liquid-drop model. 2 → b : Volume proportional to A means adding nucleons increases volume linearly. 3 → c : More than 99.9% of atomic mass is concentrated in the nucleus. 4 → d : Radius relation is obtained from scattering measurements, especially electron scattering. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Multiple mismatches.
�� Option C → Incorrect assignments.
�� Option D → Volume and mass implications are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match each nuclear property with its physical implication.

Final Logic:

Each property has a unique consequence supported by nuclear models.

🧠 Memory Trick

Density–Drop, Volume–A, Mass–Nucleus, Radius–Scattering

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Properties and Liquid Drop Model

11 If a nucleus has mass number A = 125 and R₀ = 1.2 fm, what is the order of magnitude of its nuclear density?

A 10¹⁵ kg m⁻³

B 10¹⁷ kg m⁻³

C 10¹⁹ kg m⁻³

D 10²¹ kg m⁻³

💡 Short Explanation

�� Nuclear density is nearly constant. �� Density is independent of A. �� Typical value is about 2.3 × 10¹⁷ kg m⁻³.

✅ Correct Answer Explanation

Using: R = R₀A¹ᐟ³ Volume ∝ A and nuclear mass ∝ A. Therefore, density remains approximately constant for all nuclei irrespective of mass number. The typical nuclear density is of the order of: 10¹⁷ kg m⁻³ Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option B → Too small by about two orders of magnitude.
�� Option C → Larger than the accepted nuclear density.
�� Option D → Far greater than the actual value.

🎯 Strategy

Used

Definition Recall

Application:

Recall the standard order of nuclear density.

Final Logic:

Nuclear density ≈ 10¹⁷ kg m⁻³ for all nuclei.

🧠 Memory Trick

Nuclear Density → 10¹⁷

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Density

12 Compared to ordinary matter such as water, nuclear matter:

A is extremely large in density, approximately 10¹⁴ times greater.

B has exactly the same density, behaving as a liquid drop.

C has a density 10¹⁴ times smaller due to empty space.

D varies drastically depending on the specific element being measured.

💡 Short Explanation

�� Water density ≈ 10³ kg m⁻³. �� Nuclear density ≈ 10¹⁷ kg m⁻³. �� Nuclear matter is enormously denser.

✅ Correct Answer Explanation

The density ratio is: (2.3 × 10¹⁷)/(10³) = 2.3 × 10¹⁴ Thus nuclear matter is approximately 10¹⁴ times denser than water. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Densities are vastly different.
�� Option C → Opposite of the actual relation.
�� Option D → Nuclear density is nearly constant.

🎯 Strategy

Used

Substitution

Application:

Compare nuclear density with water density.

Final Logic:

Nuclear matter is about 10¹⁴ times denser than water.

🧠 Memory Trick

17 − 3 = 14

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Density Comparison

13 Let mₙ be the mass of the nucleus and mₐ be the total mass of the atom. The value of the percentage (mₙ/mₐ) × 100 will be:

A > 99.9

B < 0.1

C 50.0

D 90.0

💡 Short Explanation

�� Most atomic mass is in the nucleus. �� Electron mass contribution is negligible. �� Nucleus contains nearly all the mass.

✅ Correct Answer Explanation

The masses of electrons are very small compared with the masses of protons and neutrons. Hence, more than 99.9% of the total atomic mass is concentrated in the nucleus. Therefore: (mₙ/mₐ) × 100 > 99.9 Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite of reality.
�� Option C → Nuclear mass is much greater than 50%.
�� Option D → Still significantly lower than the actual value.

🎯 Strategy

Used

Definition Recall

Application:

Recall the distribution of mass within the atom.

Final Logic:

Nearly all atomic mass resides in the nucleus.

🧠 Memory Trick

Mass Lives in Nucleus

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Distribution in Atom

14 Statements regarding empty space and scale:
1. The volume of the atom is roughly 10¹² times the volume of the nucleus.
2. If a nucleus is scaled to a pinhead, the atom scales to a classroom.
3. The density of the empty space region is roughly equal to the nuclear density.
4. Alpha particles scatter entirely due to interaction with the empty space.

A (1), (2)

B (1), (3)

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

D (1), (4)

💡 Short Explanation

�� Atoms are mostly empty space. �� Nuclear volume is tiny. �� Pinhead-classroom analogy illustrates scale.

✅ Correct Answer Explanation

Statement 1 is correct because the atomic volume is approximately 10¹² times the nuclear volume. Statement 2 is correct because this analogy accurately illustrates the huge size difference. Statement 3 is incorrect because empty space contains almost no mass. Statement 4 is incorrect because scattering occurs mainly due to the nucleus. Therefore, statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Statements 3 and 4 are incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using Rutherford's model.

Final Logic:

Only statements 1 and 2 are correct.

🧠 Memory Trick

Pinhead Nucleus, Classroom Atom

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Atomic Scale and Empty Space

15 Statements regarding the mathematical derivation of constant nuclear density:
1. Volume V is proportional to R³.
2. Volume V = (4/3)πR₀³A.
3. Density ρ cancels out the dependence on A.
4. Mass M varies inversely with A.

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

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

C (1), (4)

D (1), (2)

💡 Short Explanation

�� Radius follows A¹ᐟ³. �� Volume becomes proportional to A. �� Density becomes independent of A.

✅ Correct Answer Explanation

Statement 1 is correct because: V = (4/3)πR³ Statement 2 is correct because: R = R₀A¹ᐟ³ gives V = (4/3)πR₀³A Statement 3 is correct because both mass and volume are proportional to A, causing A to cancel in density calculations. Statement 4 is incorrect because nuclear mass is directly proportional to A. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Statement 4 is incorrect.
�� Option D → Omits statement 3.

🎯 Strategy

Used

Elimination

Application:

Follow the derivation of nuclear density.

Final Logic:

Mass ∝ A and Volume ∝ A ⇒ Constant Density.

🧠 Memory Trick

A Cancels Out

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Derivation of Nuclear Density

16 Given the mass of an iron nucleus (A = 56) is 55.85 u, and 1 u = 1.6605 × 10⁻²⁷ kg, what is its approximate mass in kg?

A 9.27 × 10⁻²⁶ kg

B 1.66 × 10⁻²⁷ kg

C 5.58 × 10⁻²⁵ kg

D 2.30 × 10¹⁷ kg

💡 Short Explanation

�� Convert u into kg. �� Multiply by 1.6605 × 10⁻²⁷. �� Obtain nuclear mass.

✅ Correct Answer Explanation

Mass = 55.85 × 1.6605 × 10⁻²⁷ ≈ 9.27 × 10⁻²⁶ kg Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Value of 1 u only.
�� Option C → About six times larger.
�� Option D → Represents nuclear density scale.

🎯 Strategy

Used

Substitution

Application:

Convert atomic mass units into kilograms.

Final Logic:

55.85 × 1.6605 × 10⁻²⁷ = 9.27 × 10⁻²⁶ kg.

🧠 Memory Trick

56 u ≈ 9 × 10⁻²⁶ kg

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Mass Calculation

17 Incorrect statement about neutron stars:

A They resemble a big nucleus in terms of matter compression.

B They consist of highly compressed matter.

C Their density is comparable to nuclear matter density.

D Their matter density is comparable to ordinary water.

💡 Short Explanation

�� Neutron stars are extremely dense. �� Density approaches nuclear density. �� They resemble giant nuclei.

✅ Correct Answer Explanation

Neutron stars possess densities comparable to nuclear matter (~10¹⁷ kg m⁻³), vastly greater than the density of water. Therefore, statement D is incorrect. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description.
�� Option B → Correct statement.
�� Option C → Correct statement.

🎯 Strategy

Used

Elimination

Application:

Compare neutron-star density with water density.

Final Logic:

Neutron-star density is enormously larger than water density.

🧠 Memory Trick

Neutron Star ≠ Water

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Neutron Stars

18 Statements about compressed astrophysical matter:
1. Matter in neutron stars is compressed to nuclear densities.
2. The matter in these objects retains the large empty spaces found in regular atoms.
3. It provides macroscopic evidence of nuclear density scales around 10¹⁷ kg m⁻³.
4. Their density is highly variable depending on mass number A.

A (1), (3)

B (2), (3)

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

D (3), (4)

💡 Short Explanation

�� Neutron stars contain compressed nuclear matter. �� Their density is about 10¹⁷ kg m⁻³. �� Empty atomic spaces disappear.

✅ Correct Answer Explanation

Statement 1 is correct because neutron stars reach nuclear densities. Statement 2 is incorrect because matter is highly compressed. Statement 3 is correct because observed neutron stars demonstrate nuclear-density matter on a macroscopic scale. Statement 4 is incorrect because nuclear density is nearly constant. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2.
�� Option C → Includes statements 2 and 4.
�� Option D → Includes statement 4.

🎯 Strategy

Used

Elimination

Application:

Compare neutron-star matter with ordinary atomic matter.

Final Logic:

Only statements 1 and 3 are correct.

🧠 Memory Trick

Neutron Star = Giant Nucleus

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Astrophysical Significance of Nuclear Density

19 In scattering experiments, deviations from Rutherford's formula occur because higher energy alpha particles overcome ___, while electrons are specifically useful to provide information on ___.

A Electron shells, Nuclear mass

B Nuclear forces, Nuclear mass

C Coulomb repulsion, Charge distribution

D Gravitational forces, Charge distribution

💡 Short Explanation

�� High-energy alpha particles approach more closely. �� Coulomb barrier is penetrated further. �� Electron scattering reveals charge distribution.

✅ Correct Answer Explanation

At higher energies, alpha particles can approach sufficiently close to the nucleus that deviations from pure Rutherford scattering appear after overcoming much of the Coulomb repulsion. Electron scattering experiments are used to determine nuclear charge distribution. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron shells are not responsible for the deviations.
�� Option B → Electrons do not directly measure nuclear mass.
�� Option D → Gravitational effects are negligible.

🎯 Strategy

Used

Definition Recall

Application:

Recall the roles of alpha and electron scattering.

Final Logic:

Alpha → Coulomb barrier; Electron → Charge distribution.

🧠 Memory Trick

Alpha–Barrier, Electron–Charge

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Scattering Experiments

20 The slight difference in radii determined by electron and alpha scattering is due to the fact that:

A electrons have a much larger mass than alpha particles.

B alpha particles only interact with atomic electrons.

C electron scattering senses the charge distribution whereas alpha particles sense the nuclear matter.

D electron scattering is completely unaffected by Coulomb forces.

💡 Short Explanation

�� Different probes measure different distributions. �� Electrons map charge distribution. �� Alpha particles probe matter distribution.

✅ Correct Answer Explanation

Electron scattering is sensitive to the nuclear charge distribution, while alpha-particle scattering is more sensitive to the distribution of nuclear matter. Since charge and matter distributions are not exactly identical, slightly different radii may be obtained. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electrons are much lighter than alpha particles.
�� Option B → Alpha particles interact primarily with nuclei.
�� Option D → Electron scattering is electromagnetic in nature.

🎯 Strategy

Used

Definition Recall

Application:

Identify what each scattering probe measures.

Final Logic:

Different distributions produce slightly different measured radii.

🧠 Memory Trick

Electron → Charge, Alpha → Matter

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Charge and Matter Distribution in Nuclei

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