CUET UG Physics Booster Test 3-Nuclear Dimensions and Density -

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CUET UG Physics Booster Test 3-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 3-Nuclear Dimensions and Density

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QUESTION 1 OF 20

Statements regarding the impact of higher energy alpha particles in Rutherford scattering:
1. A higher energy alpha particle will reach a smaller distance of closest approach than a 5.5 MeV one.
2. At energies high enough to overcome the distance threshold, scattering will deviate from Rutherford's prediction.
3. Deviations occur because the short-range nuclear force begins to affect scattering.
4. Rutherford's calculations are based entirely on pure Coulomb repulsion.

Answer Review

1 Statements regarding the impact of higher energy alpha particles in Rutherford scattering:
1. A higher energy alpha particle will reach a smaller distance of closest approach than a 5.5 MeV one.
2. At energies high enough to overcome the distance threshold, scattering will deviate from Rutherford's prediction.
3. Deviations occur because the short-range nuclear force begins to affect scattering.
4. Rutherford's calculations are based entirely on pure Coulomb repulsion.

A (1), (2), (3), (4)

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

C (2), (3)

D (1), (4)

💡 Short Explanation

�� Higher energy gives smaller closest approach. �� Nuclear force becomes important at short distances. �� Rutherford theory assumes only Coulomb interaction.

✅ Correct Answer Explanation

Statement 1 is correct because increasing alpha-particle energy decreases the distance of closest approach. Statement 2 is correct because sufficiently energetic alpha particles can approach the nucleus closely enough for deviations from Rutherford scattering to appear. Statement 3 is correct because short-range nuclear forces begin to influence scattering at very small separations. Statement 4 is correct because Rutherford's theory considers only Coulomb repulsion between charges. Therefore, statements 1, 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement 3, which is correct.
�� Option C → Omits statements 1 and 4.
�� Option D → Omits statements 2 and 3.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using Rutherford scattering theory.

Final Logic:

All four statements are correct.

🧠 Memory Trick

High Energy → Nuclear Force Appears

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

2 Statements evaluating Rutherford's model boundaries:
1. It successfully calculates scattering assuming pure Coulomb repulsion between positive charges.
2. It correctly models the internal strong nuclear forces of the gold nucleus.
3. It infers nuclear size from the distance at which deviations set in.
4. It inherently measures the exact boundary of nuclear matter at any energy.

A (1), (3)

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

C (2), (4)

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

💡 Short Explanation

�� Rutherford theory uses Coulomb force. �� Nuclear size is inferred indirectly. �� Strong nuclear force is outside the model.

✅ Correct Answer Explanation

Statement 1 is correct because Rutherford's model is based on Coulomb interaction. Statement 2 is incorrect because the strong nuclear force was not included in Rutherford's analysis. Statement 3 is correct because nuclear dimensions can be estimated from deviations from Rutherford scattering. Statement 4 is incorrect because Rutherford scattering does not directly determine the exact nuclear boundary. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Separate Rutherford-model assumptions from later nuclear-force concepts.

Final Logic:

Only statements 1 and 3 are correct.

🧠 Memory Trick

Rutherford = Coulomb Only

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

3 If the atomic radius scales as rₐ and nuclear radius as rₙ, and we define a scaling parameter S = rₐ/rₙ ≈ 10⁴, the ratio of their cross-sectional areas will be:

A S³

B S²

C S

D S⁻²

💡 Short Explanation

�� Area depends on radius squared. �� Radius ratio is S. �� Area ratio becomes S².

✅ Correct Answer Explanation

Cross-sectional area is proportional to r². Therefore, Area Ratio = (rₐ/rₙ)² = S² Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to volume ratio.
�� Option C → Radius ratio only.
�� Option D → Inverse relation.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use the geometric relation Area ∝ Radius².

Final Logic:

Area ratio = S².

🧠 Memory Trick

Radius² → Area

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

4 If a nucleus (R = 1.2 × 10⁻¹⁵ m) is scaled up to a sphere of radius 1.2 cm, what would be the corresponding scaled-up radius of the atom (assuming the 10⁴ scaling factor)?

A 120 m

B 12 m

C 1.2 km

D 12000 m

💡 Short Explanation

�� Atomic radius is about 10⁴ times nuclear radius. �� Scale factor remains unchanged. �� Multiply scaled nucleus radius by 10⁴.

✅ Correct Answer Explanation

Scaled nuclear radius = 1.2 cm Scaled atomic radius = 10⁴ × 1.2 cm = 12000 cm = 120 m Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Too small by factor 10.
�� Option C → Too large.
�� Option D → Incorrect conversion to metres.

🎯 Strategy

Used

Substitution

Application:

Apply the atomic-to-nuclear radius ratio.

Final Logic:

1.2 cm × 10⁴ = 120 m.

🧠 Memory Trick

1.2 cm → 120 m

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

5 Incorrect statement regarding the distance of closest approach (d):

A It provides an upper limit to the actual size of the nucleus.

B It is calculated by equating initial kinetic energy to the electrostatic potential energy at distance d.

C It gives the precise and exact value of the nuclear radius for all possible particle energies.

D Using alpha particles of higher energies than 5.5 MeV yields a smaller distance of closest approach.

💡 Short Explanation

�� Closest approach is not nuclear radius. �� It provides only an upper limit. �� Higher energy gives smaller distances.

✅ Correct Answer Explanation

The distance of closest approach is obtained by equating kinetic energy to Coulomb potential energy. It provides an estimate and an upper limit for nuclear size but not the exact nuclear radius. Therefore, statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct calculation method.
�� Option D → Correct consequence of higher energy.

🎯 Strategy

Used

Elimination

Application:

Distinguish between closest approach and actual nuclear radius.

Final Logic:

Closest approach ≠ Exact nuclear radius.

🧠 Memory Trick

Upper Limit, Not Exact Limit

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

6 When comparing alpha particles and fast electrons as probes for nuclear size:

A both interact identically with the strong nuclear force.

B fast electrons are subject to the strong nuclear force, causing major deviations.

C electron scattering maps the spatial distribution of charge, while alpha particles sense nuclear matter.

D alpha particles cannot penetrate the electron cloud at all.

💡 Short Explanation

�� Different probes reveal different properties. �� Electrons probe charge distribution. �� Alpha particles probe matter distribution.

✅ Correct Answer Explanation

Fast electrons interact electromagnetically and map nuclear charge distribution. Alpha particles are used to probe nuclear matter distribution. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Interactions are not identical.
�� Option B → Electrons are not governed by strong nuclear force in these experiments.
�� Option D → Alpha particles easily penetrate electron clouds.

🎯 Strategy

Used

Definition Recall

Application:

Recall the roles of scattering probes.

Final Logic:

Electron → Charge, Alpha → Matter.

🧠 Memory Trick

Electron-Charge, Alpha-Matter

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

7 Match List I (Isotopes) with List II (Radius relative to R₀)

List I	List II
1. ⁸Be	a. 3R₀
2. ²⁷Al	b. 5R₀
3. ⁶⁴Cu	c. 2R₀
4. ¹²⁵Te	d. 4R₀

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

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

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

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

💡 Short Explanation

�� Radius ∝ A¹ᐟ³. �� 8 = 2³. �� 27 = 3³. �� 64 = 4³ and 125 = 5³.

✅ Correct Answer Explanation

1 → c : ⁸Be → R = 2R₀ 2 → a : ²⁷Al → R = 3R₀ 3 → d : ⁶⁴Cu → R = 4R₀ 4 → b : ¹²⁵Te → R = 5R₀ Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Multiple incorrect radius assignments.
�� Option C → Incorrect cube-root matching.
�� Option D → Several mismatches.

🎯 Strategy

Used

Option Grouping

Application:

Use cube roots of mass numbers.

Final Logic:

R = R₀A¹ᐟ³.

🧠 Memory Trick

8-2, 27-3, 64-4, 125-5

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

8 Statements about R = R₀A¹ᐟ³:
1. It implies the volume of a nucleus is linearly proportional to the mass number A.
2. It demonstrates mathematically that density is independent of A.
3. It proves that nucleons are highly compressible.
4. It is a formula fitted from electron scattering experiments.

A (1), (2), (4)

B (1), (3)

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

D (1), (2)

💡 Short Explanation

�� Volume ∝ A. �� Density becomes constant. �� Formula is experimentally supported.

✅ Correct Answer Explanation

Statement 1 is correct because V ∝ R³ ∝ A. Statement 2 is correct because mass and volume are both proportional to A. Statement 3 is incorrect because nuclei exhibit near-constant density rather than high compressibility. Statement 4 is correct because the radius relation is supported by scattering measurements. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Apply consequences of R = R₀A¹ᐟ³.

Final Logic:

Volume ∝ A and density is constant.

🧠 Memory Trick

Cube Root → Constant Density

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

9 Since the total nuclear mass is approximately proportional to A and the volume is proportional to A, the ratio of mass to volume is ___, indicating that nuclei are like a drop of liquid of ___.

A Constant, Constant density

B Variable, Compressible matter

C Constant, Variable density

D Variable, Constant density

💡 Short Explanation

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

✅ Correct Answer Explanation

Since both mass and volume vary directly with A: Density = Mass/Volume remains constant. This supports the liquid-drop model of the nucleus. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Density is not variable.
�� Option C → Density is not variable.
�� Option D → Mass-to-volume ratio is constant.

🎯 Strategy

Used

Definition Recall

Application:

Use proportionality of mass and volume.

Final Logic:

A/A = Constant Density.

🧠 Memory Trick

Mass A, Volume A

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

10 In the liquid drop model, if two identical nuclei of mass number A perfectly fuse to form a nucleus of mass number 2A, the radius of the new nucleus Rnew in terms of the original radius R will be:

A 2R

B 2¹ᐟ³R

C 2²ᐟ³R

D 4R

💡 Short Explanation

�� Radius ∝ A¹ᐟ³. �� New mass number = 2A. �� Apply radius relation.

✅ Correct Answer Explanation

Original radius: R = R₀A¹ᐟ³ New radius: Rnew = R₀(2A)¹ᐟ³ = 2¹ᐟ³R₀A¹ᐟ³ = 2¹ᐟ³R Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Assumes direct proportionality.
�� Option C → Incorrect power.
�� Option D → Far larger than actual value.

🎯 Strategy

Used

Substitution

Application:

Apply R ∝ A¹ᐟ³ to the fused nucleus.

Final Logic:

Doubling A multiplies radius by 2¹ᐟ³.

🧠 Memory Trick

Double A → Cube Root 2

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

11 Given the density of nuclear matter is ρ ≈ 2.3 × 10¹⁷ kg m⁻³, what is the approximate mass of 1 mm³ (10⁻⁹ m³) of nuclear matter?

A 2.3 × 10⁸ kg

B 2.3 × 10²⁶ kg

C 2.3 × 10¹⁷ kg

D 2.3 × 10⁻⁹ kg

💡 Short Explanation

�� Mass = Density × Volume. �� Nuclear density is extremely high. �� Even a tiny volume has enormous mass.

✅ Correct Answer Explanation

Using: Mass = Density × Volume = (2.3 × 10¹⁷) × (10⁻⁹) = 2.3 × 10⁸ kg Thus, 1 mm³ of nuclear matter would have a mass of approximately 2.3 × 10⁸ kg. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect multiplication of powers.
�� Option C → Ignores volume factor.
�� Option D → Opposite power of ten.

🎯 Strategy

Used

Substitution

Application:

Substitute the given density and volume into M = ρV.

Final Logic:

2.3 × 10¹⁷ × 10⁻⁹ = 2.3 × 10⁸ kg.

🧠 Memory Trick

ρV = Mass

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

12 Statements evaluating densities:
1. A highly concentrated sample of nuclear matter would be incredibly massive compared to ordinary matter of the same volume.
2. Nuclear density is approximately 10¹⁴ times that of water.
3. The mass density of the atom does not follow the independent density rule of the nucleus.
4. The difference in density between a nucleus and an atom is due to the atomic void.

A (1), (2), (3), (4)

B (2), (3)

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

D (1), (2)

💡 Short Explanation

�� Nuclear matter is extremely dense. �� Atoms contain large empty regions. �� Nuclear density remains nearly constant.

✅ Correct Answer Explanation

Statement 1 is correct because nuclear matter is enormously denser than ordinary matter. Statement 2 is correct because nuclear density is approximately 10¹⁴ times greater than water density. Statement 3 is correct because atomic density depends strongly on atomic size and empty space, unlike nuclear density. Statement 4 is correct because most atomic volume consists of empty space, causing atomic density to be much lower than nuclear density. Therefore, statements 1, 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statements 1 and 4.
�� Option C → Omits statement 2.
�� Option D → Omits statements 3 and 4.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using the concepts of atomic and nuclear density.

Final Logic:

All four statements are correct.

🧠 Memory Trick

Nucleus Dense, Atom Empty

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

13 Incorrect statement about the mass distribution in atoms:

A The mass of electrons contributes significantly (around 10%) to the total atomic mass.

B The nucleus contains more than 99.9% of the atomic mass.

C Protons and neutrons are densely concentrated at the centre of the atom.

D The overall dimensions of a nucleus are much smaller than those of an atom.

💡 Short Explanation

�� Electron mass is negligible. �� Nucleus contains almost all atomic mass. �� Protons and neutrons form the nucleus.

✅ Correct Answer Explanation

Electrons are approximately 1836 times lighter than protons. Their total contribution to atomic mass is very small and nowhere near 10%. Therefore, statement A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct statement.
�� Option C → Correct description of nuclear structure.
�� Option D → Correct comparison of atomic and nuclear dimensions.

🎯 Strategy

Used

Elimination

Application:

Compare electron mass with nucleon masses.

Final Logic:

Atomic mass is dominated by the nucleus.

🧠 Memory Trick

Electrons Light, Nucleus Heavy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Concentration in Atoms

14 If the volume ratio of the nucleus to the atom is 10⁻¹²,

A the probability of an alpha particle interacting directly with a nucleus is extremely high.

B it indicates that electrons completely fill and have mass equal to the rest of the atomic volume.

C it clarifies why the vast majority of atomic volume is practically empty.

D it indicates that protons are loosely bound.

💡 Short Explanation

�� Nuclear volume is extremely small. �� Most atomic volume contains no concentrated matter. �� Rutherford scattering supports this conclusion.

✅ Correct Answer Explanation

Since the nucleus occupies only about 10⁻¹² of the atomic volume, almost all of the atom consists of empty space. This explains why most alpha particles pass through thin foils without significant deflection. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Direct nuclear interaction is relatively rare.
�� Option B → Electrons do not fill atomic volume with comparable mass.
�� Option D → Volume ratio gives no information about binding strength.

🎯 Strategy

Used

Definition Recall

Application:

Interpret the meaning of the nuclear-to-atomic volume ratio.

Final Logic:

Tiny nucleus ⇒ Vast atomic empty space.

🧠 Memory Trick

10⁻¹² Volume → Empty Atom

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

15 Statements regarding the constancy of nuclear density across different elements:
1. The nuclear volume increases proportionally with mass number A.
2. Different nuclei are like a drop of liquid of constant density.
3. A Uranium nucleus has a vastly higher density than a Carbon nucleus.
4. Density is heavily dependent on the atomic number Z.

A (1), (2)

B (2), (3)

C (1), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� Volume ∝ A. �� Density remains nearly constant. �� Liquid-drop model assumes constant density.

✅ Correct Answer Explanation

Statement 1 is correct because nuclear volume is proportional to A. Statement 2 is correct because nuclei behave like drops of incompressible liquid with nearly constant density. Statement 3 is incorrect because Uranium and Carbon nuclei have approximately the same density. Statement 4 is incorrect because density is nearly independent of Z. Therefore, statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Apply the concept of constant nuclear density.

Final Logic:

Density is nearly independent of nuclear size.

🧠 Memory Trick

All Nuclei, Same Density

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

16 Match List I (Steps to mathematically find nuclear density) with List II (Formulas/Values)

List I	List II
1. Compute Mass	a. Multiply atomic mass by 1.6605 × 10⁻²⁷
2. Compute Volume	b. Divide Mass by Volume
3. Convert units (u to kg)	c. Use the value A in u (e.g., 55.85 u for Fe)
4. Final Density Computation	d. Use (4/3)πR₀³A

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

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

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

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

💡 Short Explanation

�� Determine mass in u. �� Convert mass into kg. �� Compute volume and then density.

✅ Correct Answer Explanation

1 → c : Begin with the atomic mass value in u. 2 → d : Use V = (4/3)πR₀³A. 3 → a : Convert u into kg using 1.6605 × 10⁻²⁷. 4 → b : Density = Mass ÷ Volume. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect sequence of calculations.
�� Option C → Mass and volume are mismatched.
�� Option D → Incorrect density assignment.

🎯 Strategy

Used

Option Grouping

Application:

Match each calculation step with the corresponding formula.

Final Logic:

Mass → Conversion → Volume → Density.

🧠 Memory Trick

Mass → kg → Volume → Density

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

17 In extreme astrophysical environments like neutron stars, matter is heavily ___, structurally resembling a giant ___.

A Compressed, Nucleus

B Expanded, Atom

C Heated, Molecule

D Ionized, Proton

💡 Short Explanation

�� Neutron stars contain compressed matter. �� Their density approaches nuclear density. �� They resemble enormous nuclei.

✅ Correct Answer Explanation

Matter in neutron stars is compressed to extraordinary densities comparable to nuclear matter density. Hence, neutron stars can be viewed as giant nuclei. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Neutron stars are not expanded.
�� Option C → Heating alone does not explain their structure.
�� Option D → A proton is not the correct comparison.

🎯 Strategy

Used

Definition Recall

Application:

Recall the physical structure of neutron stars.

Final Logic:

Compressed matter → Giant nucleus.

🧠 Memory Trick

Neutron Star = Giant Nucleus

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

18 Statements regarding the astrophysical link to nuclear density:
1. Neutron stars have densities comparable to 2.3 × 10¹⁷ kg m⁻³.
2. They possess a density equal to that of a normal hydrogen atom.
3. Their density is equivalent to ordinary water.
4. They demonstrate that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

A (1), (2)

B (2), (4)

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

D (1), (4)

💡 Short Explanation

�� Neutron-star density is enormous. �� Comparable to nuclear density. �� Matter behaves like giant nuclear matter.

✅ Correct Answer Explanation

Statement 1 is correct because neutron-star density is of the order of 10¹⁷ kg m⁻³. Statement 2 is incorrect because hydrogen atoms are mostly empty space. Statement 3 is incorrect because water is far less dense. Statement 4 is correct because matter is compressed to nuclear-density scales. Therefore, statements 1 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2.
�� Option B → Includes statement 2.
�� Option C → Includes statements 3 and 2.

🎯 Strategy

Used

Elimination

Application:

Compare neutron-star density with atomic and ordinary matter densities.

Final Logic:

Only statements 1 and 4 are correct.

🧠 Memory Trick

Star Density = Nuclear Density

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

19 The fact that electron scattering and alpha scattering yield slightly different nuclear radii is rooted in:

A electrons being scattered by both protons and neutrons equally via the strong force.

B alpha particles sensing the exact mass of the atomic electrons.

C alpha particles sensing the nuclear matter, while electrons sense the charge distribution of the nucleus.

D electrons having a positive charge at high relativistic energies.

💡 Short Explanation

�� Different probes measure different distributions. �� Electron scattering measures charge distribution. �� Alpha scattering probes matter distribution.

✅ Correct Answer Explanation

Electron scattering experiments reveal the charge distribution inside the nucleus. Alpha-particle scattering is more closely related to the distribution of nuclear matter. Because these distributions are not exactly identical, the measured radii can differ slightly. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electrons do not probe nuclei through the strong force.
�� Option B → Alpha scattering is not used to measure electron masses.
�� Option D → Electrons remain negatively charged.

🎯 Strategy

Used

Definition Recall

Application:

Identify what each scattering probe measures.

Final Logic:

Charge radius and matter radius are slightly different.

🧠 Memory Trick

Electron = Charge, Alpha = Matter

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

20 If R₍c₎ represents the radius determined by electron scattering and R₍m₎ represents the radius determined by alpha scattering:

A R₍c₎ and R₍m₎ are found to be slightly different.

B R₍c₎ and R₍m₎ are exactly identical for all nuclei.

C R₍m₎ is entirely determined by atomic electrons.

D R₍c₎ only measures the presence of neutrons.

💡 Short Explanation

�� Different scattering methods probe different distributions. �� Charge radius differs slightly from matter radius. �� Experimental values are not exactly identical.

✅ Correct Answer Explanation

Electron scattering determines the charge radius of the nucleus, while alpha scattering is related to the matter radius. Since charge and matter distributions differ slightly, the two measured radii are usually not exactly the same. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Experimental measurements show small differences.
�� Option C → Alpha scattering is related to nuclear matter, not atomic electrons.
�� Option D → Electron scattering probes charge distribution, not only neutrons.

🎯 Strategy

Used

Definition Recall

Application:

Compare the physical quantities measured by each technique.

Final Logic:

Charge radius ≠ Matter radius exactly.

🧠 Memory Trick

Rc ≠ Rm

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

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