CUET UG Physics Booster Test 3-Early Atomic Theory -

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CUET UG Physics Booster Test 3-Early Atomic Theory

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Early Atomic Theory

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QUESTION 1 OF 20

The revelation that atoms of different elements contain identical negatively charged constituents implies that

Answer Review

1 The revelation that atoms of different elements contain identical negatively charged constituents implies that

A the mass of every element is strictly derived from the mass of its electrons

B there must exist a universal compensating positive charge structure, leading to differing atomic masses among elements

C atoms are not indivisible and share a common subatomic architecture

D all elements emit identical spectral lines under electrical discharge

💡 Short Explanation

�� Electrons are present in all atoms. �� Atoms contain subatomic particles. �� Atoms are not indivisible.

✅ Correct Answer Explanation

Thomson's discovery showed that electrons are identical constituents present in atoms of all elements. This proved that atoms are divisible and possess a common internal structure composed of subatomic particles. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because atomic mass is not determined solely by electrons.
�� Option B → Incorrect because Thomson's experiment did not establish a universal positive charge structure responsible for differing masses.
�� Option D → Incorrect because different elements have characteristic spectra.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the fundamental implication of electron discovery.

Final Logic:

Common electrons imply a common subatomic structure.

🧠 Memory Trick

Electron Everywhere → Atom Divisible

📚 Class XII – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Discovery of Electron

2 Incorrect statement concerning atomic neutrality in the context of early atomic hypothesis:

A Neutrality requires a positive charge equal in magnitude to the total electron charge.

B The continuous emission of radiation by an atom disrupts its long-term electrical neutrality.

C The identical nature of electrons across different elements necessitates balancing positive charges to maintain neutrality.

D An atom must contain positive charge to neutralize the negative charge of the electrons.

💡 Short Explanation

�� Neutrality depends on charge balance. �� Radiation emission does not determine neutrality. �� Positive charge balances electron charge.

✅ Correct Answer Explanation

Electrical neutrality depends on equal positive and negative charges within the atom. The emission of radiation does not alter the overall charge balance of an atom. Therefore, the statement that radiation disrupts electrical neutrality is incorrect. Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement about charge neutrality.
�� Option C → Correct because balancing positive charge is necessary.
�� Option D → Correct because positive charge neutralizes electrons.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement unrelated to electrical neutrality.

Final Logic:

Neutrality depends on charge balance, not radiation emission.

🧠 Memory Trick

Neutrality = Charge Balance

📚 Class XII – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Atomic Neutrality

3 Statements analyzing the uniform positive distribution:
1. It successfully explained the overall electrical neutrality of the atom.
2. It accurately predicted large-angle scattering for incident high-speed particles.
3. It assumed mass is continuously spread throughout the atomic volume.
4. It was later refuted by Geiger and Marsden's experimental data.

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

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

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

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

💡 Short Explanation

�� Thomson explained neutrality. �� Mass and charge were assumed continuous. �� Scattering results disproved the model.

✅ Correct Answer Explanation

Statement 1 is correct because Thomson's model explained electrical neutrality. Statement 2 is incorrect because it could not explain large-angle scattering. Statement 3 is correct because positive charge and mass were assumed to be spread throughout the atom. Statement 4 is correct because Geiger-Marsden results contradicted the model. Therefore, statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Includes statement 2, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Compare Thomson's predictions with Rutherford scattering results.

Final Logic:

Only statements 1, 3 and 4 are valid.

🧠 Memory Trick

Neutrality Yes, Scattering No

📚 Class XII – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Thomson's Atomic Model

4 Match List I with List II detailing model analogies and features

List I	List II
1. Watermelon seeds	a. Continuous positive charge
2. Watermelon pulp	b. Atomic vacuum in Rutherford model
3. Planets around sun	c. Embedded electrons
4. Solar system empty space	d. Orbiting electrons

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

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

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

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

💡 Short Explanation

�� Seeds represent electrons. �� Pulp represents positive charge. �� Planets represent orbiting electrons.

✅ Correct Answer Explanation

1 → c : Watermelon seeds represent embedded electrons. 2 → a : Watermelon pulp represents uniformly distributed positive charge. 3 → d : Planets around the sun represent orbiting electrons. 4 → b : Empty solar system space corresponds to atomic empty space. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of seeds and pulp.
�� Option C → Incorrect matching of pulp and planets.
�� Option D → Incorrect matching of empty space and seeds.

🎯 Strategy

Used

Option Grouping

Application:

Match each analogy with its corresponding atomic feature.

Final Logic:

Only option A correctly matches all four pairs.

🧠 Memory Trick

Seeds-Electrons, Pulp-Positive

📚 Class XII – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Thomson and Rutherford Models

5 Scattering discrepancy statements:
1. Uniform mass models could not provide the immense electrostatic force required for large deflections.
2. Alpha particles only rebound when encountering a concentrated massive positive core.
3. Electrons significantly alter the path of incoming alpha particles due to their negative charge.
4. Only about 1 in 8000 particles deflect by more than 90°.

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

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

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

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

💡 Short Explanation

�� Large deflections require concentrated charge. �� Electrons have negligible effect. �� Very few particles scatter backward.

✅ Correct Answer Explanation

Statement 1 is correct because a uniformly distributed charge cannot generate sufficient repulsive force for large deflections. Statement 2 is correct because large-angle scattering occurs near a concentrated positive nucleus. Statement 3 is incorrect because electrons are too light to significantly affect alpha particles. Statement 4 is correct because only a tiny fraction of particles were scattered through angles greater than 90°. Therefore, statements 1, 2 and 4 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 → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using Rutherford scattering observations.

Final Logic:

Only statements 1, 2 and 4 are supported experimentally.

🧠 Memory Trick

Few Backscatter, Big Nucleus

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

6 The expected deflection in Thomson's uniform model vs. the actual observed large-angle scattering in Rutherford's experiment can be described respectively as:

A Minor, Major

B Strong, Weak

C Major, Minor

D Concentrated, Uniform

💡 Short Explanation

�� Thomson predicted small deflections. �� Rutherford observed large deflections. �� Results disproved Thomson's model.

✅ Correct Answer Explanation

If positive charge were spread uniformly, alpha particles would experience only small cumulative deflections. Experimentally, however, some alpha particles underwent very large-angle scattering, indicating a concentrated nucleus. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the observed situation.
�� Option C → Opposite of experimental findings.
�� Option D → Does not compare deflection magnitude.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare theoretical prediction with observation.

Final Logic:

Predicted minor deflection, observed major deflection.

🧠 Memory Trick

Thomson Small, Rutherford Large

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Limitations of Thomson Model

7 Which of the following is correct regarding the nature of continuous wavelengths in condensed matter?

A It arises solely from the isolated discrete energy levels of individual atoms.

B It completely masks characteristic emission line spectra, making elemental identification impossible in solids.

C It is a collective effect governed by the strong interactions of an atom or molecule with its neighbours.

D It is an anomaly caused strictly by alpha-particle bombardment.

💡 Short Explanation

�� Condensed matter contains strong interactions. �� Neighboring particles affect emission. �� Continuous spectra result.

✅ Correct Answer Explanation

In solids and liquids, atoms and molecules strongly interact with neighboring particles. These interactions broaden energy states and produce continuous wavelength distributions rather than isolated spectral lines. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because isolated atoms produce line spectra.
�� Option B → Incorrect because elemental analysis can still be performed using suitable techniques.
�� Option D → Incorrect because alpha-particle bombardment is unrelated.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the physical origin of continuous spectra.

Final Logic:

Neighbor interactions create continuous emission.

🧠 Memory Trick

Neighbours → Continuum

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Continuous Spectra

8 Incorrect statement about radiation from dense gases:

A Their spectrum exhibits a continuous distribution of wavelengths.

B The atoms in a dense gas emit radiation identically to atoms in a rarefied gas glow tube.

C Interatomic interactions are strong enough to affect emission patterns.

D The emitted radiation depends on neighbour interactions rather than strictly isolated atomic properties.

💡 Short Explanation

�� Dense gases show strong interactions. �� Rarefied gases behave as isolated atoms. �� Their spectra differ.

✅ Correct Answer Explanation

Dense gases exhibit significant interatomic interactions that alter emission characteristics and often produce continuous spectra. Rarefied gases, however, emit characteristic line spectra due to isolated atomic behavior. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of dense-gas spectra.
�� Option C → Correct because interactions affect radiation.
�� Option D → Correct because neighbor interactions influence emission.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with dense-gas behavior.

Final Logic:

Dense gases and rarefied gases do not emit identically.

🧠 Memory Trick

Dense ≠ Rarefied

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Continuous and Line Spectra

9 The emission of discrete wavelengths from a rarefied gas provides strong evidence that

A classical electromagnetism perfectly describes accelerating atomic electrons

B atoms possess internal structural mechanisms that restrict emission to specific energy states

C the collisions between gas molecules randomly generate specific light frequencies

D the mass is uniformly spread throughout the atom

💡 Short Explanation

�� Discrete wavelengths indicate restrictions. �� Atomic energy states are specific. �� Spectra reveal internal structure.

✅ Correct Answer Explanation

The presence of discrete spectral lines means atoms can emit only specific amounts of energy. This strongly suggests the existence of restricted internal energy states and quantized atomic structure. Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because classical theory failed to explain line spectra.
�� Option C → Incorrect because spectral lines are not random.
�� Option D → Incorrect because spectral evidence does not imply uniform mass distribution.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect discrete spectra with atomic structure.

Final Logic:

Specific wavelengths imply specific energy states.

🧠 Memory Trick

Discrete Lines → Discrete Levels

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Atomic Spectra

10 Statements about gas excitation in glow tubes:
1. Neon signs operate by electrically exciting atoms in a rarefied state.
2. The spectrum appears as a series of bright discrete lines.
3. The spacing between atoms is small enough to merge their spectral lines continuously.
4. The radiation is considered due to individual atoms.

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

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

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

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

💡 Short Explanation

�� Neon signs contain rarefied gases. �� Line spectra are produced. �� Emission comes from individual atoms.

✅ Correct Answer Explanation

Statement 1 is correct because neon signs operate through electrical excitation of low-pressure gas. Statement 2 is correct because rarefied gases emit bright discrete spectral lines. Statement 3 is incorrect because atoms are widely separated, not closely packed. Statement 4 is correct because the emission is attributed to individual atoms. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the properties of rarefied gases.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Neon → Rarefied → Lines

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Spectra of Rarefied Gases and Glow Tubes

11 Statements about absorption and emission spectra:
1. Absorption spectra contain dark lines on a continuous background.
2. Dark lines correspond precisely to wavelengths found in emission spectra.
3. White light must pass through a gas to observe absorption lines.
4. Absorption spectra actively emit high-intensity bright lines.

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

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

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

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

💡 Short Explanation

�� Absorption spectra show dark lines. �� Dark lines match emission wavelengths. �� A continuous source is required.

✅ Correct Answer Explanation

Statement 1 is correct because absorption spectra appear as dark lines superimposed on a continuous spectrum. Statement 2 is correct because atoms absorb the same wavelengths that they emit. Statement 3 is correct because white light or a continuous spectrum source must pass through a cooler gas to produce absorption lines. Statement 4 is incorrect because absorption spectra are characterized by missing wavelengths, not bright emission lines. 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 the basic characteristics of absorption spectra.

Final Logic:

Only statements 1, 2 and 3 correctly describe absorption spectra.

🧠 Memory Trick

Absorb = Dark, Emit = Bright

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Emission and Absorption Spectra

12 If a hypothetical atomic system consists of a nucleus with y protons and an electron shell containing x electrons, what expression represents the net negative charge excess of the system?

A -(x + y)e

B (x + y)e

C (x - y)e

D (y - x)e

💡 Short Explanation

�� Positive charge = +ye. �� Negative charge = −xe. �� Net charge = (y − x)e.

✅ Correct Answer Explanation

The nucleus contributes a positive charge of +ye, while the electron cloud contributes −xe. Net charge = ye − xe = (y − x)e Therefore, option D is mathematically correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because positive and negative charges are added directly.
�� Option B → Incorrect because both charges are treated as positive.
�� Option C → Incorrect because it gives the opposite sign.

🎯 Strategy

Used

Substitution

Application:

Assign proper signs to proton and electron charges before combining them.

Final Logic:

(+ye) + (−xe) = (y − x)e.

🧠 Memory Trick

Protons Minus Electrons

📚 Class XII – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Net Atomic Charge

13 Which of the following is correct regarding the linkage between atomic structure and spectra?

A Classical mechanics successfully linked Rutherford's planetary model to the discrete hydrogen spectrum.

B The continuous change of angular velocity in classical orbits explains the fixed relative positions of emission lines.

C An accelerating classical electron would emit a continuous spectrum, contradicting the observed discrete characteristic spectrum.

D The planetary model inherently restricts electron orbits to prevent continuous emission.

💡 Short Explanation

�� Classical theory predicts continuous radiation. �� Hydrogen shows discrete spectral lines. �� This created a major theoretical problem.

✅ Correct Answer Explanation

According to classical electrodynamics, an accelerating charge continuously emits electromagnetic radiation. Since an orbiting electron is accelerating, Rutherford's model predicts continuous energy loss and a continuous spectrum. However, hydrogen exhibits discrete spectral lines. This contradiction showed that classical physics could not fully explain atomic structure. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because classical mechanics failed to explain hydrogen spectra.
�� Option B → Incorrect because continuous orbital changes would produce continuous, not discrete, spectra.
�� Option D → Incorrect because Rutherford's model imposed no restrictions on electron orbits.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare classical predictions with experimental observations.

Final Logic:

Classical continuous emission contradicts discrete spectra.

🧠 Memory Trick

Classical → Continuous, Hydrogen → Discrete

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Failure of Rutherford Model

14 Match List I with List II regarding energy states and classical predictions

List I	List II
1. Accelerated classical electron	a. Stationary state
2. Bohr's non-radiating orbit	b. Emits a discrete photon
3. Electron transition to lower state	c. Spirals into nucleus
4. Rarefied gas emission	d. Discrete characteristic spectrum

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

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

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

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

💡 Short Explanation

�� Accelerating electrons radiate energy. �� Bohr introduced stationary states. �� Transitions produce photons.

✅ Correct Answer Explanation

1 → c : A classical accelerating electron loses energy and spirals inward. 2 → a : Bohr's stationary orbit is a non-radiating state. 3 → b : Transition to a lower energy level emits a discrete photon. 4 → d : Rarefied gases produce characteristic line spectra. Thus, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect matching of classical electron and Bohr orbit.
�� Option C → Incorrect matching of transition and stationary state.
�� Option D → Multiple pairings are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Match each concept with its defining consequence.

Final Logic:

Only option B correctly matches all pairs.

🧠 Memory Trick

Spiral–Stationary–Photon–Spectrum

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Model and Spectra

15 While Balmer formulated his empirical equation in 1885, its deeper physical significance presented a challenge because

A he lacked experimental equipment to measure accurate wavelengths

B it only applied to dense matter radiation rather than atomic hydrogen

C a proper theoretical model linking internal atomic structure to these specific discrete wavelengths had not yet been developed

D the lines shifted randomly depending on the isotopes used

💡 Short Explanation

�� Balmer found a mathematical relation. �� Physical interpretation was missing. �� Bohr later provided explanation.

✅ Correct Answer Explanation

Balmer successfully described hydrogen wavelengths mathematically, but no accepted atomic model existed to explain why those wavelengths occurred. Theoretical understanding emerged later through Bohr's atomic model. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because Balmer's formula was based on accurate experimental data.
�� Option B → Incorrect because the formula specifically applies to hydrogen.
�� Option D → Incorrect because random spectral shifting was not the issue.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Separate mathematical success from physical explanation.

Final Logic:

The formula existed before the theory explaining it.

🧠 Memory Trick

Balmer Before Bohr

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Balmer Formula

16 Hydrogen's atomic composition and the nature of its resulting emission spectrum are respectively characterized as:

A Complex, Simple

B Simple, Complex

C Simple, Continuous

D Complex, Continuous

💡 Short Explanation

�� Hydrogen contains one proton and one electron. �� It is structurally simple. �� Its spectrum contains many specific lines.

✅ Correct Answer Explanation

Hydrogen is the simplest atom, containing only one proton and one electron. Despite this simplicity, it exhibits a complex spectrum consisting of many discrete wavelengths arranged in spectral series. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because the spectrum is not simple.
�� Option C → Incorrect because hydrogen does not produce a continuous spectrum.
�� Option D → Incorrect because hydrogen is structurally simple.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare atomic structure with spectral behavior.

Final Logic:

Simple atom, complex spectrum.

🧠 Memory Trick

Simple Atom → Rich Spectrum

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Hydrogen Spectrum

17 In the Geiger-Marsden experiment, what is the maximum upper limit calculated for the radius of a gold nucleus (Z=79) based on the distance of closest approach for a 7.7 MeV alpha-particle?

A 3.0 × 10⁻¹⁴ m

B 5.3 × 10⁻¹¹ m

C 1.5 × 10¹¹ m

D 2.1 × 10⁻⁷ m

💡 Short Explanation

�� Closest approach estimates nuclear size. �� Nuclear radius is extremely small. �� It is much smaller than atomic size.

✅ Correct Answer Explanation

Using the distance of closest approach calculation for a 7.7 MeV alpha particle and a gold nucleus, Rutherford's analysis gives an upper limit of approximately 3.0 × 10⁻¹⁴ m for the nuclear radius. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents an atomic-scale dimension rather than a nuclear scale.
�� Option C → Unrealistically large value.
�� Option D → Much larger than known nuclear dimensions.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare typical nuclear dimensions with atomic dimensions.

Final Logic:

Nuclear radii are of the order of 10⁻¹⁴ m.

🧠 Memory Trick

Nucleus ≈ 10⁻¹⁴ m

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

18 Incorrect statement regarding Rutherford's assumptions during the scattering experiment analysis:

A The gold nucleus remains completely stationary throughout the scattering process.

B Alpha-particles suffer not more than one scattering during passage through the thin foil.

C The atomic electrons significantly alter the trajectory of the heavy alpha particles.

D The alpha-particle does not physically touch or penetrate the gold nucleus at its distance of closest approach.

💡 Short Explanation

�� Alpha particles are heavy. �� Electron influence is negligible. �� Scattering is dominated by the nucleus.

✅ Correct Answer Explanation

Because alpha particles are much more massive than electrons, the electrons exert only a negligible influence on their trajectories. Rutherford's analysis considered nuclear interactions as the dominant cause of scattering. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Valid approximation used in the analysis.
�� Option B → Valid assumption for a thin foil.
�� Option D → Correct description of closest approach.

🎯 Strategy

Used

Odd One Out

Application:

Identify the assumption inconsistent with Rutherford's model.

Final Logic:

Electrons cannot significantly deflect heavy alpha particles.

🧠 Memory Trick

Alpha Heavy, Electron Light

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

19 Which of the following is correct regarding the theoretical conclusions drawn from the discovery of the nucleus?

A It established that the mass of the atom is uniformly distributed.

B It created a crisis in classical physics because the resulting planetary model predicted unstable atoms that should spiral inward.

C It confirmed that alpha particles are identical to electrons in mass.

D It immediately explained why hydrogen emits a discrete series of spectral lines.

💡 Short Explanation

�� Rutherford proposed orbiting electrons. �� Classical theory predicts energy loss. �� Stable atoms could not be explained.

✅ Correct Answer Explanation

In Rutherford's model, orbiting electrons are accelerating charges and should continuously radiate energy. Consequently, they would spiral into the nucleus, making atoms unstable. This conflict became a major problem for classical physics. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because mass is concentrated in the nucleus.
�� Option C → Incorrect because alpha particles are much heavier than electrons.
�� Option D → Incorrect because Rutherford's model did not explain discrete spectra.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare Rutherford's model with classical electrodynamics.

Final Logic:

The model explained scattering but not stability.

🧠 Memory Trick

Rutherford → Stability Problem

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Limitations of Rutherford Model

20 Statements about the broader implications of Rutherford's early work:
1. He discovered an isotope of radon called thoron.
2. He estimated the approximate size of the nucleus to be much smaller than the atom.
3. He proposed that noble gases cannot be ionized.
4. He created the modern theory of radioactivity alongside Soddy.

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

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

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

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

💡 Short Explanation

�� Rutherford discovered thoron. �� He estimated nuclear size. �� He worked with Soddy on radioactivity.

✅ Correct Answer Explanation

Statement 1 is correct because Rutherford identified thoron, a radioactive isotope of radon. Statement 2 is correct because scattering experiments showed the nucleus is much smaller than the atom. Statement 3 is incorrect because noble gases can be ionized under suitable conditions. Statement 4 is correct because Rutherford and Soddy developed the modern theory of radioactivity. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option B → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using Rutherford's established contributions.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Thoron–Nucleus–Soddy

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford's Contributions to Atomic Theory

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