CUET UG Physics Booster Test 3- Bohr Model Postulates -

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CUET UG Physics Booster Test 3- Bohr Model Postulates

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3- Bohr Model Postulates

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QUESTION 1 OF 20

Which is an incorrect statement about the conflict between classical theory and atomic stability?

Answer Review

1 Which is an incorrect statement about the conflict between classical theory and atomic stability?

A Classical electromagnetic theory governs large-scale phenomena successfully but fails at the atomic scale.

B An electron moving in a circular orbit undergoes centripetal acceleration, demanding continuous radiation.

C The loss of energy through continuous radiation would decrease the frequency of revolution continuously.

D Classical mechanics and electromagnetism are sufficient to understand the structure of atoms.

💡 Short Explanation

�� Classical theory fails for atomic systems. �� It cannot explain atomic stability. �� It cannot explain line spectra.

✅ Correct Answer Explanation

Classical mechanics and classical electromagnetic theory successfully explain many macroscopic phenomena, but they fail to explain atomic stability and discrete atomic spectra. This failure led to the development of quantum theory. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Circular motion implies centripetal acceleration.
�� Option C → Energy loss causes shrinking orbit and changing frequency.

🎯 Strategy

Used

Concept Validation

Application:

Identify the statement inconsistent with experimental observations.

Final Logic:

Classical theory alone cannot explain atomic structure.

🧠 Memory Trick

Classical Works Big, Fails Small

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

2 If classical theory held true and an electron's energy decreased continuously, its velocity v and radius r would mathematically change such that the ratio (v² × r) would be proportional to:

A e²/(4πε₀m)

B e²/(2πε₀m)

C m/(4πε₀e²)

D e/(4πε₀m²)

💡 Short Explanation

�� Electrostatic force provides centripetal force. �� Equate Coulomb force and centripetal force. �� Rearrangement gives v²r relation.

✅ Correct Answer Explanation

For a hydrogen atom, mv²/r = e²/(4πε₀r²) Multiplying both sides by r²: mv²r = e²/(4πε₀) Therefore, v²r = e²/(4πε₀m) Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Options B, C and D do not follow from the force balance equation.

🎯 Strategy

Used

Formula Application

Application:

Use Coulomb force = centripetal force.

Final Logic:

v²r = e²/(4πε₀m)

🧠 Memory Trick

v²r = Coulomb Constant Part

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

3 Choose the correct statements regarding the inward spiral trajectory (Classical):
1. The centripetal force is provided by the electrostatic attraction.
2. Radiation causes total energy E to become increasingly negative.
3. The radius of the orbit gradually approaches infinity.
4. It contradicts the experimental observation of stable matter.

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

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

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

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

💡 Short Explanation

�� Electrostatic attraction provides centripetal force. �� Energy loss shrinks the orbit. �� Stable matter contradicts classical prediction.

✅ Correct Answer Explanation

Statement 1 is correct because Coulomb attraction provides the centripetal force. Statement 2 is correct because continuous radiation decreases total energy. Statement 3 is incorrect because the orbit shrinks toward the nucleus, not infinity. Statement 4 is correct because atoms are experimentally stable. Therefore, statements (1), (2) and (4) are correct.

❓ Why Other Options Are Incorrect

�� All other options include statement 3, which is false.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement individually.

Final Logic:

Orbit radius decreases, not increases.

🧠 Memory Trick

Energy Loss → Smaller Orbit

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Failure of Classical Theory

4 The collapse of the classical Rutherford atom is inevitable because:

A the electron-electron electric force interaction disrupts the stable orbit entirely

B an accelerating charged particle must emit energy, violating the assumption of a dynamically stable, non-radiating orbit

C the gravitational force between the nucleus and the electron is too weak to sustain the orbit

D the nucleus absorbs the electron's kinetic energy during its revolution

💡 Short Explanation

�� Accelerating charge emits radiation. �� Electron loses energy continuously. �� Orbit collapses inward.

✅ Correct Answer Explanation

According to classical electromagnetic theory, every accelerating charge radiates energy. Since an orbiting electron is continuously accelerated, it must radiate energy, lose orbital energy, and spiral into the nucleus. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the reason for collapse.
�� Option C → Atomic binding is electrostatic.
�� Option D → No such absorption process occurs.

🎯 Strategy

Used

Cause-Effect Analysis

Application:

Identify the fundamental reason for instability.

Final Logic:

Radiation by accelerating charge causes collapse.

🧠 Memory Trick

Acceleration → Radiation → Collapse

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Classical Difficulty

5 Choose the correct statements concerning the frequency emitted by a classical collapsing atom:
1. The emitted frequency matches the electron's mechanical frequency of revolution.
2. Since the electron's angular velocity increases as it spirals inward, the emitted frequency increases continuously.
3. The result would be a continuous emission spectrum containing all possible wavelengths up to a limit.
4. This theoretical continuous spectrum perfectly explained the glow tube emissions of rarefied gases.

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

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

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

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

💡 Short Explanation

�� Classical theory links emitted frequency to revolution frequency. �� Orbit shrinks continuously. �� Continuous spectrum is predicted.

✅ Correct Answer Explanation

Statements 1, 2 and 3 follow directly from classical theory. Statement 4 is incorrect because rarefied gases experimentally show line spectra, not continuous spectra. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� All other options contain statement 4.

🎯 Strategy

Used

Theory vs Observation

Application:

Compare classical prediction with experiments.

Final Logic:

Classical theory predicts continuous spectra.

🧠 Memory Trick

Continuous Orbit → Continuous Spectrum

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Spectral Contradiction

6 Match List I with List II (Observation/Theory)

List I	List II
1. Condensed matter radiation	a. Continuous distribution of wavelengths
2. Rarefied gases in a flame	b. Discrete wavelengths (line spectrum)
3. Classical accelerating electron	c. Continuous frequency change during spiral
4. Bohr model transitions	d. Frequency defined by ΔE/h

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

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

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

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

💡 Short Explanation

�� Condensed matter → continuous spectrum. �� Rarefied gases → line spectrum. �� Classical spiral → continuously changing frequency. �� Bohr → ΔE/h.

✅ Correct Answer Explanation

List I — List II 1. Condensed matter radiation — a. Continuous distribution of wavelengths 2. Rarefied gases in a flame — b. Discrete wavelengths (line spectrum) 3. Classical accelerating electron — c. Continuous frequency change during spiral 4. Bohr model transitions — d. Frequency defined by ΔE/h Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� They contain incorrect pairings.

🎯 Strategy

Used

Direct Matching

Application:

Match concepts with defining properties.

Final Logic:

Only option A satisfies all pairings.

🧠 Memory Trick

Condensed → Continuous, Gas → Lines

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

7 The fundamental contrast between Rutherford's orbits and Bohr's orbits is that Rutherford's are radiatively _______, while Bohr's are radiatively _______.

A Stable, Unstable

B Unstable, Stable

C Quantised, Continuous

D Stable, Stable

💡 Short Explanation

�� Rutherford orbits radiate. �� Bohr stationary orbits do not. �� Stability is restored in Bohr's model.

✅ Correct Answer Explanation

Rutherford's model predicts continuous radiation and instability. Bohr introduced stationary non-radiating orbits, making atoms stable. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� They contradict the basic difference between the two models.

🎯 Strategy

Used

Comparison

Application:

Compare orbital behavior.

Final Logic:

Rutherford unstable, Bohr stable.

🧠 Memory Trick

Rutherford Radiates, Bohr Behaves

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr's First Postulate

8 The concept of "stationary states" in Bohr's model:

A implies the electron is physically at rest.

B means the electron revolves but does not emit radiant energy, keeping total energy constant.

C represents states where the electron continuously absorbs energy to stay in orbit.

D only applies to electrons outside the atom.

💡 Short Explanation

�� Electron keeps moving. �� No radiation occurs. �� Energy remains fixed.

✅ Correct Answer Explanation

A stationary state does not mean the electron is motionless. It means the electron moves in an allowed orbit without emitting radiation and retains a definite energy. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron is not at rest.
�� Option C → No continuous absorption required.
�� Option D → Incorrect interpretation.

🎯 Strategy

Used

Definition Recall

Application:

Recall Bohr's first postulate.

Final Logic:

Stationary state = constant energy orbit.

🧠 Memory Trick

Stationary Energy, Not Stationary Electron

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Stationary States

9 According to de Broglie's explanation of Bohr's second postulate, an electron moving in the 3rd orbit (n = 3) forms a standing wave. If the de Broglie wavelength is λ, the circumference of this orbit is:

A 1.5 λ

B 9 λ

C 3 λ

D 2π λ

💡 Short Explanation

�� Standing wave condition: 2πr = nλ �� For n = 3: 2πr = 3λ

✅ Correct Answer Explanation

According to de Broglie: 2πrn = nλ For n = 3, 2πr = 3λ Hence the circumference equals 3λ. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� They do not satisfy the standing wave condition.

🎯 Strategy

Used

Formula Substitution

Application:

Use 2πr = nλ.

Final Logic:

For n = 3, circumference = 3λ.

🧠 Memory Trick

Orbit Number = Number of Wavelengths

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: de Broglie Explanation

10 Which is an incorrect statement regarding the quantisation rule and de Broglie's explanation?

A The quantised electron orbits are due to the wave nature of the electron.

B Only resonant standing waves can persist in the circular orbit.

C Waves with non-integral wavelengths interfere constructively and survive.

D The condition 2πrn = nλ mathematically leads to L = nh/2π.

💡 Short Explanation

�� Stable orbits require standing waves. �� Non-integral wavelengths cancel out. �� Only resonance survives.

✅ Correct Answer Explanation

For a stable orbit: 2πr = nλ Only integral numbers of wavelengths produce standing waves. Non-integral wavelengths lead to destructive interference and cannot persist. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct.
�� Option B → Correct.
�� Option D → Correct derivation of Bohr quantisation.

🎯 Strategy

Used

Wave Condition Analysis

Application:

Apply standing wave requirement.

Final Logic:

Non-integral wavelengths are not stable.

🧠 Memory Trick

Whole Waves Stay, Fractional Waves Fade

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: de Broglie Interpretation of Bohr Model

11 During a transition from state n = 3 to n = 2, the energy of the emitted photon in a hydrogen atom is:

A 13.6 × (1/2 − 1/3) eV

B 13.6 × (1/4 − 1/9) eV

C 13.6 × (1/9 − 1/4) eV

D 13.6 × (1/3 − 1/2) eV

💡 Short Explanation

�� Hydrogen energy levels are quantized. �� Photon energy equals energy difference. �� Transition is from n = 3 to n = 2.

✅ Correct Answer Explanation

For hydrogen: En = −13.6/n² eV E3 = −13.6/9 eV E2 = −13.6/4 eV Photon energy emitted: hν = E3 − E2 = (−13.6/9) − (−13.6/4) = 13.6 × (1/4 − 1/9) eV Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uses incorrect energy dependence.
�� Option C → Gives negative photon energy.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Substitution

Application:

Apply Bohr energy formula directly.

Final Logic:

hν = 13.6 × (1/4 − 1/9) eV.

🧠 Memory Trick

Emission = Lower Orbit Term − Higher Orbit Term

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Energy Transitions

12 In Bohr's model, the frequency of the emitted photon during a transition:

A matches exactly with the electron's frequency of revolution in the final orbit

B matches exactly with the electron's frequency of revolution in the initial orbit

C is not directly connected to the mechanical frequency of revolution of the electron in its orbit (except for very large quantum numbers)

D depends on the electron's velocity but not its energy levels

💡 Short Explanation

�� Photon frequency depends on energy difference. �� Orbit frequency and photon frequency are generally different. �� Agreement occurs only for large n.

✅ Correct Answer Explanation

Bohr's frequency condition states: hν = Ei − Ef Thus emitted frequency depends on the difference in energy levels and not directly on the electron's orbital frequency. Only for very large quantum numbers does correspondence approximately occur. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not generally true.
�� Option B → Not generally true.
�� Option D → Energy levels determine frequency.

🎯 Strategy

Used

Conceptual Analysis

Application:

Distinguish orbital frequency from photon frequency.

Final Logic:

Photon frequency is governed by ΔE/h.

🧠 Memory Trick

Photon Frequency = Energy Difference Frequency

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Frequency Condition

13 The absorption spectrum of a material:
1. shows bright lines on a dark background.
2. contains dark lines corresponding to frequencies where photons have exactly the energy needed to excite an electron.
3. is independent of the atom's energy states.
4. forms a completely continuous dark band.
Choose correct:

A (1)

B (2)

C (3)

D (4)

💡 Short Explanation

�� Absorption removes specific wavelengths. �� Missing wavelengths appear dark. �� Energy matches atomic transitions.

✅ Correct Answer Explanation

When white light passes through a cooler gas, atoms absorb photons whose energies exactly match allowed electronic transitions. These missing wavelengths appear as dark lines in the spectrum. Hence statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes emission spectrum.
�� Option C → Absorption depends on energy levels.
�� Option D → Absorption spectrum contains discrete dark lines.

🎯 Strategy

Used

Odd One Out

Application:

Differentiate absorption from emission spectra.

Final Logic:

Dark lines correspond to absorbed transition energies.

🧠 Memory Trick

Absorption = Missing Lines

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

14 Choose the correct statements related to energy, frequency and transitions:
1. Ei − Ef = hν applies for emission.
2. Ei + hν = Ef applies for absorption, where nf > ni.
3. Bohr's model can accurately predict the relative intensities of the emitted frequencies.
4. Atomic energies are often expressed in electron volts rather than joules.

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

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

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

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

💡 Short Explanation

�� Emission obeys hν = Ei − Ef. �� Absorption obeys Ei + hν = Ef. �� Energy levels are commonly expressed in eV.

✅ Correct Answer Explanation

Statement 1 is correct. Statement 2 is correct. Statement 3 is incorrect because Bohr's model cannot explain spectral line intensities. Statement 4 is correct because atomic energies are commonly measured in electron volts. Therefore, statements (1), (2) and (4) are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 3.
�� Option B → Includes incorrect statement 3.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Elimination

Application:

Identify the known limitation of Bohr's model.

Final Logic:

Bohr explains frequencies but not intensities.

🧠 Memory Trick

Bohr → Frequencies Yes, Intensities No

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

15 Choose the correct statements about the quantised orbit radius:
1. The radius is proportional to the square of the principal quantum number.
2. The smallest possible radius corresponds to n = 1.
3. The spacing between successive orbits (rₙ₊₁ − rₙ) is constant.
4. The radius expression includes the mass of the electron in the denominator.

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

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

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

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

💡 Short Explanation

�� rn ∝ n². �� Ground state has minimum radius. �� Orbit spacing is not uniform.

✅ Correct Answer Explanation

For hydrogen: rn = (ε₀ h² n²)/(π m e²) Therefore: Statement 1 is correct. Statement 2 is correct. Statement 3 is incorrect because orbit spacing increases with n. Statement 4 is correct. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� All include statement 3, which is false.

🎯 Strategy

Used

Formula Verification

Application:

Use Bohr radius formula.

Final Logic:

rn ∝ n² and contains m in denominator.

🧠 Memory Trick

Radius Grows as n²

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Orbit Radius

16 If the Bohr radius a₀ = 5.3 × 10⁻¹¹ m, what is the radius of the second excited state of a hydrogen atom?

A 1.06 × 10⁻¹⁰ m

B 2.12 × 10⁻¹⁰ m

C 4.77 × 10⁻¹⁰ m

D 8.48 × 10⁻¹⁰ m

💡 Short Explanation

�� Second excited state means n = 3. �� Radius varies as n². �� r₃ = 9a₀.

✅ Correct Answer Explanation

Second excited state: n = 3 r₃ = n²a₀ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to n = 2.
�� Option B → Incorrect value.
�� Option D → Too large.

🎯 Strategy

Used

Substitution

Application:

Apply rₙ = n²a₀.

Final Logic:

r₃ = 9a₀ = 4.77 × 10⁻¹⁰ m.

🧠 Memory Trick

Second Excited = n³? No, n = 3

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Radius Calculation

17 Match List I (Energy expressions in Bohr orbit) with List II (Dependencies in terms of constants)

List I	List II
1. Total Energy E	a. −e²/(8πε₀r)
2. Potential Energy U	b. −e²/(4πε₀r)
3. Kinetic Energy K	c. e²/(8πε₀r)
4. Radius r	d. ε₀h²n²/(πme²)

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

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

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

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

💡 Short Explanation

�� E = K + U. �� K positive. �� U negative.

✅ Correct Answer Explanation

List I — List II 1. Total Energy E — a. −e²/(8πε₀r) 2. Potential Energy U — b. −e²/(4πε₀r) 3. Kinetic Energy K — c. e²/(8πε₀r) 4. Radius r — d. ε₀h²n²/(πme²) Thus option A is correct.

❓ Why Other Options Are Incorrect

�� They contain incorrect formula pairings.

🎯 Strategy

Used

Formula Matching

Application:

Recall standard Bohr orbit expressions.

Final Logic:

Only option A satisfies all relations.

🧠 Memory Trick

U = −2K and E = −K

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy Expressions

18 As the principal quantum number n increases, the absolute value of energy becomes _______, and the total energy E becomes progressively _______.

A Smaller, Larger (less negative)

B Larger, Smaller (more negative)

C Smaller, Smaller (more negative)

D Larger, Larger (less negative)

💡 Short Explanation

�� En = −13.6/n² eV. �� Magnitude decreases with n. �� Energy approaches zero.

✅ Correct Answer Explanation

As n increases: |En| = 13.6/n² decreases continuously. The energy becomes less negative and approaches zero. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Opposite trends are given.

🎯 Strategy

Used

Formula Analysis

Application:

Use En = −13.6/n².

Final Logic:

Higher n means weaker binding.

🧠 Memory Trick

Higher n → Nearer Zero

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy Levels

19 The ground state of a hydrogen atom is defined such that:

A the electron has zero kinetic energy and is at rest near the nucleus

B the energy is at its maximum possible negative value and the electron revolves in the smallest radius

C the energy is 0 eV and the electron is completely removed from the nucleus

D the atom freely emits radiation without any external excitation

💡 Short Explanation

�� Ground state corresponds to n = 1. �� Lowest energy state. �� Smallest orbit radius.

✅ Correct Answer Explanation

For hydrogen: E1 = −13.6 eV This is the most negative allowed energy and corresponds to the smallest orbit radius. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron is moving.
�� Option C → Describes ionized state.
�� Option D → Ground state does not radiate.

🎯 Strategy

Used

Definition Recall

Application:

Identify properties of n = 1 state.

Final Logic:

Ground state = lowest energy, smallest orbit.

🧠 Memory Trick

Ground = n1 = −13.6 eV

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Ground State

20 Which is an incorrect statement regarding the higher energy states (excited states)?

A At room temperature, most hydrogen atoms are automatically in the highest excited states.

B The highest energy state corresponds to n = ∞ and has an energy of 0 eV.

C The energies of the excited states come closer and closer together as n increases.

D 13.6 eV of energy must be supplied to ionise an electron from the ground state.

💡 Short Explanation

�� Most atoms remain in ground state. �� Excited states require energy input. �� Energy levels converge at large n.

✅ Correct Answer Explanation

At room temperature, thermal energy is insufficient to place most hydrogen atoms in highly excited states. Most atoms remain in the ground state. Statements B, C and D are correct. Therefore, option A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct; E∞ = 0 eV.
�� Option C → Energy levels crowd together at large n.
�� Option D → Hydrogen ionization energy is 13.6 eV.

🎯 Strategy

Used

Extreme Word Filter

Application:

Check for unrealistic absolute claims.

Final Logic:

Most hydrogen atoms are not automatically highly excited.

🧠 Memory Trick

Room Temperature → Ground State Dominates

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Excited States and Ionization Energy

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