CUET UG Physics Booster Test 2- Bohr Model Postulates -

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CUET UG Physics Booster Test 2- Bohr Model Postulates

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2- Bohr Model Postulates

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QUESTION 1 OF 20

According to classical electromagnetic theory applied to atoms:

Answer Review

1 According to classical electromagnetic theory applied to atoms:

A an accelerating charged particle emits radiation continuously.

B an object moving in a circle is not accelerated.

C an accelerating charged particle absorbs radiation continuously.

D the acceleration is tangential in nature.

💡 Short Explanation

�� Accelerating charges emit electromagnetic radiation. �� Circular motion involves acceleration. �� This led to the instability problem of Rutherford's atom.

✅ Correct Answer Explanation

According to classical electromagnetic theory, any accelerating charged particle continuously emits electromagnetic radiation. Since an electron moving around the nucleus undergoes centripetal acceleration, it should continuously radiate energy. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Circular motion always involves centripetal acceleration.
�� Option C → Classical theory predicts emission, not continuous absorption.
�� Option D → The acceleration is centripetal (towards the centre), not tangential.

🎯 Strategy

Used

Elimination

Application:

Reject statements inconsistent with circular motion and electromagnetic theory.

Final Logic:

Accelerating charges continuously emit electromagnetic radiation.

🧠 Memory Trick

Accelerate → Radiate

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

2 Match List I with List II for a classical revolving electron:

List I	List II
1. Nature of force	a. Continuous energy decrease
2. Source of acceleration	b. Coulomb's Law of force
3. Electromagnetic theory consequence	c. Fall into nucleus
4. End result	d. Change in velocity direction (circular motion)

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

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

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

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

💡 Short Explanation

�� Coulomb force provides attraction. �� Circular motion causes acceleration. �� Radiation causes energy loss.

✅ Correct Answer Explanation

1 → b : Force between nucleus and electron is Coulomb force. 2 → d : Circular motion changes velocity direction, causing acceleration. 3 → a : Radiation leads to continuous energy decrease. 4 → c : Energy loss causes the electron to fall into the nucleus. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of all major concepts.
�� Option C → Acceleration source matched incorrectly.
�� Option D → Nature of force and acceleration mismatched.

🎯 Strategy

Used

Option Grouping

Application:

Match each concept with its physical consequence.

Final Logic:

Only option A correctly links all classical predictions.

🧠 Memory Trick

Coulomb → Acceleration → Radiation → Collapse

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

3 In a classical atom model with an inward spiralling electron:

A The electron's angular velocity and frequency change continuously, changing the emitted light frequency

B The electron's angular velocity remains constant, but the orbital radius decreases continuously

C The electron's frequency of revolution remains constant throughout the spiral

D The electron's energy loss has no effect on its angular velocity

💡 Short Explanation

�� Orbit continuously shrinks. �� Orbital frequency changes continuously. �� Emitted frequency also changes continuously.

✅ Correct Answer Explanation

As the electron loses energy, its orbit shrinks. Consequently, its angular velocity and frequency of revolution continuously change. Since classical theory assumes emitted radiation frequency equals orbital frequency, the emitted light frequency also changes continuously. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Angular velocity does not remain constant.
�� Option C → Revolution frequency changes continuously.
�� Option D → Energy loss directly affects orbital motion.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the consequences of continuous energy loss.

Final Logic:

Changing orbit produces continuously changing frequency.

🧠 Memory Trick

Shrinking Orbit → Changing Frequency

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

4 Which is an incorrect statement about the radiation collapse of the classical atom?

A The continuous decrease in energy implies the electron orbit shrinks.

B The inward spiral happens because an accelerating charge must emit electromagnetic waves.

C This model successfully explains the stability of matter.

D Rutherford's model imitates the sun-planet system but involves charged objects.

💡 Short Explanation

�� Classical model predicts instability. �� Electron continuously loses energy. �� Matter should collapse according to the model.

✅ Correct Answer Explanation

The major failure of Rutherford's model was its inability to explain atomic stability. Classical theory predicts that electrons radiate continuously and spiral into the nucleus. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct consequence of energy loss.
�� Option B → Correct classical prediction.
�� Option D → Correct description of Rutherford's model.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting the known failure of the model.

Final Logic:

The classical model fails to explain stability.

🧠 Memory Trick

Classical Atom = Unstable Atom

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

5 Choose the correct statements about continuous frequency emission:
1. Revolving electrons emit electromagnetic waves with a frequency equal to their frequency of revolution.
2. As the orbit shrinks, the frequency of revolution changes.
3. This results in the emission of a continuous spectrum.
4. This classical prediction perfectly matched experimental observations.

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

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

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

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

💡 Short Explanation

�� Orbital frequency changes continuously. �� Emitted frequency changes continuously. �� Continuous spectrum is predicted.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Classical theory predicts that emitted radiation has the same frequency as electron revolution. Since the revolution frequency changes continuously as the orbit shrinks, a continuous spectrum is expected. Statement 4 is incorrect because experiments showed line spectra.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Remove the statement contradicting experimental evidence.

Final Logic:

Continuous spectrum prediction disagreed with observation.

🧠 Memory Trick

Changing Orbit → Continuous Spectrum

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

6 The classical model predicted a _______ spectrum, while the actually observed spectrum of atomic hydrogen is a _______ spectrum.

A Line, Continuous

B Continuous, Line

C Continuous, Continuous

D Line, Line

💡 Short Explanation

�� Classical theory predicts continuous frequencies. �� Hydrogen shows discrete lines. �� This motivated Bohr's theory.

✅ Correct Answer Explanation

Classical theory predicts a continuous spectrum because orbital frequency changes continuously. Experimentally, hydrogen exhibits a line spectrum consisting of discrete wavelengths. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reversed order.
�� Option C → Observed hydrogen spectrum is not continuous.
�� Option D → Classical prediction is not line spectrum.

🎯 Strategy

Used

Odd One Out

Application:

Compare theoretical prediction with experimental observation.

Final Logic:

Predicted continuous, observed line spectrum.

🧠 Memory Trick

Classical = Continuous, Hydrogen = Line

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

7 Choose the correct statements about Bohr's stable orbits:
1. They are a radical departure from established principles of classical electromagnetism.
2. They are governed exclusively by gravitational forces.
3. They resolve the classical difficulty of the inward spiral.
4. Electrons in these orbits do not emit radiant energy.

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

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

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

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

💡 Short Explanation

�� Stable orbits are non-radiating. �� They contradict classical expectations. �� They prevent atomic collapse.

✅ Correct Answer Explanation

Statements 1, 3 and 4 are correct. Bohr introduced stationary orbits in which electrons do not radiate energy. This was a major departure from classical theory and solved the instability problem. Statement 4 is incorrect because electrostatic attraction governs atomic motion.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2.
�� Option B → Includes statement 2.
�� Option D → Includes statement 2.

🎯 Strategy

Used

Elimination

Application:

Remove the statement involving gravitational forces.

Final Logic:

Atomic orbits are governed by electrostatic attraction.

🧠 Memory Trick

Bohr Orbit = Stable & Non-Radiating

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

8 If an atom is in a stationary state n, the energy associated with it represents:

A (−13.6 / n²) eV

B (13.6 × n²) eV

C (−13.6 × n²) eV

D (13.6 / n²) eV

💡 Short Explanation

�� Energy is quantized. �� Depends on n². �� Negative sign indicates binding.

✅ Correct Answer Explanation

For hydrogen: En = −13.6/n² eV This gives the energy of the nth stationary state. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong sign and dependence.
�� Option C → Incorrect n² dependence.
�� Option D → Missing negative sign.

🎯 Strategy

Used

Formula Recall

Application:

Recall Bohr energy expression.

Final Logic:

En = −13.6/n² eV.

🧠 Memory Trick

Energy → −13.6/n²

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

9 What is the angular momentum of an electron in the n = 2 state of a hydrogen atom?
(h = 6.63 × 10⁻³⁴ J s, π = 3.14)

A 1.05 × 10⁻³⁴ kg m²/s

B 2.11 × 10⁻³⁴ kg m²/s

C 4.22 × 10⁻³⁴ kg m²/s

D 3.15 × 10⁻³⁴ kg m²/s

💡 Short Explanation

�� Use Bohr quantization rule. �� L = nh/2π. �� Substitute n = 2.

✅ Correct Answer Explanation

L = nh/(2π) L = (2 × 6.63 × 10⁻³⁴)/(2 × 3.14) L = 13.26 × 10⁻³⁴ / 6.28 L = 2.11 × 10⁻³⁴ kg m²/s Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds approximately to n = 1.
�� Option C → Twice the correct value.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

Substitution

Application:

Apply L = nh/2π.

Final Logic:

Substitution gives 2.11 × 10⁻³⁴ kg m²/s.

🧠 Memory Trick

n = 2 ⇒ L = 2h/2π

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Quantization of Angular Momentum

10 Which is an incorrect statement about the quantisation rule?

A Louis de Broglie provided an explanation for this rule using particle waves.

B It states that L = nh/2π.

C The rule was derived directly from classical mechanics.

D It implies that only certain discrete orbits are dynamically allowed.

💡 Short Explanation

�� Quantization was introduced as a postulate. �� Classical mechanics cannot derive it. �� De Broglie later explained it.

✅ Correct Answer Explanation

Bohr introduced angular momentum quantization as a postulate. It was not derived from classical mechanics. Later, de Broglie's matter-wave theory provided a physical explanation. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct quantization condition.
�� Option D → Correct implication of quantization.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with the history of Bohr's theory.

Final Logic:

Quantization was postulated, not classically derived.

🧠 Memory Trick

Bohr Postulated, de Broglie Explained

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bohr Quantisation Rule

11 In an atomic gas at low pressure passing an electric current, when an electron falls back to a state of lower energy:

A it emits a photon, creating an emission line spectrum with specific wavelengths only

B it absorbs a photon, creating an absorption line spectrum

C it emits a continuous range of wavelengths

D it loses its mass as an electromagnetic wave

💡 Short Explanation

�� Electrons transition between discrete energy levels. �� Downward transition emits a photon. �� Specific energy differences produce line spectra.

✅ Correct Answer Explanation

According to Bohr's third postulate, when an electron moves from a higher energy state to a lower energy state, it emits a photon whose energy equals the energy difference between the two states. hν = Ei − Ef Since only certain energy levels are allowed, only specific wavelengths are emitted, producing an emission line spectrum. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Absorption occurs during upward transitions.
�� Option C → Atomic gases produce line spectra, not continuous spectra.
�� Option D → Electrons do not lose mass during emission.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate electron transitions to emission spectra.

Final Logic:

Downward transition produces discrete photon emission.

🧠 Memory Trick

Fall Down → Photon Out

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

12 Transition from higher to lower energy causes _______, whereas transition from lower to higher energy requires _______.

A Absorption, Emission

B Emission, Absorption

C Ionisation, Emission

D Absorption, Ionisation

💡 Short Explanation

�� Downward transition releases energy. �� Upward transition requires energy. �� Energy exchange occurs through photons.

✅ Correct Answer Explanation

When an electron moves from a higher energy state to a lower energy state, energy is released as a photon (emission). When an electron moves from a lower energy state to a higher energy state, energy must be supplied through photon absorption. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reverses the processes.
�� Option C → Ionisation is not required for ordinary transitions.
�� Option D → Incorrect description of energy transfer.

🎯 Strategy

Used

Odd One Out

Application:

Identify the correct sequence of energy release and gain.

Final Logic:

Higher → Lower = Emission; Lower → Higher = Absorption.

🧠 Memory Trick

Down = Emit, Up = Absorb

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

13 If an electron transitions from orbit ni to nf (where ni > nf), the energy of the emitted photon is proportional to:

A (1/nf² − 1/ni²)

B (1/ni² − 1/nf²)

C (ni² − nf²)

D (nf² − ni²)

💡 Short Explanation

�� Hydrogen energy levels vary as 1/n². �� Photon energy equals level difference. �� Final lower orbit has larger energy magnitude.

✅ Correct Answer Explanation

For hydrogen: En = −13.6/n² eV Therefore, hν = Ei − Ef = [−13.6/ni²] − [−13.6/nf²] = 13.6[(1/nf²) − (1/ni²)] Hence photon energy is proportional to: (1/nf² − 1/ni²) Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives the negative of the required expression.
�� Option C → Incorrect dependence.
�� Option D → Negative quantity for emission.

🎯 Strategy

Used

Substitution

Application:

Substitute Bohr energy expression into hν = Ei − Ef.

Final Logic:

Photon energy depends on the difference of inverse squares.

🧠 Memory Trick

Emission = Lower Level Term − Higher Level Term

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

14 Choose the correct statements regarding the relationship hν = Ei − Ef:
1. h is the Planck's constant.
2. The frequency ν emitted depends directly on the energy difference.
3. This incorporates Einstein and Planck's quantum concepts into the atomic theory.
4. The frequency of the emitted photon equals the electron's frequency of revolution.

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

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

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

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

💡 Short Explanation

�� Photon energy equals energy difference. �� Frequency depends on ΔE. �� Bohr used quantum concepts.

✅ Correct Answer Explanation

Statement 1 is correct because h is Planck's constant. Statement 2 is correct since: ν = (Ei − Ef)/h Statement 3 is correct because Bohr incorporated Planck's quantization and Einstein's photon concept. Statement 4 is incorrect because emitted photon frequency is determined by energy difference, not electron revolution frequency.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 4.
�� Option C → Includes incorrect statement 4.
�� Option D → Includes incorrect statement 4.

🎯 Strategy

Used

Elimination

Application:

Remove the statement based on classical frequency prediction.

Final Logic:

Photon frequency depends on energy difference, not orbital frequency.

🧠 Memory Trick

Photon Frequency = ΔE/h

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

15 Match List I with List II for Bohr radius rn dependencies:

List I	List II
1. Mass of electron (m)	a. Directly proportional to n²
2. Principal quantum number (n)	b. Inversely proportional to m
3. Planck constant (h)	c. Directly proportional to h²
4. Permittivity (ε₀)	d. Directly proportional to ε₀

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

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

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

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

💡 Short Explanation

�� Radius depends on n². �� Radius is inversely proportional to mass. �� Radius depends on h² and ε₀.

✅ Correct Answer Explanation

Bohr radius relation: rn = (ε₀n²h²)/(πme²) Therefore: 1 → b 2 → a 3 → c 4 → d Thus option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching.
�� Option C → Incorrect matching.
�� Option D → Incorrect matching.

🎯 Strategy

Used

Option Grouping

Application:

Match each parameter using the Bohr radius formula.

Final Logic:

Only option A satisfies all dependencies.

🧠 Memory Trick

Radius ∝ ε₀h²n²/m

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

16 The Bohr radius a₀ of the lowest energy state:

A corresponds to n = 2

B is exactly 5.3 × 10⁻¹⁰ m

C corresponds to n = 1

D represents the maximum distance an electron can reach

💡 Short Explanation

�� Ground state has n = 1. �� Bohr radius belongs to ground state. �� Value is 5.3 × 10⁻¹¹ m.

✅ Correct Answer Explanation

The Bohr radius a₀ is the radius of the first orbit (ground state). a₀ = 5.3 × 10⁻¹¹ m Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → n = 2 is first excited state.
�� Option B → Incorrect value (one order larger).
�� Option D → Not a maximum distance.

🎯 Strategy

Used

Formula Recall

Application:

Recall definition of Bohr radius.

Final Logic:

Bohr radius corresponds to n = 1.

🧠 Memory Trick

a₀ = First Orbit

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

17 Choose the correct statements regarding total energy expressions:
1. Total energy is the sum of kinetic and electrostatic potential energy.
2. Total energy E = −e²/(8πε₀r).
3. The potential energy is positive while kinetic is negative.
4. Kinetic energy K = e²/(8πε₀r).

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

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

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

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

💡 Short Explanation

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

✅ Correct Answer Explanation

Statement 1 is correct because total energy equals kinetic plus potential energy. Statement 2 is correct: E = −e²/(8πε₀r) Statement 4 is correct: K = e²/(8πε₀r) Statement 3 is incorrect because kinetic energy is positive and potential energy is negative.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify the sign convention error.

Final Logic:

K > 0 and U < 0.

🧠 Memory Trick

K Positive, U Negative

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy of Electron in Orbit

18 If the total energy E of an electron in an atom were positive:

A The electron would not follow a closed orbit around the nucleus

B The electron would spiral into the nucleus faster

C The electron would be more tightly bound to the nucleus

D The atom would emit radiation continuously

💡 Short Explanation

�� Positive energy means unbound state. �� Electron can escape. �� Closed orbit is impossible.

✅ Correct Answer Explanation

A bound electron has negative total energy. If total energy becomes zero or positive, the electron is no longer bound to the nucleus and cannot remain in a closed orbit. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Positive energy does not imply spiralling inward.
�� Option C → Positive energy means less bound, not more.
�� Option D → Not the defining consequence.

🎯 Strategy

Used

Definition Recall

Application:

Interpret the physical meaning of positive energy.

Final Logic:

Positive total energy corresponds to an unbound electron.

🧠 Memory Trick

Positive E = Escape

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bound and Unbound States

19 What is the energy required to excite an electron in a hydrogen atom from its ground state (n = 1) to its first excited state (n = 2)?
(Ground state energy = −13.6 eV)

A 3.4 eV

B 10.2 eV

C 12.09 eV

D 13.6 eV

💡 Short Explanation

�� E₁ = −13.6 eV. �� E₂ = −3.4 eV. �� Excitation energy = difference.

✅ Correct Answer Explanation

For hydrogen: E₂ = −13.6/2² = −3.4 eV Required energy: ΔE = E₂ − E₁ = (−3.4) − (−13.6) = 10.2 eV Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equals magnitude of E₂.
�� Option C → Corresponds to n = 3 excitation.
�� Option D → Ionization energy from ground state.

🎯 Strategy

Used

Substitution

Application:

Use Bohr energy formula.

Final Logic:

10.2 eV is needed for n = 1 → n = 2.

🧠 Memory Trick

1 → 2 = 10.2 eV

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

20 Which is an incorrect statement about atomic excitation?

A The ionisation energy of the hydrogen atom is 13.6 eV.

B An atom in an excited state can emit a photon and fall to a lower energy state.

C 12.09 eV is required to excite the hydrogen atom to the second excited state (n = 3).

D As excitation increases, the minimum energy required to free the electron from the excited atom increases.

💡 Short Explanation

�� Higher excited states are less tightly bound. �� Ionization energy decreases with n. �� Excited electrons are easier to remove.

✅ Correct Answer Explanation

As the electron moves to higher energy levels, its total energy approaches zero. Therefore, less energy is required to remove it completely from the atom. For example: E₃ = −13.6/9 = −1.51 eV Only 1.51 eV is needed to ionize from n = 3. Thus option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct; hydrogen ionization energy is 13.6 eV.
�� Option B → Correct description of de-excitation.
�� Option C → Correct because E₃ − E₁ = 12.09 eV.

🎯 Strategy

Used

Extreme Word Filter

Application:

Check how ionization energy changes with increasing excitation.

Final Logic:

Higher excitation means lower ionization energy.

🧠 Memory Trick

Higher n → Easier Ionization

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

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