CUET UG Physics Booster Test 3- Spectral Series and Duality -

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CUET UG Physics Booster Test 3- Spectral Series and Duality

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3- Spectral Series and Duality

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QUESTION 1 OF 20

Consider the emission spectrum of hydrogen resulting from transitions to various levels. Which statements are analytically true?
1. In a classical model, an accelerating electron spiralling inward would emit a continuous spectrum.
2. According to Bohr's postulate, the frequency of the emitted photon strictly equals (Ei − Ef)/h.
3. For transitions between very large quantum numbers (n to n−1), the frequency of the emitted spectral line coincides with the frequency of revolution.
4. The frequency of the emitted photon is always exactly equal to the frequency of revolution of the electron in its initial orbit.

Answer Review

1 Consider the emission spectrum of hydrogen resulting from transitions to various levels. Which statements are analytically true?
1. In a classical model, an accelerating electron spiralling inward would emit a continuous spectrum.
2. According to Bohr's postulate, the frequency of the emitted photon strictly equals (Ei − Ef)/h.
3. For transitions between very large quantum numbers (n to n−1), the frequency of the emitted spectral line coincides with the frequency of revolution.
4. The frequency of the emitted photon is always exactly equal to the frequency of revolution of the electron in its initial orbit.

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

B (2)

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

D (1), (4)

💡 Short Explanation

�� Classical theory predicts continuous radiation. �� Bohr frequency condition gives photon frequency. �� Correspondence principle applies at large n.

✅ Correct Answer Explanation

Statement 1 is correct because a classical accelerating electron would continuously lose energy and emit a continuous spectrum. Statement 2 is correct because Bohr's frequency condition states that the emitted photon frequency is given by ν = (Ei − Ef)/h. Statement 3 is correct due to Bohr's correspondence principle, according to which transitions between very large quantum numbers produce frequencies approaching the orbital frequency. Statement 4 is incorrect because the emitted photon frequency is generally determined by the energy difference between two states and is not always equal to the orbital frequency of the initial orbit. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Check each statement using Bohr's postulates and correspondence principle.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Classical → Continuous, Bohr → Discrete

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

2 When analyzing different transitions in a hydrogen atom, lines that are spaced progressively closer together correspond to transitions involving _______ energy levels, because the absolute energy difference between adjacent levels _______ as n increases.

A Lower, Increases

B Higher, Decreases

C Lower, Decreases

D Higher, Increases

💡 Short Explanation

�� Energy levels crowd together at large n. �� Adjacent energy differences decrease. �� Spectral lines become closer.

✅ Correct Answer Explanation

For hydrogen, En = −13.6/n² eV. As n increases, the spacing between adjacent energy levels becomes smaller. Therefore, transitions involving higher energy levels produce spectral lines that are progressively closer together. This explains the convergence of spectral series. Hence, the correct combination is Higher energy levels and Decreases.

❓ Why Other Options Are Incorrect

�� Option A → Energy spacing does not increase at lower levels.
�� Option C → Closely spaced lines occur at higher, not lower, levels.
�� Option D → Energy difference decreases rather than increases.

🎯 Strategy

Used

Concept Application

Application:

Use the hydrogen energy-level formula.

Final Logic:

Higher n gives smaller level spacing and closer spectral lines.

🧠 Memory Trick

Higher n → Narrow Gap

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

3 Correct statements concerning the transmitted light through a cold, rarefied atomic hydrogen gas:
1. It contains dark lines known collectively as the absorption spectrum.
2. Photons with exact energy differences between a lower and available higher state are actively absorbed.
3. Since most hydrogen atoms are in the ground state at room temperature, the gas primarily absorbs photons of discrete energies like 10.2 eV, 12.09 eV, etc.
4. The transmitted light visually appears as bright spectral lines superimposed on a completely dark background.

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

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

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

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

💡 Short Explanation

�� Absorption produces dark lines. �� Only specific photon energies are absorbed. �� Ground-state atoms dominate absorption.

✅ Correct Answer Explanation

Statement 1 is correct because absorption spectra consist of dark lines in a continuous spectrum. Statement 2 is correct because atoms absorb only photons whose energies match allowed transitions. Statement 3 is correct because most hydrogen atoms are in the ground state and absorb photons corresponding to excitation from n = 1. Statement 4 is incorrect because bright lines on a dark background represent an emission spectrum, not an absorption spectrum. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4 and excludes statement 2.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Distinguish absorption spectra from emission spectra.

Final Logic:

Only statements 1, 2 and 3 describe absorption correctly.

🧠 Memory Trick

Absorption = Dark Lines

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

4 The precise physical origin of dark lines in the absorption spectrum of a rarefied gas indicates that:

A the gas atoms continuously emit radiation that cancels the incident white light via destructive interference.

B all incoming photons are elastically scattered exclusively at 90-degree angles away from the detector.

C specific frequencies from the continuous spectrum are selectively absorbed to excite electrons to higher energy states.

D the isolated atoms completely convert all incident electromagnetic waves into measurable mass.

💡 Short Explanation

�� Absorption occurs at specific frequencies. �� Electrons move to higher levels. �� Missing frequencies appear as dark lines.

✅ Correct Answer Explanation

Dark absorption lines arise because atoms absorb photons whose energies exactly match the energy differences between allowed electronic states. The absorbed photons excite electrons to higher levels, removing those frequencies from the transmitted light. Therefore, option C correctly explains the origin of absorption lines.

❓ Why Other Options Are Incorrect

�� Option A → Dark lines are not produced by destructive interference.
�� Option B → Absorption is not equivalent to all photons being scattered at 90°.
�� Option D → Electromagnetic radiation is not converted entirely into mass.

🎯 Strategy

Used

Elimination

Application:

Identify the actual mechanism behind absorption spectra.

Final Logic:

Selective absorption of specific frequencies produces dark lines.

🧠 Memory Trick

Dark Line = Missing Photon

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

5 Why is the line spectrum of an element essentially considered its unique identifying fingerprint?

A Because all elements contain the exact identical number of electrons regardless of atomic number.

B Because there is an intimate relationship between the unique internal structure of a given atom and the discrete wavelengths of radiation it selectively emits.

C Because the continuous spectrum emitted by heated condensed matter is perfectly identical for all elements.

D Because gas molecules randomly alter their specific emission wavelengths upon thermal collision.

💡 Short Explanation

�� Each element has unique energy levels. �� Unique transitions produce unique wavelengths. �� Spectra identify elements.

✅ Correct Answer Explanation

Every element possesses a unique electronic structure and hence a unique set of allowed energy levels. The wavelengths emitted during transitions between these levels form a characteristic line spectrum. This makes the spectrum a fingerprint for identifying the element.

❓ Why Other Options Are Incorrect

�� Option A → Different elements have different numbers of electrons.
�� Option C → Continuous spectra do not explain line-spectrum uniqueness.
�� Option D → Spectral lines arise from quantized transitions, not random collisions.

🎯 Strategy

Used

Concept Application

Application:

Relate atomic structure to spectral lines.

Final Logic:

Unique energy levels produce unique spectral fingerprints.

🧠 Memory Trick

Unique Levels = Unique Lines

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Line Spectra of Elements

6 Given En = −13.6/n² eV, the maximum unique wavelength (λmax) corresponding to a transition out of the first excited state (n=2) down to the ground state (n=1) is derived from the energy difference ΔE = hc/λmax. The exact analytical expression for ΔE in joules is:

A 13.6 × (3/4) × 1.6 × 10⁻¹⁹

B 13.6 × (1/2) × 1.6 × 10⁻¹⁹

C −13.6 × (3/4) × 1.6 × 10⁻¹⁹

D 13.6 × (1/4) × 1.6 × 10⁻¹⁹

💡 Short Explanation

�� E₁ = −13.6 eV. �� E₂ = −3.4 eV. �� ΔE = 10.2 eV.

✅ Correct Answer Explanation

ΔE = E₂ − E₁ = (−3.4) − (−13.6) = 10.2 eV = 13.6 × (3/4) eV Converting to joules: ΔE = 13.6 × (3/4) × 1.6 × 10⁻¹⁹ J Thus option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives only half of 13.6 eV.
�� Option C → Energy emitted is positive, not negative.
�� Option D → Represents 3.4 eV, not 10.2 eV.

🎯 Strategy

Used

Substitution

Application:

Substitute n = 1 and n = 2 into the energy formula.

Final Logic:

Energy difference equals 10.2 eV.

🧠 Memory Trick

2 → 1 = 10.2 eV

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

7 If the orbital radius of an electron in a hydrogen atom is 5.3 × 10⁻¹¹ m in the ground state (n=1), what is the exact absolute energy required to remove the electron completely (Ef = 0) if it is currently orbiting in the n=2 excited state?

A 13.6 eV

B 10.2 eV

C 3.40 eV

D 1.51 eV

💡 Short Explanation

�� Ionisation energy depends on present state. �� E₂ = −3.4 eV. �� Need 3.4 eV to reach zero energy.

✅ Correct Answer Explanation

The total energy of hydrogen in the n-th orbit is En = −13.6/n² eV For n = 2, E₂ = −13.6/4 = −3.4 eV To remove the electron completely, it must be brought to E = 0 eV. Required energy = 0 − (−3.4) = 3.4 eV Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ionisation energy from ground state.
�� Option B → Excitation energy from n = 1 to n = 2.
�� Option D → Energy of n = 3 bound state magnitude.

🎯 Strategy

Used

Substitution

Application:

Use En = −13.6/n² eV directly.

Final Logic:

Ionisation energy equals magnitude of bound-state energy.

🧠 Memory Trick

Ionise from n = 2 → 3.4 eV

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

8 Match the calculated total energy state in a hydrogen atom to its physical significance.

List I	List II
1. −13.6 eV	a. Energy of a completely removed electron at rest
2. −3.40 eV	b. Energy required to excite a ground-state electron to the first excited state
3. 0 eV	c. Minimum energy to ionise from ground state (expressed as a bound state value)
4. 10.2 eV	d. First excited state bound energy

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

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

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

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

💡 Short Explanation

�� −13.6 eV is ground-state energy. �� −3.4 eV is first excited state. �� 0 eV corresponds to free electron.

✅ Correct Answer Explanation

1 → c : −13.6 eV represents the ground-state bound energy and its magnitude equals the ionisation energy. 2 → d : −3.40 eV is the first excited-state energy. 3 → a : 0 eV corresponds to a completely free electron. 4 → b : 10.2 eV is the excitation energy from n = 1 to n = 2. Thus, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiple incorrect assignments.
�� Option C → Ground state and free-electron energies are mismatched.
�� Option D → Incorrect pairing of first excited state and free electron.

🎯 Strategy

Used

Option Grouping

Application:

Match each energy value with its physical meaning.

Final Logic:

Only option B correctly matches all four quantities.

🧠 Memory Trick

Ground, Excited, Free, Excitation

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

9 Incorrect statement about discrete transitions from highly excited states directly to the ground state (n=1):

A The minimum energy of an emitted photon in this specific series of transitions is 10.2 eV.

B The frequency of the emitted photon continuously changes as the electron slowly spirals into the nucleus.

C A single photon is emitted whose exact energy equals the discrete difference between the initial and final states.

D These energy transitions account for distinct bright lines rather than a completely continuous spectrum.

💡 Short Explanation

�� Bohr orbits are stable. �� Transitions are discrete. �� Spectra are line spectra.

✅ Correct Answer Explanation

Bohr's model states that electrons occupy stable stationary states and do not spiral into the nucleus. Radiation is emitted only during transitions between allowed energy levels. Therefore, the emitted frequency is discrete and not continuously varying. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct for the first Lyman transition.
�� Option C → Correctly states Bohr's frequency condition.
�� Option D → Hydrogen produces discrete spectral lines.

🎯 Strategy

Used

Elimination

Application:

Identify the statement inconsistent with Bohr theory.

Final Logic:

Continuous spiralling radiation belongs to classical theory, not Bohr theory.

🧠 Memory Trick

Bohr = Jumps, Not Spirals

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

10 When an electron in a hydrogen atom falls to the n=3 state from a theoretically infinite quantum number (n → ∞), the energy of the emitted photon approaches a limiting mathematical value. This precise limit corresponds exactly to:

A the total ionisation energy of the hydrogen atom from its ground state.

B the energy magnitude of the n=3 state, which is 1.51 eV.

C the energy difference strictly between n=2 and n=1.

D zero, since infinitely large orbits possess infinite bound energy.

💡 Short Explanation

�� E∞ = 0 eV. �� E₃ = −1.51 eV. �� Limit equals |E₃|.

✅ Correct Answer Explanation

For n → ∞, E∞ = 0 eV For n = 3, E₃ = −13.6/9 = −1.51 eV Energy emitted: ΔE = 0 − (−1.51) = 1.51 eV Thus the series limit for transitions ending at n = 3 equals 1.51 eV. Therefore option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Refers to ground-state ionisation energy (13.6 eV).
�� Option C → Represents 10.2 eV transition energy.
�� Option D → E∞ equals zero, not infinite bound energy.

🎯 Strategy

Used

Substitution

Application:

Use the hydrogen energy-level formula for n = 3 and n = ∞.

Final Logic:

Series limit equals the ionisation energy from the final level.

🧠 Memory Trick

Paschen Limit = |E₃|

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Hydrogen Spectral Series Limits

11 Correct statements about the wave-particle dual nature implication for Bohr orbits:
1. Bohr's specific stable orbits are a direct mathematical consequence of the electron behaving as a particle wave that forms resonant standing waves.
2. The de Broglie wavelength strictly equals h/mvn when the electron speed is significantly less than the speed of light.
3. The wave-particle duality completely and entirely replaces the classical electrostatic force in determining physical orbit stability.
4. Only resonant standing waves uniquely avoid rapid destructive interference upon cyclic reflection within the orbit.

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

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

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

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

💡 Short Explanation

�� Electron exhibits wave nature. �� Stable orbits correspond to standing waves. �� Coulomb force still governs orbital motion.

✅ Correct Answer Explanation

Statement 1 is correct because de Broglie's hypothesis explains Bohr's stable orbits as standing matter waves. Statement 2 is correct because for non-relativistic speeds, λ = h/mv. Statement 3 is incorrect because wave-particle duality does not replace the electrostatic attraction between the electron and nucleus. Statement 4 is correct because only standing-wave configurations remain self-consistent and avoid destructive interference. 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:

Check each statement using de Broglie's interpretation of Bohr orbits.

Final Logic:

Only statements 1, 2 and 4 are valid.

🧠 Memory Trick

Standing Wave = Stable Orbit

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: de Broglie Interpretation of Bohr Orbits

12 Substituting the classical velocity expression vn = e / √(4πε₀mrn) into the fundamental de Broglie wavelength equation λ = h/mvn, we find λ strictly depends on the radius rn as:

A λ is proportional to √rn

B λ is proportional to rn²

C λ is proportional to 1/rn

D λ is proportional to rn

💡 Short Explanation

�� λ = h/mv �� v ∝ 1/√r �� Therefore λ ∝ √r

✅ Correct Answer Explanation

Given v ∝ 1/√rn and λ = h/mv Therefore, λ ∝ 1/v ∝ √rn Hence, the de Broglie wavelength varies as the square root of the orbital radius. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → No quadratic dependence exists.
�� Option C → Wavelength increases, not decreases, with radius.
�� Option D → Dependence is square root, not linear.

🎯 Strategy

Used

Substitution

Application:

Substitute the velocity-radius relation into the de Broglie equation.

Final Logic:

λ varies directly as √rn.

🧠 Memory Trick

v ↓ ⇒ λ ↑

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: de Broglie Wavelength

13 In analogizing de Broglie particle waves on a circular orbit to mechanical waves vibrating on a fixed string:
1. Both macroscopic and microscopic systems excite exactly a single wavelength when randomly disturbed.
2. Both require defined nodes at the boundary ends, translating to exactly an integer number of full wavelengths fitting the circumference.
3. Both necessarily result in a continuously changing broad spectrum of electromagnetic radiation.
4. Neither physical system relies fundamentally on the geometric dimensions of the string or orbit.

A (1)

B (2)

C (2), (3)

D (1), (4)

💡 Short Explanation

�� Standing waves require boundary conditions. �� Integer wavelengths fit the path. �� Geometry determines allowed modes.

✅ Correct Answer Explanation

Statement 2 is correct because standing waves are formed only when an integral number of wavelengths fit the allowed length or circumference. Statement 1 is incorrect because multiple harmonics can be excited. Statement 3 is incorrect because standing waves do not necessarily produce electromagnetic radiation. Statement 4 is incorrect because the dimensions of the string or orbit determine the allowed wavelengths. Therefore, only statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statements 1 and 4 are incorrect.

🎯 Strategy

Used

Elimination

Application:

Compare standing waves on strings with de Broglie standing waves.

Final Logic:

Only statement 2 correctly describes both systems.

🧠 Memory Trick

Integer λ Fits

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Standing Matter Waves

14 Statements conceptually concerning the standing wave relation 2πrn = nλ:
1. It dictates that rn can dynamically take any continuous fractional value.
2. It leads directly to Bohr's specific quantisation condition of angular momentum.
3. It demonstrates why hypothetical orbits with non-integer wavelength fractions rapidly decay to zero amplitude.
4. It implies that the total distance travelled by the electron wave down the orbit and back is a perfect integer number of wavelengths.

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

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

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

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

💡 Short Explanation

�� Circumference must contain integer wavelengths. �� Leads to angular momentum quantisation. �� Non-resonant waves cancel out.

✅ Correct Answer Explanation

Statement 1 is incorrect because quantisation restricts allowed radii. Statement 2 is correct since 2πr = nλ directly leads to mvr = nh/2π. Statement 3 is correct because non-integer wavelength fitting causes destructive interference. Statement 4 is correct because the orbital path accommodates an integral number of wavelengths. Therefore, statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 1, which is incorrect.
�� Option C → Includes statement 1, which is incorrect.
�� Option D → Includes statement 1, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Apply the standing-wave condition.

Final Logic:

Only statements 2, 3 and 4 follow from 2πr = nλ.

🧠 Memory Trick

Integer λ → Quantisation

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: de Broglie Explanation of Bohr Quantisation

15 If the orbital radius of a hydrogen atom in the ground state (n=1) is 5.3 × 10⁻¹¹ m, what is the exact analytical quantum condition required for the angular momentum in the second excited state (n=3)?

A 3h/π

B 1.5h/π

C h/2π

D 9h/2π

💡 Short Explanation

�� Second excited state means n = 3. �� Bohr condition: L = nh/2π. �� L = 3h/2π = 1.5h/π.

✅ Correct Answer Explanation

Bohr's quantisation condition is L = mvr = nh/2π For n = 3, L = 3h/2π = 1.5h/π Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Twice the correct value.
�� Option C → Corresponds to n = 1.
�� Option D → Corresponds to n = 9.

🎯 Strategy

Used

Substitution

Application:

Substitute n = 3 into Bohr's angular momentum equation.

Final Logic:

Second excited state corresponds to n = 3.

🧠 Memory Trick

n = 3 ⇒ 3h/2π

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

16 The core physical reason an atom does not collapse continuously as originally predicted by classical electromagnetic theory is because the electron strictly resides in _______ states where its particle wave forms a completely _______ wave.

A Continuous, Traveling

B Spiral, Resonant

C Accelerated, Continuous

D Stationary, Standing

💡 Short Explanation

�� Electrons occupy stationary states. �� Matter waves form standing waves. �� Stable configurations persist.

✅ Correct Answer Explanation

Modern quantum interpretation explains atomic stability through stationary states associated with standing matter waves. In such states, electrons do not radiate continuously as predicted classically. Therefore, the correct pair is Stationary, Standing.

❓ Why Other Options Are Incorrect

�� Option A → Continuous travelling waves do not explain atomic stability.
�� Option B → Spiral motion is contrary to stable atomic structure.
�� Option C → Accelerated continuous motion reflects classical collapse.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the terminology associated with atomic stability.

Final Logic:

Stable atoms require stationary states and standing waves.

🧠 Memory Trick

Stationary = Stable

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Standing Matter Waves and Atomic Stability

17 Match the atomic system with its theoretical compatibility with Bohr's original model.

List I	List II
1. Hydrogen atom (Z = 1)	a. Fails completely due to electron-electron interactions
2. Helium atom (2 electrons)	b. Hydrogenic model, gross features predicted correctly
3. Singly ionised helium (He⁺)	c. Hydrogenic model, Z = 2
4. Doubly ionised lithium (Li²⁺)	d. Hydrogenic model, Z = 3

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

💡 Short Explanation

�� Bohr model works for one-electron systems. �� Helium has electron-electron interactions. �� Hydrogenic ions follow Bohr predictions.

✅ Correct Answer Explanation

1 → b : Hydrogen atom is the ideal Bohr system. 2 → a : Helium contains two electrons, causing electron-electron interactions neglected by Bohr theory. 3 → c : He⁺ is a one-electron hydrogenic ion with Z = 2. 4 → d : Li²⁺ is a one-electron hydrogenic ion with Z = 3. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Hydrogen and helium are mismatched.
�� Option C → He⁺ is incorrectly matched.
�� Option D → Hydrogen atom is incorrectly assigned.

🎯 Strategy

Used

Option Grouping

Application:

Identify which systems contain only one electron.

Final Logic:

Only option A correctly matches all systems.

🧠 Memory Trick

One Electron = Bohr Works

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Success and Limitations of Bohr Model

18 Incorrect statement about the theoretical transition likelihoods in Bohr's model:

A Bohr's model correctly and inherently calculates the exact probability of transitions leading to specific intensity variations.

B Experimental observations definitively show that some discrete transitions are highly more favoured than others.

C The simple model fundamentally does not explain why some visible frequencies have notably weak intensities.

D The model is essentially a limited classical trajectory picture selectively combined with some modern quantum postulates.

💡 Short Explanation

�� Bohr model predicts energies. �� It does not predict intensities. �� Transition probabilities require quantum mechanics.

✅ Correct Answer Explanation

A major limitation of Bohr's model is that it cannot predict transition probabilities or relative spectral line intensities. These quantities are explained by quantum mechanics through selection rules and wave functions. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Experimentally observed spectral intensities vary.
�� Option C → This is a known limitation of Bohr theory.
�� Option D → Correctly describes the hybrid nature of the model.

🎯 Strategy

Used

Elimination

Application:

Identify a known limitation of Bohr's theory.

Final Logic:

Intensity prediction lies beyond the Bohr model.

🧠 Memory Trick

Bohr Gives Energy, Not Intensity

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

19 When attempting to cleanly extend the Bohr model to complex multi-electron atoms, the theoretical analysis failed primarily because:

A classical physics is entirely sufficient for multi-electron atoms without needing quantum mechanics.

B the model rigidly omits the complex electrical forces between multiple electrons, which are highly comparable in magnitude to the electron-nucleus force.

C complex atoms technically do not possess a centrally positively charged, massive nucleus.

D the gravitational force significantly overwhelms the electrical force in much heavier elements.

💡 Short Explanation

�� Multi-electron atoms involve repulsion. �� Electron-electron interaction is significant. �� Bohr model ignores this effect.

✅ Correct Answer Explanation

Bohr's model successfully describes hydrogen-like atoms with one electron. In multi-electron atoms, electron-electron repulsion significantly affects energy levels, and Bohr's theory cannot account for these interactions. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Classical physics alone cannot explain atomic structure.
�� Option C → Multi-electron atoms still possess a nucleus.
�� Option D → Electrical forces dominate gravitational forces.

🎯 Strategy

Used

Elimination

Application:

Identify the major limitation in extending Bohr's model.

Final Logic:

Electron-electron interactions are the key difficulty.

🧠 Memory Trick

Many Electrons = Many Interactions

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

20 How does modern quantum mechanics explicitly redefine the basic conceptual framework of the Bohr model?

A It mathematically confirms that electrons strictly orbit in perfectly precise planet-like classical trajectories.

B It completely replaces precise discrete orbits with three-dimensional regions where an electron may be found with a large mathematical probability.

C It removes the fundamental concept of quantised discrete energy states entirely from physics.

D It states that total energy depends exclusively on all four quantum numbers (n, l, m, s) even for a theoretically pure Coulomb potential.

💡 Short Explanation

�� Electrons are described by wave functions. �� Orbitals replace fixed orbits. �� Probability replaces exact trajectories.

✅ Correct Answer Explanation

Modern quantum mechanics describes electrons through wave functions and probability distributions. Instead of definite Bohr orbits, electrons occupy orbitals, which are regions of high probability of finding the electron. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Exact classical trajectories are not accepted.
�� Option C → Quantised energy levels remain fundamental.
�� Option D → For hydrogen-like atoms, energy primarily depends on n in a pure Coulomb field.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement consistent with modern quantum mechanics.

Final Logic:

Orbitals and probability distributions replace fixed Bohr orbits.

🧠 Memory Trick

Orbit → Orbital

📚 Class XII – NCERT Physics Part II – Chapter: Dual Nature of Radiation and Matter – Topic: Quantum Mechanical Picture of the Atom

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