CUET UG Physics Booster Test 2- Spectral Series and Duality -

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CUET UG Physics Booster Test 2- 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 2- Spectral Series and Duality

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QUESTION 1 OF 20

When a hydrogen atom makes a transition from the second excited state (n=3) to the ground state (n=1), what is the exact energy of the emitted photon?

Answer Review

1 When a hydrogen atom makes a transition from the second excited state (n=3) to the ground state (n=1), what is the exact energy of the emitted photon?

A 10.2 eV

B 12.09 eV

C 1.51 eV

D 3.40 eV

💡 Short Explanation

�� Energy levels in hydrogen are quantized. �� Emitted photon energy equals the difference between energy levels. �� For n=3 → n=1, a large energy gap exists.

✅ Correct Answer Explanation

For hydrogen: Eₙ = -13.6/n² eV E₁ = -13.6 eV E₃ = -13.6/9 = -1.51 eV Photon energy emitted: ΔE = E₃ - E₁ ΔE = (-1.51) - (-13.6) ΔE = 12.09 eV Therefore, the emitted photon has energy 12.09 eV.

❓ Why Other Options Are Incorrect

�� Option A → 10.2 eV corresponds to the n=2 transition energy from the ground state.
�� Option C → 1.51 eV is the magnitude of the energy of the n=3 state itself.
�� Option D → 3.40 eV is the magnitude of the n=2 state energy.

🎯 Strategy

Used

Substitution

Application:

Substitute n=1 and n=3 into the hydrogen energy formula.

Final Logic:

Photon energy = E₃ − E₁ = 12.09 eV.

🧠 Memory Trick

3 → 1 gives 12.09 eV

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

2 Match List I with List II regarding spectral emission observations:

List I	List II
1. Heated rarefied gases	a. Series of discrete bright lines
2. Condensed matter and dense gases	b. Continuous distribution of wavelengths
3. Atomic hydrogen	c. Set of lines with fixed relative positions
4. Emission spectrum generally	d. Bright lines on a dark background

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

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

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

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

💡 Short Explanation

�� Rarefied gases give line spectra. �� Dense matter gives continuous spectra. �� Hydrogen has fixed spectral line positions.

✅ Correct Answer Explanation

List I — List II 1. Heated rarefied gases — a. Series of discrete bright lines 2. Condensed matter and dense gases — b. Continuous distribution of wavelengths 3. Atomic hydrogen — c. Set of lines with fixed relative positions 4. Emission spectrum generally — d. Bright lines on a dark background Thus, option B is correct.

❓ Why Other Options Are Incorrect

�� Options A, C and D contain incorrect pairings between radiation sources and spectral characteristics.

🎯 Strategy

Used

Option Grouping

Application:

Match known spectrum types with their properties.

Final Logic:

Rarefied gas → Line spectrum; Dense matter → Continuous spectrum.

🧠 Memory Trick

Rare Gas = Lines, Dense Matter = Continuous

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

3 Incorrect statement about the absorption process and transmitted light:

A When white light passes through a gas, dark lines appear in the transmitted light spectrum.

B Dark lines correspond exactly to the discrete wavelengths found in the emission line spectrum.

C The absorption process occurs when an atom emits a photon to jump to a higher energy state.

D Atoms absorb photons that have precisely the energy needed for transitions to higher states.

💡 Short Explanation

�� Absorption requires photon intake. �� Emission occurs during downward transitions. �� Dark lines indicate absorbed frequencies.

✅ Correct Answer Explanation

An atom moves to a higher energy state by absorbing a photon whose energy matches the energy gap between two states. Emission occurs when an electron falls to a lower state. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of absorption spectra.
�� Option B → Absorption and emission lines occur at the same wavelengths.
�� Option D → Correct statement of photon absorption.

🎯 Strategy

Used

Elimination

Application:

Identify the statement that reverses absorption and emission.

Final Logic:

Higher state transition requires absorption, not emission.

🧠 Memory Trick

Up = Absorb, Down = Emit

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

4 Statements regarding dark lines in atomic spectra:
1. They indicate the specific frequencies that have been absorbed by atoms of the gas.
2. They appear in the continuous spectrum when transmitted light is analysed with a spectrometer.
3. They are produced spontaneously when electrons fall to lower energy levels.
4. They form what is known as the absorption spectrum of the material.

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

B (1), (3)

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

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

💡 Short Explanation

�� Dark lines indicate absorption. �� They appear in transmitted continuous light. �� Absorption spectrum contains dark lines.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Statement 3 is incorrect because electrons falling to lower levels produce emission lines, not absorption lines. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Statement 3 is incorrect.
�� Options B, C and D include statement 3.

🎯 Strategy

Used

Statement Verification

Application:

Check each statement using absorption spectrum concepts.

Final Logic:

Dark lines arise from absorption, not spontaneous emission.

🧠 Memory Trick

Dark Line = Absorbed Light

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

5 Because each element is associated with a characteristic spectrum of radiation, analyzing the emission line spectra of a material can serve as a type of fingerprint for gas identification, which strictly implies that:

A atoms of all elements have perfectly identical energy level spacings.

B the spacing between atoms in dense gases is too small to differentiate spectral lines.

C each element emits specific discrete wavelengths uniquely related to its internal atomic structure.

D the continuous spectrum of a gas alone identifies its atomic number easily.

💡 Short Explanation

�� Every element has unique energy levels. �� Unique energy levels produce unique spectra. �� Spectra act as fingerprints.

✅ Correct Answer Explanation

The characteristic emission spectrum of an element arises from transitions between its unique energy levels. Since these levels differ from element to element, the emitted wavelengths are unique. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Energy level spacings differ among elements.
�� Option B → Does not explain fingerprint identification.
�� Option D → Continuous spectra are not unique fingerprints.

🎯 Strategy

Used

Conceptual Matching

Application:

Link spectral lines to atomic structure.

Final Logic:

Unique energy levels → Unique spectral lines.

🧠 Memory Trick

Unique Atom = Unique Spectrum

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

6 In Bohr's theory, which explains the unique discrete lines emitted by atomic hydrogen, the specific wavelengths can be deduced from the energy differences between allowed states. The mathematical expression for total energy Eₙ is:

A -me⁴ / [8ε₀²n²h²]

B -me² / (8ε₀nh)

C me⁴ / [4ε₀²n²h²]

D -m²e⁴ / (8ε₀n²h²)

💡 Short Explanation

�� Bohr derived quantized energy levels. �� Energy varies as 1/n². �� Energy is negative for bound states.

✅ Correct Answer Explanation

The Bohr model gives: Eₙ = -me⁴ / (8ε₀²h²n²) This formula predicts hydrogen energy levels and explains spectral lines through energy differences. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect dependence on constants.
�� Option C → Positive sign is incorrect.
�� Option D → Wrong power of mass.

🎯 Strategy

Used

Formula Recall

Application:

Use standard Bohr energy expression.

Final Logic:

Hydrogen energy ∝ -1/n².

🧠 Memory Trick

Bohr Energy = Negative 1/n²

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

7 Statements about the ionisation energy of the hydrogen atom:
1. It is the minimum energy required to free the electron from the ground state.
2. Its value is exactly 13.6 eV.
3. The theoretical prediction disagrees heavily with the experimental value.
4. It corresponds to the energy required to reach n = ∞.

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

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

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

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

💡 Short Explanation

�� Ionisation removes the electron completely. �� Ground-state ionisation energy is 13.6 eV. �� Final state corresponds to n = ∞.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Statement 3 is incorrect because Bohr's prediction agrees well with experimental results for hydrogen. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Statement 3 is false.
�� Options B, C and D include statement 3.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement using hydrogen energy levels.

Final Logic:

Ionisation = n=1 → n=∞.

🧠 Memory Trick

13.6 eV Frees Electron

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

8 When a 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature:

A The electrons can be excited up to the n=3 state, as 12.09 eV is required to reach it.

B The electrons will permanently escape the atom, as it exceeds 13.6 eV.

C The ground state energy of hydrogen shifts to positive values.

D The ionisation energy dynamically drops below 10.2 eV to accommodate the collision.

💡 Short Explanation

�� Excitation to n=3 requires 12.09 eV. �� Ionisation requires 13.6 eV. �� 12.5 eV is insufficient for ionisation.

✅ Correct Answer Explanation

Energy needed: n=1 → n=3 ΔE = 13.6(1 − 1/9) = 12.09 eV Since 12.5 eV > 12.09 eV but < 13.6 eV, excitation to n=3 is possible, but ionisation is not. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ionisation requires 13.6 eV.
�� Option C → Ground state energy does not change.
�� Option D → Ionisation energy remains fixed.

🎯 Strategy

Used

Substitution

Application:

Compare supplied energy with excitation and ionisation energies.

Final Logic:

12.09 eV < 12.5 eV < 13.6 eV.

🧠 Memory Trick

12.09 → n=3, 13.6 → Ionisation

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

9 When an electron falls from an excited state to the ground state (n=1), a characteristic series of lines is emitted. As the principal quantum number n of the excited state increases, the minimum energy required to subsequently free the electron _______ and the energy gap to the ground state _______.

A Increases, Increases

B Increases, Decreases

C Decreases, Decreases

D Decreases, Increases

💡 Short Explanation

�� Outer states are less tightly bound. �� Ionisation energy decreases with n. �� Energy gap to ground increases.

✅ Correct Answer Explanation

As n increases: |Eₙ| becomes smaller. Hence ionisation energy from that level decreases. However, the difference between Eₙ and E₁ increases and approaches 13.6 eV. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Options A, B and C contradict hydrogen energy-level behavior.

🎯 Strategy

Used

Conceptual Analysis

Application:

Observe behavior of Eₙ = -13.6/n².

Final Logic:

Higher n → Easier ionisation, larger drop to ground.

🧠 Memory Trick

Higher n = Looser Electron

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

10 When analyzing transitions that drop back to lower energy states, the exact difference of energy is carried away by a photon. Since nf and ni are both restricted to integers, this mathematically ensures that:

A the atom will spiral into the nucleus within a fraction of a second.

B the light is radiated in various continuous, overlapping frequencies.

C light is radiated solely in various discrete frequencies.

D the angular momentum remains completely independent of n.

💡 Short Explanation

�� Energy levels are quantized. �� Only specific transitions occur. �� Photon frequencies are discrete.

✅ Correct Answer Explanation

Since nᵢ and nf can only take integer values, the allowed energy differences are discrete. Using: hν = Eᵢ − Ef Only specific values of ν are possible. Therefore, line spectra are produced.

❓ Why Other Options Are Incorrect

�� Option A → Contradicts Bohr model.
�� Option B → Continuous frequencies are not predicted.
�� Option D → Angular momentum depends on n.

🎯 Strategy

Used

Conceptual Matching

Application:

Use quantized energy-level theory.

Final Logic:

Discrete energy levels → Discrete frequencies.

🧠 Memory Trick

Discrete E → Discrete ν

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

11 Incorrect statement about the wave nature of electrons:

A Louis de Broglie argued that the electron in its circular orbit must be seen as a particle wave.

B The wave nature of electrons was experimentally verified by Davisson and Germer in 1927.

C Particle waves can lead to standing waves under precise resonant conditions.

D Waves of any arbitrary wavelength will survive upon reflection and dynamically form standing waves in the orbit.

💡 Short Explanation

�� Electron exhibits wave nature. �� Standing waves require resonance. �� Arbitrary wavelengths cannot persist.

✅ Correct Answer Explanation

According to de Broglie, electrons behave as matter waves. Only those wavelengths that satisfy the standing-wave condition can persist in an orbit. Waves with arbitrary wavelengths undergo destructive interference and disappear. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement of de Broglie's hypothesis.
�� Option B → Correct; Davisson-Germer verified electron diffraction.
�� Option C → Standing waves arise under resonant conditions.

🎯 Strategy

Used

Elimination

Application:

Identify the statement violating standing-wave conditions.

Final Logic:

Only resonant wavelengths survive.

🧠 Memory Trick

No Resonance = No Orbit

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

12 Statements concerning particle waves within Bohr orbits:
1. A standing wave is successfully formed only when the total orbital distance is an integral number of wavelengths.
2. The momentum p of the electron is given by mv assuming the speed is much less than light.
3. The corresponding de Broglie wavelength (λ) is equal to h/mv.
4. All fractional, continuous wavelengths form perfectly resonant standing waves.

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

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

C (1), (2), (3)

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

💡 Short Explanation

�� Standing waves require integral wavelengths. �� p = mv for non-relativistic motion. �� λ = h/p.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. For stable orbits: 2πr = nλ Also, p = mv λ = h/p = h/mv Statement 4 is incorrect because fractional wavelengths do not form stable standing waves. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Statement 4 is false.
�� Options A, B and D include statement 4.

🎯 Strategy

Used

Statement Verification

Application:

Check each statement using de Broglie theory.

Final Logic:

Integral wavelengths alone survive.

🧠 Memory Trick

Integer λ = Stable Orbit

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Matter Waves

13 For an electron in a circular resonant standing wave orbit, the quantum condition linking the full orbit circumference to the electron's momentum p is:

A 2πrₙ = n(h/p)

B 2πrₙ = n(p/h)

C rₙ = n[h/(2πp²)]

D 2πp = n(h/rₙ)

💡 Short Explanation

�� λ = h/p. �� Standing-wave condition is 2πr = nλ. �� Combine both relations.

✅ Correct Answer Explanation

Using: λ = h/p and 2πrₙ = nλ Substituting λ: 2πrₙ = n(h/p) Thus option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal relation is incorrect.
�� Option C → Incorrect derivation.
�� Option D → Dimensionally incorrect.

🎯 Strategy

Used

Substitution

Application:

Insert λ = h/p into standing-wave condition.

Final Logic:

2πr = n(h/p)

🧠 Memory Trick

Circumference = nλ

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

14 What specific condition ensures that an electron's wave does not quickly drop to zero amplitude due to self-interference in a circular orbit?

A The orbit circumference must be an odd multiple of half-wavelengths.

B The orbit circumference must precisely equal one, two, or any integral whole number of the de Broglie wavelength.

C The classical orbital velocity must match the speed of light perfectly.

D The principal quantum number n must be fractional.

💡 Short Explanation

�� Stable orbit requires resonance. �� Integral wavelengths fit the orbit. �� Destructive interference is avoided.

✅ Correct Answer Explanation

Stable standing waves occur only when: 2πr = nλ where n is an integer. This ensures constructive interference and prevents the wave amplitude from vanishing. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the general condition.
�� Option C → Speed need not equal c.
�� Option D → n must be an integer.

🎯 Strategy

Used

Conceptual Matching

Application:

Apply standing-wave criterion.

Final Logic:

Integral wavelengths produce stable resonance.

🧠 Memory Trick

Whole λ = Whole Orbit

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Standing Waves

15 Match List I with List II for the derivation of angular momentum from the de Broglie hypothesis:

List I	List II
1. λ	a. h/p
2. 2πrₙ	b. nλ
3. p	c. mv
4. mvrₙ	d. nh/2π

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

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

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

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

💡 Short Explanation

�� λ = h/p. �� p = mv. �� L = mvr = nh/2π.

✅ Correct Answer Explanation

List I — List II 1. λ — a. h/p 2. 2πrₙ — b. nλ 3. p — c. mv 4. mvrₙ — d. nh/2π Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Other options contain incorrect mathematical pairings.

🎯 Strategy

Used

Option Grouping

Application:

Match standard de Broglie relations.

Final Logic:

λ → p → Standing Wave → Quantised Angular Momentum.

🧠 Memory Trick

λ = h/p, L = nh/2π

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Quantisation from Matter Waves

16 Correct statements about why only certain quantised states persist without radiating:
1. Waves with non-integral wavelengths interfere with themselves and their amplitudes drop to zero.
2. The electron waves resonate, allowing the orbit to maintain dynamically stable standing waves.
3. The integer n in the angular momentum formula represents the number of full wavelengths fitting into the circumference.
4. The continuous emission spectrum matches these persistent non-radiating states exactly.

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

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

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

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

💡 Short Explanation

�� Standing waves explain quantisation. �� Resonance creates stability. �� n counts wavelengths.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Statement 4 is incorrect because stable Bohr states are associated with discrete energy levels, not a continuous emission spectrum. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Statement 4 is incorrect.
�� Options B, C and D include statement 4.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement using standing-wave theory.

Final Logic:

Quantisation arises from resonance, not continuous spectra.

🧠 Memory Trick

n = Number of Waves

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Standing-Wave Interpretation

17 The Bohr model is strictly applicable only to _______ atoms and entirely fails to account for the electrical forces between _______ in more complex atoms.

A Multi-electron, Protons

B Complex, Neutrons

C Hydrogenic, Electrons

D Neutral, Nucleons

💡 Short Explanation

�� Bohr model works for one-electron systems. �� Electron-electron interactions are ignored. �� Multi-electron atoms require quantum mechanics.

✅ Correct Answer Explanation

Bohr's model accurately describes hydrogen and hydrogen-like ions containing a single electron. It neglects electron-electron repulsion, making it unsuitable for complex atoms. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multi-electron atoms are not described accurately.
�� Option B → Neutron interactions are not the issue.
�� Option D → Nucleon interactions are irrelevant here.

🎯 Strategy

Used

Concept Recall

Application:

Identify the known limitation of Bohr's theory.

Final Logic:

Hydrogenic atoms only.

🧠 Memory Trick

Bohr = One Electron

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

18 If the hydrogen atom corresponds to atomic number Z = 1, what is the atomic number Z for which Bohr's model successfully predicts transition frequencies but still fundamentally fails to explain the relative intensity variations of the spectral lines?

A Z = 1

B Z = 2

C Z = 3

D Z = 4

💡 Short Explanation

�� Bohr predicts hydrogen frequencies well. �� Intensity prediction remains unsuccessful. �� Limitation exists even for hydrogen.

✅ Correct Answer Explanation

Bohr's model successfully predicts the spectral frequencies of hydrogen (Z = 1). However, it cannot explain the relative intensities of spectral lines, even in hydrogen. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Options B, C and D refer to atoms where additional limitations arise.

🎯 Strategy

Used

Conceptual Analysis

Application:

Separate frequency prediction from intensity prediction.

Final Logic:

Hydrogen frequencies yes, intensities no.

🧠 Memory Trick

Bohr: Position Yes, Strength No

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

19 Statements regarding the true picture of multi-electron complex atoms:
1. It firmly requires a new and radical theory based on Quantum Mechanics.
2. A given energy state is characterised generally by four quantum numbers (n, l, m, s).
3. Bohr's single quantum number n provides a complete and flawless picture for multi-electron atoms.
4. The electron-electron electrical interaction is comparable in magnitude to the electron-nucleus electrical force.

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

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

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

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

💡 Short Explanation

�� Multi-electron atoms need quantum mechanics. �� Four quantum numbers describe states. �� Electron-electron interactions are important.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Statement 3 is incorrect because a single quantum number cannot completely describe multi-electron atoms. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Statement 3 is false.
�� Options B, C and D include statement 3.

🎯 Strategy

Used

Statement Verification

Application:

Check modern quantum mechanical description.

Final Logic:

Complex atoms require more than n alone.

🧠 Memory Trick

Complex Atom = Four Quantum Numbers

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

20 Incorrect statement regarding classical Bohr orbits versus modern quantum mechanics:

A In strict quantum mechanics, Bohr's specific trajectories are retained as exact physical tracks for the electron.

B Bohr's exact orbital picture was ultimately found inconsistent with the uncertainty principle.

C In modern theory, Bohr's discrete orbits are replaced by regions where the electron may be found with large probability.

D A pure Coulomb potential, as in a hydrogen atom, allows the energy to depend solely on the principal quantum number n.

💡 Short Explanation

�� Exact trajectories are abandoned. �� Orbitals replace orbits. �� Probability interpretation is used.

✅ Correct Answer Explanation

Quantum mechanics does not permit exact electron trajectories. Instead, electrons are described by wavefunctions and probability distributions called orbitals. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Correct consequence of uncertainty principle.
�� Option C → Correct description of orbitals.
�� Option D → Correct for hydrogen-like atoms.

🎯 Strategy

Used

Elimination

Application:

Identify the statement contradicting quantum mechanics.

Final Logic:

Orbit ≠ Orbital.

🧠 Memory Trick

Bohr Orbit → Quantum Orbital

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Quantum Mechanical Model of Atom

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