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Atoms: Class 12 Physics Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

When a gas discharge tube containing hydrogen gas is connected to a high-voltage source, the hydrogen atoms get excited and emit light. This emitted light, when passed through a prism, produces a line spectrum — a series of distinct coloured lines rather than a continuous band of colours. Each line corresponds to a specific transition of an electron between two energy levels of the hydrogen atom. The visible lines form what is known as the Balmer series.

A science student reads that the hydrogen atom emits specific colours of light when electricity is passed through hydrogen gas. She finds that one of these colours lies in the visible region and corresponds to a photon of wavelength 656 nm.

(i) Name the spectral series to which this wavelength belongs and identify the quantum numbers n₁ and n₂ of the transition responsible for it.
(ii) Calculate the energy of the emitted photon in eV. (Given: h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)
(iii) Using Bohr's formula Eₙ = −13.6/n² eV, verify that the difference in energy levels matches the photon energy calculated in part (ii).
(iv) Why does a hydrogen atom emit light only at specific wavelengths and not a continuous spectrum? (Answer in one or two sentences.)

Show answer
(i) The spectral series is the Balmer series.

The Balmer series corresponds to transitions where the electron falls to the n₁ = 2 level from higher levels.

For λ = 656 nm (the red line, H<sub>α</sub>), the transition is from n₂ = 3 → n₁ = 2.

∴ n₁ = 2, n₂ = 3.

(ii) Energy of the emitted photon:

By Planck's quantum formula, E = hν = hc/λ

Substituting:
E = (6.63 × 10<sup>−34</sup> J s × 3 × 10<sup>8</sup> m/s) / (656 × 10<sup>−9</sup> m)

E = (19.89 × 10<sup>−26</sup>) / (656 × 10<sup>−9</sup>)

E = 3.032 × 10<sup>−19</sup> J

Converting to eV:
E = (3.032 × 10<sup>−19</sup>) / (1.6 × 10<sup>−19</sup>)

∴ E ≈ 1.90 eV

(iii) Using Bohr's formula Eₙ = −13.6/n² eV:

Energy of n = 3 level:
E₃ = −13.6 / 3² = −13.6 / 9 = −1.51 eV

Energy of n = 2 level:
E₂ = −13.6 / 2² = −13.6 / 4 = −3.40 eV

Energy difference:
ΔE = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV

This matches the photon energy calculated in part (ii) (≈ 1.90 eV), confirming the transition.

∴ The Bohr energy levels correctly account for the emitted photon energy.

(iv) According to Bohr's postulate, electrons in a hydrogen atom can only occupy discrete (quantised) energy levels. A photon is emitted only when an electron makes a transition from a higher to a lower energy level, and its energy equals exactly the difference between those two levels. Since only specific energy differences are possible, only specific wavelengths are emitted — giving a line spectrum, not a continuous spectrum.
Q2Case-based4 marks

A school student observes a bright red line (λ ≈ 656 nm) in the hydrogen emission spectrum using a discharge tube. She incorrectly identifies it as the n = 3 → n = 1 transition. The ground state energy of the hydrogen atom is −13.6 eV. Use Bohr's model throughout.

A school student reads that the hydrogen atom emits light when an electron 'falls' from a higher energy level to a lower one. Curious, she sets up a discharge tube containing hydrogen gas and observes a bright red line in the visible spectrum. She identifies this as the transition from n = 3 to n = 1.

(i) Identify the error in her identification. State which series this red line actually belongs to and write the correct transition responsible for it.
(ii) Calculate the energy of the photon emitted in the correct transition you identified in part (i). (Ground state energy of hydrogen atom = −13.6 eV)
(iii) Find the angular momentum of the electron in the orbit from which it falls (higher level of the correct transition).
(iv) If the student now increases the voltage across the discharge tube such that the electron in the hydrogen atom is completely removed from n = 2 level, what minimum energy (in eV) must the photon supply?

Show answer
(i) Identification of error and correct transition:

The red line (λ ≈ 656 nm) lies in the VISIBLE region of the spectrum.

According to Bohr's model, transitions that end at n = 1 (Lyman series) emit photons in the ULTRAVIOLET region — NOT in the visible region.

∴ The student's identification is INCORRECT.

The red line in the visible region belongs to the BALMER SERIES (transitions ending at n = 2).

The correct transition responsible for the red line (Hα line) is: n = 3 → n = 2.

(ii) Energy of the photon emitted (n = 3 → n = 2):

By Bohr's formula, energy of nth orbit:
En = −13.6 / n² eV

E₃ = −13.6 / (3)² = −13.6 / 9 = −1.511 eV

E₂ = −13.6 / (2)² = −13.6 / 4 = −3.4 eV

Energy of emitted photon:
E = E₃ − E₂ = (−1.511) − (−3.4)

E = −1.511 + 3.4

∴ Energy of emitted photon = 1.89 eV

(iii) Angular momentum of electron in n = 3 orbit:

By Bohr's quantisation condition:
mvr = nh / 2π

For n = 3:
L₃ = 3h / 2π

L₃ = (3 × 6.63 × 10⁻³⁴) / (2π)

L₃ = (19.89 × 10⁻³⁴) / (6.283)

∴ L₃ = 3.165 × 10⁻³⁴ J s ≈ 3.17 × 10⁻³⁴ J s

(iv) Minimum energy to ionise the atom from n = 2:

Ionisation means removing the electron completely, i.e., taking it to E = 0 (n → ∞).

Minimum energy required = 0 − E₂ = 0 − (−3.4 eV)

∴ Minimum energy required = 3.4 eV
Q3Case-based4 marks

Bohr's model of the hydrogen atom gives the energy of the electron in the nth orbit as Eₙ = −13.6/n² eV. The ground state (n = 1) energy is −13.6 eV. When radiation of appropriate frequency falls on the atom, the electron can absorb a photon and jump to a higher orbit. The reverse process — falling back to a lower orbit — releases a photon of energy equal to the difference in orbital energies.

A school student reads that the hydrogen atom in its ground state has an electron revolving in the first Bohr orbit of radius 0.529 Å. The student is curious about what happens when the electron absorbs exactly 12.09 eV of energy.

(i) Identify the orbit to which the electron gets excited. Show your working using Bohr's energy formula.

(ii) When the electron returns to the ground state, it emits a photon. Calculate the wavelength of this photon. (Given: h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

(iii) To which spectral series does this emitted photon belong, and in which region of the electromagnetic spectrum does it lie?

(iv) The student claims: 'If the intensity of the incident radiation is doubled, the electron will jump to an even higher orbit.' Is the student correct? Give a reason.

Show answer
(i) Identifying the excited orbit

By Bohr's energy formula: Eₙ = −13.6 / n² eV

Energy in ground state: E₁ = −13.6 eV

After absorbing 12.09 eV:
E_final = E₁ + 12.09 = −13.6 + 12.09 = −1.51 eV

Setting −13.6 / n² = −1.51:
n² = 13.6 / 1.51 = 9.006 ≈ 9
∴ n = 3

∴ The electron gets excited to the n = 3 (second excited) orbit.

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(ii) Wavelength of the emitted photon (transition n = 3 → n = 1)

Energy of emitted photon:
E = E₃ − E₁ = (−1.51) − (−13.6) = 12.09 eV

Converting to joules:
E = 12.09 × 1.6 × 10⁻¹⁹ = 1.934 × 10⁻¹⁸ J

Using E = hc / λ:
λ = hc / E
λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.934 × 10⁻¹⁸)
λ = (1.989 × 10⁻²⁵) / (1.934 × 10⁻¹⁸)

∴ λ ≈ 1.028 × 10⁻⁷ m = 102.8 nm

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(iii) Spectral series and region

The electron falls from n = 3 to the ground state n = 1.
All transitions ending at n = 1 belong to the Lyman series.

∴ The emitted photon belongs to the Lyman series and lies in the ultraviolet (UV) region of the electromagnetic spectrum.

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(iv) Effect of doubling intensity

The student's claim is incorrect.

According to Bohr's model and the photon picture of light (quantum theory), the excitation of an electron to a higher orbit depends solely on the energy (frequency) of individual photons — not on the intensity of radiation.

Intensity determines the number of photons per unit area per second. Doubling the intensity doubles the number of photons, but does not change the energy of each photon.

Since no individual photon now carries more than 12.09 eV, no electron can jump beyond n = 3.

∴ The student is incorrect; the electron will still be excited only to n = 3 (or not at all if no photon has sufficient energy).
Q4Case-based4 marks

Rutherford's α-particle scattering experiment revealed the nuclear model of the atom. In this experiment, α-particles are fired at a thin metal foil, and their scattering patterns are studied. When an α-particle is fired directly (head-on) at a nucleus, it decelerates due to electrostatic repulsion and momentarily stops at a distance called the distance of closest approach (r₀), where all its kinetic energy is converted into electrostatic potential energy: KE = (1/4πε₀)(2Ze²/r₀).

A school physics club sets up a simulation of Rutherford's α-particle scattering experiment. They fire α-particles of kinetic energy 1.6 MeV at a gold (Z = 79) target foil and record the scattering pattern.

(i) An α-particle is fired head-on at the gold nucleus. Using the concept of distance of closest approach, calculate the distance of closest approach for this α-particle. (Given: 1/4πε₀ = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C, kinetic energy of α-particle = 1.6 MeV)

(ii) The students replace the gold target with a silver (Z = 47) target, keeping the kinetic energy of the α-particles the same. Will the distance of closest approach increase, decrease, or remain the same? Justify your answer.

(iii) A student claims: 'Most α-particles pass straight through the foil with little or no deflection.' Is this claim correct? What does this observation tell us about the structure of the atom?

(iv) Another student argues: 'If we double the kinetic energy of the α-particles used with the gold target, the distance of closest approach will also double.' Is this argument correct? Give a reason.

Show answer
(i) At the distance of closest approach, the entire kinetic energy of the α-particle is converted into electrostatic potential energy.

By energy conservation:
KE = (1/4πε₀) · (2Ze²/r₀)

Solving for r₀:
r₀ = (1/4πε₀) · (2Ze²/KE)

Substituting values (KE = 1.6 MeV = 1.6 × 10⁶ × 1.6 × 10⁻¹⁹ J = 2.56 × 10⁻¹³ J):
r₀ = (9 × 10⁹) × (2 × 79 × (1.6 × 10⁻¹⁹)²) / (2.56 × 10⁻¹³)
r₀ = (9 × 10⁹ × 158 × 2.56 × 10⁻³⁸) / (2.56 × 10⁻¹³)
r₀ = (9 × 10⁹ × 158 × 10⁻³⁸) / 10⁻¹³
r₀ = 9 × 158 × 10⁻¹⁶
r₀ = 1422 × 10⁻¹⁶ m

∴ r₀ ≈ 1.42 × 10⁻¹³ m

(ii) The distance of closest approach is given by:
r₀ = (1/4πε₀) · (2Ze²/KE)

Since r₀ ∝ Z, and silver has Z = 47 < Z = 79 (gold), the distance of closest approach will decrease when gold is replaced by silver (with the same kinetic energy).

∴ The distance of closest approach will decrease for the silver target.

(iii) Yes, the student's claim is correct. Most α-particles pass straight through (or are deflected by very small angles) because the atom is mostly empty space, and the positive charge (nucleus) is concentrated in an extremely small, dense region at the centre. Only those α-particles that pass very close to the nucleus experience a strong repulsive force and are deflected through large angles. This observation established the nuclear model of the atom: the atom has a tiny, massive, positively charged nucleus surrounded by mostly empty space in which electrons revolve.

(iv) No, the argument is incorrect. The distance of closest approach is given by:
r₀ = (1/4πε₀) · (2Ze²/KE)

Since r₀ ∝ 1/KE, if the kinetic energy is doubled, the distance of closest approach will be halved (not doubled).

∴ Doubling the kinetic energy reduces r₀ to half its original value.
Q5Case-based4 marks

Hydrogen gas in a discharge tube emits light that, when passed through a prism, produces distinct spectral lines. These lines arise due to electrons in excited hydrogen atoms making transitions between allowed energy levels as described by Bohr's model. The Balmer series corresponds to all transitions ending at n = 2 and lies partly in the visible region. One prominent line in this series appears at 486 nm.

A science student reads that the light emitted by hydrogen gas in a discharge tube produces a series of spectral lines. She observes that one particular line in the Balmer series has a wavelength of 486 nm (blue-green light). Using this information, answer the following questions:

(i) Identify the transition (n₂ → n₁) responsible for producing a wavelength of 486 nm in the Balmer series.

(ii) Using Bohr's model, calculate the energy (in eV) of the photon emitted during this transition. (Given: R = 1.097 × 10⁷ m⁻¹, hc = 1240 eV·nm)

(iii) The student increases the temperature of the gas significantly. Will the intensity of this spectral line increase or decrease? Give one reason.

(iv) If the electron in a hydrogen atom is in the n = 4 orbit, how many different spectral lines can be emitted when it de-excites to the ground state? Name the series to which the line emitted during the n = 4 → n = 1 transition belongs.

Show answer
(i) Identifying the transition:

For the Balmer series, the lower level is n₁ = 2.

Using the Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²)

→ 1/(486 × 10⁻⁹) = 1.097 × 10⁷ × (1/4 − 1/n₂²)

→ 2.057 × 10⁶ = 1.097 × 10⁷ × (1/4 − 1/n₂²)

→ 1/4 − 1/n₂² = 0.1875

→ 1/n₂² = 0.25 − 0.1875 = 0.0625

→ n₂² = 16

∴ n₂ = 4

∴ The transition responsible is n = 4 → n = 2 (i.e., n₂ = 4 to n₁ = 2).

(ii) Energy of the emitted photon:

By Planck's quantum condition, the energy of a photon is:
E = hc/λ

→ E = 1240 eV·nm / 486 nm

→ E = 2.552 eV

∴ Energy of the emitted photon = 2.55 eV (approximately).

(iii) Effect of increasing temperature on intensity:

The intensity of this spectral line will increase.

Because at higher temperatures, more hydrogen atoms are thermally excited to higher energy levels (n ≥ 4). As more atoms undergo the n = 4 → n = 2 transition per unit time, more photons of this wavelength are emitted, increasing the intensity of the 486 nm line.

(iv) Number of spectral lines from n = 4:

When an electron de-excites from n = 4 to the ground state (n = 1), the number of possible spectral lines is given by:

Number of lines = n(n − 1)/2

→ Number of lines = 4 × 3 / 2 = 6

∴ 6 different spectral lines can be emitted.

The transition n = 4 → n = 1 ends at n = 1, which belongs to the Lyman series (lies in the ultraviolet region).
Q6Case-based4 marks

James Webb Space Telescope observations of distant galaxies involve studying redshifted hydrogen spectral lines. To interpret such spectra, knowledge of Bohr's model applied to hydrogen-like ions is essential. For a one-electron ion of nuclear charge Ze, the radius of the n-th Bohr orbit is r_n = (n²a₀)/Z and the total energy is E_n = −(Z²×13.6)/n² eV, where a₀ = 0.529 Å is the Bohr radius. The Rydberg formula for such ions gives: 1/λ = RZ²(1/n₁² − 1/n₂²), where R = 1.097×10⁷ m⁻¹.

A science student reads that James Webb Space Telescope has detected a distant galaxy whose hydrogen spectral lines are significantly redshifted. To understand this, she revisits Bohr's model. Consider a hypothetical one-electron ion with nuclear charge Ze (where Z is the atomic number). She notes that in such an ion, the radius of the n-th orbit is r_n = (n²a₀)/Z and the energy of the electron in the n-th orbit is E_n = −(Z²×13.6)/n² eV.

(i) Using Bohr's model, write the expression for the speed of the electron in the n-th orbit of this ion. On what factors does it depend? (1 mark)

(ii) The student observes that the wavelength of the first line of the Lyman series (n = 2 → n = 1) of He⁺ (Z = 2) is λ₁. A singly ionised lithium atom Li²⁺ (Z = 3) also emits radiation for the same transition (n = 2 → n = 1). Find the ratio λ₁/λ₂, where λ₂ is the wavelength of the corresponding line in Li²⁺. (1 mark)

(iii) The student notices that the ionisation energy of He⁺ from its ground state is four times that of hydrogen. Using the given formula E_n = −(Z²×13.6)/n² eV, verify this statement. (1 mark)

(iv) In the context of Bohr's model, explain why the total energy of the electron in any orbit is negative, and what this implies about the stability of the atom. (1 mark)

Show answer
(i) Speed of electron in the n-th Bohr orbit of a hydrogen-like ion:

By Bohr's postulate, the necessary centripetal force is provided by the Coulomb electrostatic attraction:

(mv²)/r_n = kZe²/r_n²

→ mv² = kZe²/r_n

Using the quantisation of angular momentum: mvr_n = nh/2π → r_n = nh/(2πmv)

Substituting and simplifying:

∴ v_n = Ze²/(2ε₀nh) = (Z/n) × e²/(2ε₀h)

The speed v_n depends on: (i) the atomic number Z (v_n ∝ Z) and (ii) the principal quantum number n (v_n ∝ 1/n). It does NOT depend on the mass of the electron or the orbit radius directly.

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(ii) Ratio of wavelengths λ₁/λ₂ for the n = 2 → n = 1 transition in He⁺ and Li²⁺:

By the Rydberg formula for hydrogen-like ions:

1/λ = RZ²(1/n₁² − 1/n₂²)

For He⁺ (Z = 2), transition n = 2 → n = 1:
1/λ₁ = R(2)²(1/1² − 1/2²) = 4R(1 − 1/4) = 4R × (3/4) = 3R

For Li²⁺ (Z = 3), same transition:
1/λ₂ = R(3)²(1/1² − 1/2²) = 9R × (3/4) = (27R)/4

∴ λ₁/λ₂ = (1/λ₂)/(1/λ₁) = [(27R/4)] / [3R] = (27/4) / 3 = 27/12

∴ λ₁/λ₂ = 9/4

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(iii) Verification that ionisation energy of He⁺ is four times that of hydrogen:

The ionisation energy from the ground state (n = 1) is the energy required to remove the electron from n = 1 to n = ∞, i.e., E_ionisation = 0 − E₁ = |E₁|.

For hydrogen (Z = 1):
E₁(H) = −(1² × 13.6)/1² = −13.6 eV
→ Ionisation energy of H = 13.6 eV

For He⁺ (Z = 2):
E₁(He⁺) = −(2² × 13.6)/1² = −(4 × 13.6) = −54.4 eV
→ Ionisation energy of He⁺ = 54.4 eV

Ratio = 54.4 / 13.6 = 4

∴ The ionisation energy of He⁺ from the ground state is indeed four times that of hydrogen. (Verified)

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(iv) Physical significance of the negative total energy:

In Bohr's model, the total energy of the electron is the sum of its kinetic energy (K, always positive) and potential energy (U, negative, since the electron is bound to the nucleus by attractive Coulomb force).

It can be shown that K = −E_total and U = 2E_total, so |U| > K, making the total energy negative.

The negative value of total energy means the electron is in a bound state — it does not have enough energy to escape the nucleus. Energy must be supplied from outside to ionise the atom (bring E to zero or above).

∴ The negative total energy implies that the electron is bound to the nucleus, and this is precisely the condition for the stability of the atom. The atom is stable because the electron cannot escape spontaneously.
Q7Case-based4 marks

Bohr's quantisation condition, originally proposed for electrons in atoms, is a universal principle applicable to any circular orbit. For a particle of mass m moving with speed v in a circular orbit of radius r, Bohr's condition states that the angular momentum is an integral multiple of h/2π. When this condition is applied to macroscopic objects such as satellites, the resulting quantum number n turns out to be astronomically large, which is the reason classical mechanics works perfectly for such objects.

A science student reads that a satellite of mass 500 kg revolves around the Earth in a circular orbit of radius 8.0 × 10⁶ m with an orbital speed of 7.1 × 10³ m/s. Inspired by Bohr's model of the hydrogen atom, the student applies Bohr's quantisation condition to this satellite–Earth system.

(i) State Bohr's quantisation condition for angular momentum.
(ii) Calculate the quantum number n that characterises the satellite's orbit. (h = 6.63 × 10⁻³⁴ J s)
(iii) What does the extremely large value of n imply about the satellite's motion?
(iv) If an electron in a hydrogen atom has the same angular momentum as this satellite, write the expression for its total energy in that orbit.

Show answer
(i) Bohr's Quantisation Condition:
Bohr's quantisation condition states that the angular momentum of a particle revolving in a circular orbit is an integral multiple of h/2π.

∴ L = mvr = nh/2π, where n = 1, 2, 3, …

(ii) Calculation of quantum number n:

By Bohr's quantisation condition:
n = mvr/(h/2π) = 2πmvr/h

Substituting the values:
m = 500 kg, v = 7.1 × 10³ m/s, r = 8.0 × 10⁶ m, h = 6.63 × 10⁻³⁴ J s

n = (2π × 500 × 7.1 × 10³ × 8.0 × 10⁶) / (6.63 × 10⁻³⁴)

Numerator: 2π × 500 × 7.1 × 10³ × 8.0 × 10⁶
= 2 × 3.14 × 500 × 7.1 × 10³ × 8.0 × 10⁶
= 6.28 × 500 × 56.8 × 10⁹
= 6.28 × 2.84 × 10¹³
= 1.784 × 10¹⁴

n = (1.784 × 10¹⁴) / (6.63 × 10⁻³⁴)

∴ n ≈ 2.69 × 10⁴⁷

(iii) Physical implication of very large n:
The quantum number n ≈ 10⁴⁷ is an astronomically large number. This implies that the energy levels of the satellite are so closely spaced (nearly continuous) that the discrete, quantised nature of the orbit is completely undetectable. The transition between adjacent levels (n → n − 1) releases negligible energy. Therefore, for macroscopic objects like satellites, quantum effects are imperceptible and classical mechanics applies perfectly. This is consistent with Bohr's Correspondence Principle.

(iv) Expression for total energy of an electron with the same angular momentum:
If an electron has angular momentum L = mvr = nh/2π (same n as the satellite), then by Bohr's model for hydrogen:

Total energy: Eₙ = −13.6/n² eV

∴ Eₙ = −13.6/n² eV, where n ≈ 2.69 × 10⁴⁷

(Since n is extremely large, Eₙ ≈ 0 eV, meaning the electron is essentially free — at the ionisation limit.)
Q8MCQ1 mark

In Bohr's model of the hydrogen atom, the radius of the first orbit (n = 1) is a₀ = 0.529 Å. What is the radius of the third orbit (n = 3)?

Show answer
Option (B) is correct.

Explanation: By Bohr's model, the radius of the n-th orbit is given by rₙ = n² a₀.

For n = 3: r₃ = (3)² × 0.529 Å = 9 × 0.529 Å

∴ r₃ = 4.761 Å
Q9MCQ1 mark

The radius of the first orbit (n = 1) of the hydrogen atom is a₀ = 0.529 Å. What is the radius of the third orbit (n = 3)?

Show answer
Option (B) is correct.

Explanation: By Bohr's model, the radius of the n-th orbit is given by rₙ = n² a₀.

For n = 3: r₃ = (3)² × 0.529 Å = 9 × 0.529 Å

∴ r₃ = 4.761 Å
Q10MCQ1 mark

The total energy of an electron in the second excited state (n = 3) of a hydrogen atom is:

Show answer
Option (C) is correct.

Explanation: By Bohr's formula, the total energy of an electron in the n-th orbit of hydrogen is

Eₙ = −13.6 / n² eV

The second excited state corresponds to n = 3.

∴ E₃ = −13.6 / 3² = −13.6 / 9 ≈ −1.51 eV
Q11MCQ1 mark

In the Bohr model of hydrogen atom, the speed of the electron in the n<sup>th</sup> orbit is proportional to:

Show answer
Option (C) is correct.

Explanation: By Bohr's quantisation condition, the angular momentum of the electron in the n<sup>th</sup> orbit is:

mv<sub>n</sub>r<sub>n</sub> = nh/2π

Since r<sub>n</sub> ∝ n<sup>2</sup>, substituting gives v<sub>n</sub> ∝ n/n<sup>2</sup>, i.e., v<sub>n</sub> ∝ 1/n.

∴ The speed of the electron in the n<sup>th</sup> orbit is proportional to 1/n.
Q12MCQ1 mark

According to Bohr's model, the radius of the first orbit of hydrogen atom is a₀. What is the radius of the third orbit?

Show answer
Option (C) is correct.

Explanation: According to Bohr's model, the radius of the n-th orbit is given by rₙ = n²a₀, where a₀ is the radius of the first orbit (n = 1).

For n = 3: r₃ = (3)²a₀ = 9a₀.
Q13MCQ1 mark

The kinetic energy of an electron in the ground state of the hydrogen atom is:

Show answer
Option (B) is correct.

Explanation: By Bohr's model, the total energy of an electron in the n-th orbit is Eₙ = −13.6/n² eV. The kinetic energy is always positive and equals +13.6/n² eV, while the potential energy equals −27.2/n² eV (twice the total energy in magnitude, and negative).

For the ground state n = 1:
∴ Kinetic energy = +13.6 eV.
Q14Short Answer2 marks

Using Bohr's second postulate, derive an expression for the speed of an electron in the n-th orbit of a hydrogen atom. In which orbit is the electron moving fastest?

Show answer
Bohr's second postulate states that the angular momentum of an electron in a stationary orbit is an integral multiple of h/2π:

mvr = nh/2π … (1)

For circular motion, the electrostatic force provides the centripetal force:

ke²/r² = mv²/r → r = ke²/mv² … (2)

Substituting (2) into (1):

mv(ke²/mv²) = nh/2π

→ ke²/v = nh/2π

∴ vₙ = ke²/nh·2π = e²/4πε₀ · 2π/nh = e²/2ε₀nh

Or equivalently: vₙ = e²/(2ε₀nh)

Since vₙ ∝ 1/n, the speed decreases as n increases.

∴ The electron moves fastest in the first orbit (n = 1), i.e., the ground state.
Q15Short Answer2 marks

Name the spectral series of hydrogen that lies in the ultraviolet region. Using the Rydberg formula, find the ratio of the shortest to the longest wavelength of this series.

Show answer
The spectral series of hydrogen that lies in the ultraviolet region is the Lyman series (transitions to n = 1).

By the Rydberg formula:

1/λ = R(1/n₁² − 1/n₂²)

For the Lyman series, n₁ = 1 and n₂ = 2, 3, 4, … ∞.

For the longest wavelength (λ_max): n₂ = 2
1/λ_max = R(1/1² − 1/2²) = R(1 − 1/4) = 3R/4
→ λ_max = 4/3R

For the shortest wavelength (λ_min): n₂ = ∞
1/λ_min = R(1/1² − 0) = R
→ λ_min = 1/R

∴ λ_min/λ_max = (1/R)/(4/3R) = 3/4

∴ The ratio of the shortest to the longest wavelength of the Lyman series is 3 : 4.
Q16Short Answer2 marks

In Bohr's model of the hydrogen atom, the radius of the first orbit (n = 1) is a₀ = 0.529 Å. Find the radius of the third orbit (n = 3) and the speed of the electron in that orbit, given that the speed in the first orbit is v₁ = 2.18 × 10⁶ m/s.

Show answer
By Bohr's model, the radius of the n-th orbit is given by:

rₙ = n² a₀

For n = 3:

r₃ = (3)² × 0.529 Å = 9 × 0.529 Å

∴ r₃ = 4.761 Å ≈ 4.76 Å

The speed of the electron in the n-th orbit is given by:

vₙ = v₁ / n

For n = 3:

v₃ = (2.18 × 10⁶) / 3

∴ v₃ = 0.727 × 10⁶ m/s ≈ 7.27 × 10⁵ m/s
Q17Short Answer3 marks

A science student is studying spectral emission from hydrogen-like atoms. She observes that when a hydrogen atom absorbs energy and the electron gets excited to the n = 4 orbit, it subsequently de-excites and emits photons of various wavelengths.

(i) How many different spectral lines can be emitted when the electron de-excites from n = 4 to the ground state? Give the formula used.

(ii) The student notices that one of the emitted photons corresponds to the transition n = 4 → n = 2. Identify the spectral series to which this line belongs. Calculate the wavelength of this photon.
(Given: Rydberg constant R = 1.097 × 10⁷ m⁻¹)

(iii) She now replaces hydrogen with a singly ionised helium atom (He⁺, Z = 2) in the same experiment. The electron in He⁺ is also excited to n = 4. Will the wavelength of the photon emitted for the analogous n = 4 → n = 2 transition in He⁺ be longer or shorter than that in hydrogen? Justify without detailed calculation.

(iv) The radius of the n = 1 orbit in hydrogen is 0.529 Å. What is the radius of the n = 4 orbit in He⁺ (Z = 2)?

Show answer
(i) Number of spectral lines emitted when electron de-excites from n = 4:

The formula for the number of spectral lines possible from the nᵗʰ orbit is:

Number of lines = n(n − 1) / 2

Here n = 4:

Number of lines = 4(4 − 1) / 2 = 4 × 3 / 2 = 6

∴ 6 different spectral lines can be emitted.

[The transitions are: 4→1, 4→2, 4→3, 3→1, 3→2, 2→1]

(ii) Spectral series identification and wavelength calculation:

The transition n = 4 → n = 2 falls in the Balmer series (transitions to n₁ = 2 lie in the visible region — Balmer series).

By Rydberg formula:

1/λ = R (1/n₁² − 1/n₂²)

where n₁ = 2, n₂ = 4, R = 1.097 × 10⁷ m⁻¹

1/λ = 1.097 × 10⁷ × (1/2² − 1/4²)

1/λ = 1.097 × 10⁷ × (1/4 − 1/16)

1/λ = 1.097 × 10⁷ × (4/16 − 1/16)

1/λ = 1.097 × 10⁷ × (3/16)

1/λ = 1.097 × 10⁷ × 0.1875

1/λ = 2.057 × 10⁶ m⁻¹

∴ λ = 1 / (2.057 × 10⁶) ≈ 4.86 × 10⁻⁷ m = 486 nm

∴ The wavelength of the n = 4 → n = 2 photon in hydrogen is 486 nm (blue-green line of the Balmer series).

(iii) Wavelength comparison for He⁺ (Z = 2) vs. hydrogen (Z = 1):

For a hydrogen-like atom with atomic number Z, the Rydberg formula modifies to:

1/λ = RZ² (1/n₁² − 1/n₂²)

For He⁺, Z = 2, so 1/λ is multiplied by Z² = 4 compared to hydrogen for the same transition.

Since 1/λ is larger for He⁺, λ is smaller (shorter wavelength).

∴ The wavelength of the n = 4 → n = 2 photon in He⁺ is shorter than that in hydrogen (approximately λ_H / 4 ≈ 121.5 nm, in the UV region).

Physical reason: The nuclear charge Z = 2 in He⁺ pulls the electron more strongly, causing larger energy gaps between orbits, so the emitted photon carries more energy and hence has a shorter wavelength.

(iv) Radius of n = 4 orbit in He⁺:

The radius of the nᵗʰ orbit in a hydrogen-like atom is given by:

rₙ = (n² / Z) × a₀

where a₀ = 0.529 Å (Bohr radius), n = 4, Z = 2

r₄ = (4² / 2) × 0.529 Å

r₄ = (16 / 2) × 0.529 Å

r₄ = 8 × 0.529 Å

∴ r₄ = 4.232 Å ≈ 4.23 Å
Q18Short Answer3 marks

A muonic hydrogen atom is a bound system in which a muon (a particle with the same charge as an electron but mass m_μ = 207 mₑ, where mₑ = 9.1 × 10⁻³¹ kg) replaces the electron orbiting a proton. Using Bohr's model adapted for this system:
(i) Derive an expression for the radius of the n-th orbit of the muonic hydrogen atom in terms of the Bohr radius a₀ (= 0.529 Å). Hence find the radius of its first orbit (n = 1). [2 marks]
(ii) The muon transitions from the n = 2 orbit to the n = 1 orbit. Calculate the energy of the photon emitted. (Ground state energy of ordinary hydrogen = −13.6 eV) [2 marks]

Show answer
PART (i) — Radius of n-th orbit of muonic hydrogen [2 marks]

According to Bohr's first postulate, the electron (here, the muon) moves in a stable circular orbit where the Coulomb attraction provides the centripetal force, and by Bohr's quantisation postulate, angular momentum is quantised:

m_μ v r_n = n(h/2π) … (1)

Coulomb force provides centripetal force:

ke²/r_n² = m_μ v²/r_n … (2)

From (1): v = nh/(2π m_μ r_n). Substituting into (2):

ke²/r_n² = m_μ/r_n · [nh/(2π m_μ r_n)]²

ke²/r_n = n²h²/(4π² m_μ r_n²)

→ r_n = n²h²/(4π² m_μ ke²)

For ordinary hydrogen: a₀ = h²/(4π² mₑ ke²) = 0.529 Å

→ r_n (muonic) = n² · h²/(4π² m_μ ke²) = n² · a₀ · (mₑ/m_μ)

∴ r_n = n² · a₀/207

For n = 1:

r₁ = a₀/207 = 0.529 Å / 207

∴ r₁ = 2.56 × 10⁻³ Å = 2.56 × 10⁻¹³ m

(This is about 207 times smaller than ordinary hydrogen's first Bohr orbit.)

---

PART (ii) — Energy of photon emitted in n=2 → n=1 transition [2 marks]

The energy of the n-th orbit is given by Bohr's model:

E_n = −(m_particle · ke⁴ · 4π²)/(2n²h²) = −E₁/n²

For ordinary hydrogen, the ground-state energy E₁ = −13.6 eV corresponds to mass mₑ.
For the muonic atom, E₁(muonic) scales with the particle mass:

E₁(muonic) = E₁(H) × (m_μ/mₑ) = −13.6 eV × 207

E₁(muonic) = −2815.2 eV ≈ −2.815 × 10³ eV

Energy of n = 2 orbit:
E₂(muonic) = E₁(muonic)/4 = −2815.2/4 = −703.8 eV

Energy of photon emitted (n = 2 → n = 1):
E_photon = E₂ − E₁ = −703.8 − (−2815.2)

E_photon = 2815.2 − 703.8 = 2111.4 eV

∴ E_photon ≈ 2.11 × 10³ eV = 2.11 keV

(This photon lies in the X-ray region of the electromagnetic spectrum, far higher in energy than the 10.2 eV Lyman-α photon of ordinary hydrogen — a direct consequence of the muon's much greater mass.)
Q19Short Answer3 marks

A research team studying hydrogen-like ions observes spectral emission lines from a doubly ionised lithium atom (Li²⁺, Z = 3). One of the observed photons has a wavelength of 11.4 nm (in the UV region). Using this data, answer the following:

(i) Using Bohr's model, write the expression for the energy of the n-th orbit of a hydrogen-like ion with atomic number Z. Calculate the energy (in eV) of the ground state of Li²⁺.

(ii) Identify between which two energy levels (n_i → n_f) this UV photon (λ = 11.4 nm) was emitted. (Take R = 1.097 × 10⁷ m⁻¹, hc = 1242 eV·nm)

(iii) The same research team now replaces Li²⁺ with He⁺ (Z = 2) and finds that a certain spectral line of He⁺ has exactly the same wavelength as the first line of the Lyman series of hydrogen. Verify whether this is possible. If yes, find the transition in He⁺ that produces this wavelength.

Show answer
MARKING SCHEME (4 marks total: 1 + 1½ + 1½)

─────────────────────────────────────────
(i) [1 mark]
─────────────────────────────────────────

By Bohr's model, the energy of the n-th orbit of a hydrogen-like ion with atomic number Z is:

E_n = −13.6 × Z² / n² eV

For Li²⁺ (Z = 3), ground state (n = 1):

E₁ = −13.6 × (3)² / (1)²
= −13.6 × 9

∴ E₁(Li²⁺) = −122.4 eV

─────────────────────────────────────────
(ii) [1½ marks]
─────────────────────────────────────────

The energy of the emitted photon:

E_photon = hc / λ = 1242 eV·nm / 11.4 nm

∴ E_photon ≈ 108.9 eV

By Bohr's frequency condition, the transition energy equals the photon energy:

E_ni − E_nf = E_photon

13.6 × Z² (1/n_f² − 1/n_i²) = 108.9 eV

Since the photon is in the UV region and is emitted, the transition ends at a lower level. For Li²⁺ (Z = 3), checking n_f = 1:

13.6 × 9 × (1/1² − 1/n_i²) = 108.9

122.4 × (1 − 1/n_i²) = 108.9

1 − 1/n_i² = 108.9 / 122.4 = 0.8897

1/n_i² = 1 − 0.8897 = 0.1103

n_i² ≈ 9.07 → n_i ≈ 3

∴ The transition is n_i = 3 → n_f = 1 (i.e., the first line of the Lyman-equivalent series of Li²⁺).

─────────────────────────────────────────
(iii) [1½ marks]
─────────────────────────────────────────

First line of Lyman series in hydrogen (n = 2 → n = 1):

Using 1/λ = R(1/n_f² − 1/n_i²):

1/λ_H = 1.097 × 10⁷ × (1/1² − 1/2²)
= 1.097 × 10⁷ × (3/4)
= 8.228 × 10⁶ m⁻¹

For a transition n_i → n_f in He⁺ (Z = 2), the wavelength condition gives:

1/λ_He = R × Z² × (1/n_f² − 1/n_i²)
= R × 4 × (1/n_f² − 1/n_i²)

Setting λ_He = λ_H:

4 × (1/n_f² − 1/n_i²) = 1/1² − 1/2² = 3/4

1/n_f² − 1/n_i² = 3/16

Trying n_f = 2, n_i = 4:

1/4 − 1/16 = 4/16 − 1/16 = 3/16 ✓

∴ Yes, this is possible. The transition n_i = 4 → n_f = 2 in He⁺ produces exactly the same wavelength as the first line (n = 2 → n = 1) of the Lyman series of hydrogen.

(This is consistent with the general result: for Z = 2, scaling both quantum numbers by Z maps hydrogen Lyman lines to He⁺ Balmer lines.)
Q20Short Answer3 marks

In the Rutherford α-particle scattering experiment, an α-particle of kinetic energy 5.0 MeV is fired head-on towards a gold nucleus (Z = 79). (i) State the principle used to find the distance of closest approach. (ii) Derive an expression for the distance of closest approach (r₀). (iii) Calculate the distance of closest approach for this α-particle.

Show answer
(i) Principle:
The principle of conservation of energy is used. At the distance of closest approach, the entire initial kinetic energy of the α-particle is converted into electrostatic potential energy of the system.

(ii) Derivation of distance of closest approach (r₀):
Let the α-particle (charge +2e) be fired head-on towards a nucleus of atomic number Z (charge +Ze).

Initially, the α-particle is far away from the nucleus, so its potential energy is zero and it possesses only kinetic energy K.E. = ½mv².

At the distance of closest approach r₀, the α-particle momentarily comes to rest (v = 0), so its kinetic energy is completely converted to electrostatic potential energy.

By conservation of energy:
K.E. = Potential energy at r₀

½mv² = (1/4πε₀) · (2e)(Ze) / r₀

→ r₀ = (1/4πε₀) · 2Ze² / (½mv²)

∴ r₀ = (1/4πε₀) · (2Ze²) / K.E.

(iii) Calculation:
Given: K.E. = 5.0 MeV = 5.0 × 1.6 × 10⁻¹³ J = 8.0 × 10⁻¹³ J, Z = 79, e = 1.6 × 10⁻¹⁹ C, 1/4πε₀ = 9 × 10⁹ N m² C⁻²

Using r₀ = (1/4πε₀) · (2Ze²) / K.E.

r₀ = (9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²) / (8.0 × 10⁻¹³)

→ Numerator: 9 × 10⁹ × 158 × 2.56 × 10⁻³⁸
= 9 × 158 × 2.56 × 10⁻²⁹
= 3,637.9 × 10⁻²⁹
≈ 3.638 × 10⁻²⁶

→ r₀ = 3.638 × 10⁻²⁶ / 8.0 × 10⁻¹³

∴ r₀ ≈ 4.55 × 10⁻¹⁴ m

The distance of closest approach of the α-particle to the gold nucleus is ∴ r₀ ≈ 4.55 × 10⁻¹⁴ m.
Q21Short Answer3 marks

A science student reads about the Bohr model of the hydrogen atom and comes across the following data for spectral series:

| Series | Transition ends at n = | Region of spectrum |
|------------|------------------------|--------------------|
| Lyman | 1 | Ultraviolet |
| Balmer | 2 | Visible |
| Paschen | 3 | Infrared |

Using this information, answer the following:
(i) An electron in a hydrogen atom falls from n = 4 to n = 2. Calculate the wavelength of the emitted photon. (Given: Rydberg constant R = 1.097 × 10⁷ m⁻¹)
(ii) The same student claims: 'If the electron jumps from n = 5 to n = 1, the emitted photon will have a longer wavelength than the photon emitted in part (i).' Is the student correct? Justify your answer with a calculation or reasoning.
(iii) An atom absorbs a photon of wavelength 97.2 nm and an electron is excited from n = 1. Identify the series this transition belongs to, and find the final energy level n of the electron. (Use R = 1.097 × 10⁷ m⁻¹)
(iv) State one way in which Bohr's model successfully explains atomic spectra, and one limitation of the model.

Show answer
(i) By Rydberg's formula for hydrogen spectrum:

1/λ = R (1/n₁² − 1/n₂²)

Here n₁ = 2, n₂ = 4 (Balmer series, visible region).

→ 1/λ = 1.097 × 10⁷ × (1/2² − 1/4²)
→ 1/λ = 1.097 × 10⁷ × (1/4 − 1/16)
→ 1/λ = 1.097 × 10⁷ × (4 − 1)/16
→ 1/λ = 1.097 × 10⁷ × 3/16
→ 1/λ = 1.097 × 10⁷ × 0.1875
→ 1/λ = 2.057 × 10⁶ m⁻¹

∴ λ = 1 / (2.057 × 10⁶) ≈ 4.86 × 10⁻⁷ m = 486 nm

(ii) The student is INCORRECT.

For n = 5 → n = 1 (Lyman series):

1/λ' = R (1/1² − 1/5²) = 1.097 × 10⁷ × (1 − 1/25)
→ 1/λ' = 1.097 × 10⁷ × 24/25
→ 1/λ' = 1.053 × 10⁷ m⁻¹
∴ λ' = 9.5 × 10⁻⁸ m = 95 nm

Since 95 nm < 486 nm, the photon emitted in n = 5 → n = 1 has a SHORTER wavelength (higher energy) than the photon in part (i).

Reason: A larger energy difference (n = 5 → n = 1) corresponds to a photon of higher frequency and shorter wavelength, since E = hc/λ.

(iii) Using Rydberg's formula for absorption from n₁ = 1:

1/λ = R (1/n₁² − 1/n²)

→ 1/(97.2 × 10⁻⁹) = 1.097 × 10⁷ × (1/1 − 1/n²)
→ 1.029 × 10⁷ = 1.097 × 10⁷ × (1 − 1/n²)
→ (1 − 1/n²) = 1.029/1.097 = 0.938
→ 1/n² = 1 − 0.938 = 0.062 ≈ 1/16
∴ n² = 16 → n = 4

Since the transition is from n = 1 to n = 4 (ground state to higher level), it belongs to the Lyman series (transitions from/to n = 1, ultraviolet region).

∴ The electron is excited to the n = 4 energy level, and the transition belongs to the Lyman series.

(iv) Success: Bohr's model successfully explains the discrete line spectrum of hydrogen — the model predicts that electrons can only occupy fixed energy orbits, and emission/absorption of photons occurs only when an electron transitions between these orbits, giving rise to specific spectral lines that match observed wavelengths.

Limitation: Bohr's model fails to explain the spectra of atoms with more than one electron (multi-electron atoms), nor does it account for the fine structure (splitting) of spectral lines observed under high-resolution instruments or in the presence of magnetic fields.
Q22Short Answer3 marks

A research team is studying the spectra of two hydrogen-like ions: He⁺ (Z = 2) and Li²⁺ (Z = 3). They observe that a particular spectral line of He⁺ has a wavelength of 164 nm (in the ultraviolet region). They then look for the corresponding transition (same quantum numbers n₂ → n₁) in Li²⁺.

(i) Using Bohr's model, identify the transition (give the values of n₂ and n₁) responsible for the 164 nm line in He⁺. [2]
(ii) For the same transition in Li²⁺, calculate the wavelength of the emitted photon. Will it fall in the same spectral region (UV) as He⁺? [2]

Show answer
Part (i): Identifying the transition in He⁺ [2 marks]

By Bohr's model, the energy of the n-th orbit for a hydrogen-like ion of atomic number Z is:

Eₙ = −13.6 × Z²/n² eV

The wavelength of the emitted photon for a transition n₂ → n₁ (n₂ > n₁) is given by the generalised Rydberg formula:

1/λ = RZ²(1/n₁² − 1/n₂²)

where R = 1.097 × 10⁷ m⁻¹ and Z = 2 for He⁺.

From the given data: λ = 164 nm = 164 × 10⁻⁹ m

→ 1/λ = 1/(164 × 10⁻⁹) = 6.098 × 10⁶ m⁻¹

Substituting:

6.098 × 10⁶ = 1.097 × 10⁷ × 4 × (1/n₁² − 1/n₂²)

→ (1/n₁² − 1/n₂²) = 6.098 × 10⁶ / (4.388 × 10⁷)

→ (1/n₁² − 1/n₂²) ≈ 0.139

Testing n₁ = 2, n₂ = 4:
1/4 − 1/16 = 0.25 − 0.0625 = 0.1875 ✗

Testing n₁ = 2, n₂ = 3:
1/4 − 1/9 = 0.25 − 0.111 = 0.139 ✓

∴ The transition responsible for the 164 nm line in He⁺ is n₂ = 3 → n₁ = 2 (the Balmer series equivalent for He⁺).

---

Part (ii): Wavelength for the same transition (3 → 2) in Li²⁺ [2 marks]

For Li²⁺, Z = 3. Using the Rydberg formula:

1/λ = RZ²(1/n₁² − 1/n₂²)

→ 1/λ = 1.097 × 10⁷ × (3)² × (1/4 − 1/9)

→ 1/λ = 1.097 × 10⁷ × 9 × (5/36)

→ 1/λ = 1.097 × 10⁷ × 9 × 0.1389

→ 1/λ = 1.097 × 10⁷ × 1.25

→ 1/λ = 1.371 × 10⁷ m⁻¹

→ λ = 1/(1.371 × 10⁷)

∴ λ ≈ 72.9 nm ≈ 73 nm

Since λ ≈ 73 nm < 164 nm and lies well below 400 nm, it falls in the ultraviolet (UV) region — specifically in the extreme/vacuum UV, which is at a shorter wavelength (higher energy) than the corresponding He⁺ line.

Physical reasoning: For the same transition, wavelength ∝ 1/Z². Since Z increases from 2 (He⁺) to 3 (Li²⁺), the photon energy increases and the wavelength decreases. The ratio λ(Li²⁺)/λ(He⁺) = Z²(He⁺)/Z²(Li²⁺) = 4/9, giving λ = 164 × 4/9 ≈ 73 nm, consistent with the above.

∴ The 3 → 2 transition in Li²⁺ emits a photon of wavelength ≈ 73 nm, which lies in the UV region (extreme UV), at a shorter wavelength than the corresponding He⁺ line.
Q23Short Answer3 marks

Using Bohr's postulates for the hydrogen atom, derive an expression for the wavelength of radiation emitted when an electron makes a transition from the n₂ orbit to the n₁ orbit (n₂ > n₁). Hence, identify the spectral series obtained when n₁ = 2 and state the region of the electromagnetic spectrum to which it belongs.

Show answer
Bohr's frequency condition states that when an electron transitions from a higher energy orbit n₂ to a lower energy orbit n₁, the energy of the emitted photon equals the difference in the orbital energies:

hν = Eₙ₂ − Eₙ₁

The total energy of the electron in the nᵗʰ orbit of hydrogen is:

Eₙ = −13.6 / n² eV

By Bohr’s third postulate, the allowed energy levels are:

Eₙ₂ − Eₙ₁ = −13.6/n₂² − (−13.6/n₁²) = 13.6 × (1/n₁² − 1/n₂²) eV

Since E = hν = hc/λ, substituting:

hc/λ = 13.6 × (1/n₁² − 1/n₂²) eV

→ 1/λ = (13.6 × e)/(hc) × (1/n₁² − 1/n₂²)

Recognising the constant as the Rydberg constant R = 1.097 × 10⁷ m⁻¹:

∴ 1/λ = R (1/n₁² − 1/n₂²), where n₂ > n₁

This is the Rydberg formula for the wavelength of the emitted spectral line.

For n₁ = 2 and n₂ = 3, 4, 5, …:

1/λ = R (1/2² − 1/n₂²), n₂ = 3, 4, 5, …

∴ This gives the Balmer series, which lies in the visible region of the electromagnetic spectrum.
Q24Short Answer3 marks

Using Bohr's postulates, derive an expression for the radius of the n<sup>th</sup> orbit of the electron in a hydrogen atom. Hence show that the orbital radii are proportional to the square of the principal quantum number.

Show answer
Bohr's postulates used:
(i) The electron moves in circular orbits around the nucleus under the Coulomb force of attraction.
(ii) The angular momentum of the electron is quantised: mvr = nh/2π, where n = 1, 2, 3, …

Derivation:

Let the electron of mass m<sub>e</sub> and charge −e revolve in the n<sup>th</sup> circular orbit of radius r<sub>n</sub> with speed v<sub>n</sub>.

The Coulomb force provides the centripetal force:

ke²/r<sub>n</sub>² = m<sub>e</sub>v<sub>n</sub>²/r<sub>n</sub> … (1)

where k = 1/4πε₀ = 9×10⁹ N m² C⁻².

From Bohr's quantisation condition:

m<sub>e</sub>v<sub>n</sub>r<sub>n</sub> = nh/2π … (2)

→ v<sub>n</sub> = nh/(2πm<sub>e</sub>r<sub>n</sub>) … (3)

Substituting (3) in (1):

ke²/r<sub>n</sub>² = m<sub>e</sub>/r<sub>n</sub> × [nh/(2πm<sub>e</sub>r<sub>n</sub>)]²

ke²/r<sub>n</sub>² = n²h²/(4π²m<sub>e</sub>r<sub>n</sub>³)

Rearranging for r<sub>n</sub>:

r<sub>n</sub> = n²h²/(4π²m<sub>e</sub>ke²)

Substituting constants: h = 6.63×10⁻³⁴ Js, m<sub>e</sub> = 9.1×10⁻³¹ kg, k = 9×10⁹ N m² C⁻², e = 1.6×10⁻¹⁹ C:

r<sub>n</sub> = n² × a₀

where a₀ = h²/(4π²m<sub>e</sub>ke²) = 0.529 Å is the Bohr radius (radius of the first orbit, n = 1).

∴ r<sub>n</sub> = n² × 0.529 Å

Since a₀ is a collection of constants, r<sub>n</sub> ∝ n².

Hence, the orbital radii of the electron in hydrogen are proportional to the square of the principal quantum number.
Q25Short Answer3 marks

State the two postulates of Bohr's atomic model that resolved the instability problem of Rutherford's model. Using Bohr's second postulate and the condition for circular orbit, derive an expression for the radius of the n-th orbit (r_n) of hydrogen atom. Show that r_n ∝ n².

Show answer
Bohr's Postulates (1 mark — ½ each):

(i) An electron revolves around the nucleus only in certain discrete, non-radiating circular orbits called stationary orbits, in which the centripetal force is provided by the Coulomb attraction.

(ii) An electron can revolve only in those orbits in which its angular momentum is an integral multiple of h/2π, i.e.,

mvr = nh/2π, n = 1, 2, 3, …

(These postulates resolve Rutherford's instability: the electron does not radiate while in a stationary orbit, so it does not spiral inward.)

─────────────────────────────────────
Derivation of r_n (1 mark):
─────────────────────────────────────

Let an electron of mass m<sub>e</sub> and charge e revolve in the n-th orbit of radius r_n with speed v_n around a nucleus of charge +e (hydrogen, Z = 1).

Condition 1 — Centripetal force = Coulomb force:

m<sub>e</sub>v_n²/r_n = (1/4πε₀) · e²/r_n²

→ m<sub>e</sub>v_n² = e²/(4πε₀ r_n) … (1)

Condition 2 — Bohr's quantisation of angular momentum:

m<sub>e</sub>v_n r_n = nh/2π

→ v_n = nh/(2π m<sub>e</sub> r_n) … (2)

Substituting (2) in (1):

m<sub>e</sub> · [nh/(2π m<sub>e</sub> r_n)]² = e²/(4πε₀ r_n)

→ n²h²/(4π² m<sub>e</sub> r_n²) = e²/(4πε₀ r_n)

→ r_n = n²h²·4πε₀ / (4π² m<sub>e</sub> e²)

→ r_n = n² · (ε₀h²)/(π m<sub>e</sub> e²)

Writing the Bohr radius a₀ = ε₀h²/(π m<sub>e</sub>e²) = 0.529 Å:

∴ r_n = n² a₀

─────────────────────────────────────
Proof that r_n ∝ n² (1 mark):
─────────────────────────────────────

From the derived expression:

r_n = n² · (ε₀h²)/(π m<sub>e</sub>e²)

Since ε₀, h, m<sub>e</sub>, e are all universal constants, the quantity in brackets is a fixed constant (= a₀).

∴ r_n ∝ n²

This shows that the radius of successive Bohr orbits increases as the square of the principal quantum number, confirming that the orbits are NOT equally spaced.

For hydrogen: r₁ = 0.529 Å, r₂ = 4 × 0.529 Å, r₃ = 9 × 0.529 Å, …
Q26Short Answer3 marks

A hydrogen atom in its ground state absorbs a photon and the electron jumps to the third excited state (n = 4).

(i) What is the energy of the photon absorbed? (Express in eV.)
(ii) If the electron now makes a direct transition back to the ground state, what is the wavelength of the emitted photon? (Given: R = 1.097 × 10⁷ m⁻¹)
(iii) To which spectral series does this emitted photon belong, and in which region of the electromagnetic spectrum does it lie?
(iv) A student claims: 'Since the electron absorbed only one photon to reach n = 4, it must emit only one photon when returning to n = 1.' Is the student correct? Justify.

Show answer
(i) Energy of the photon absorbed:

By Bohr's model, the energy of the n-th level is:

E<sub>n</sub> = −13.6 / n² eV

E<sub>1</sub> = −13.6 / 1² = −13.6 eV

E<sub>4</sub> = −13.6 / 4² = −13.6 / 16 = −0.85 eV

Energy of absorbed photon:
hν = E<sub>4</sub> − E<sub>1</sub> = (−0.85) − (−13.6)

∴ hν = 12.75 eV

──────────────────────────────
(ii) Wavelength of the emitted photon (n = 4 → n = 1):

By the Rydberg formula:
1/λ = R (1/n<sub>1</sub>² − 1/n<sub>2</sub>²)

Here n<sub>1</sub> = 1, n<sub>2</sub> = 4:

1/λ = 1.097 × 10⁷ × (1/1² − 1/4²)

1/λ = 1.097 × 10⁷ × (1 − 1/16)

1/λ = 1.097 × 10⁷ × (15/16)

1/λ = 1.097 × 10⁷ × 0.9375

1/λ = 1.028 × 10⁷ m⁻¹

∴ λ = 1 / (1.028 × 10⁷) ≈ 9.73 × 10⁻⁸ m = 97.3 nm

──────────────────────────────
(iii) Spectral series and region:

Since the transition is to n<sub>1</sub> = 1 (ground state), the emitted photon belongs to the Lyman series.
The wavelength 97.3 nm lies in the ultraviolet (UV) region of the electromagnetic spectrum.

──────────────────────────────
(iv) Evaluation of the student's claim:

The student is NOT correct.

When an electron is excited by absorbing one photon, it reaches a higher energy level. However, while returning to the ground state, the electron may take any available downward path — it can return directly (n = 4 → n = 1, emitting one photon) OR it can cascade through intermediate levels (e.g., n = 4 → n = 3 → n = 2 → n = 1), emitting multiple photons.

The number of photons emitted depends on the path taken during de-excitation, not on the number of photons absorbed. Therefore, the atom can emit one, two, or three photons while returning to n = 1.
Q27Short Answer3 marks

A physics teacher sets up a demonstration in which a beam of alpha particles (each of charge +2e and mass 6.64 × 10⁻²⁷ kg) is fired at a thin gold foil. A student observes that most alpha particles pass straight through, a few are deflected by small angles, and very rarely one bounces back at nearly 180°.

(i) What does the observation that most alpha particles pass straight through the gold foil indicate about the structure of the atom? (1 mark)

(ii) An alpha particle fired head-on toward a gold nucleus (Z = 79) comes to a momentary rest at a distance of 4.0 × 10⁻¹⁴ m from the centre of the nucleus. Calculate the kinetic energy (in MeV) of the alpha particle. (2 marks)

(iii) If the same experiment were repeated using a silver foil (Z = 47) instead of gold (Z = 79), how would the distance of closest approach change for an alpha particle of the same kinetic energy? Justify your answer in one sentence. (1 mark)

Show answer
(i) Most alpha particles pass straight through the gold foil because an atom is mostly empty space, and the positive charge is concentrated in a very small, dense region called the nucleus. (1 mark)

(ii) At the distance of closest approach, the entire kinetic energy of the alpha particle is converted into electrostatic potential energy.

By conservation of energy:

KE = U = (1/4πε₀) · (q₁ q₂) / r₀

Here q₁ = 2e (alpha particle), q₂ = Ze = 79e (gold nucleus), r₀ = 4.0 × 10⁻¹⁴ m

Substituting values:

KE = (9 × 10⁹ N m² C⁻²) × (2 × 1.6 × 10⁻¹⁹ C) × (79 × 1.6 × 10⁻¹⁹ C) / (4.0 × 10⁻¹⁴ m)

KE = (9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²) / (4.0 × 10⁻¹⁴)

KE = (9 × 10⁹ × 158 × 2.56 × 10⁻³⁸) / (4.0 × 10⁻¹⁴)

KE = (9 × 158 × 2.56 × 10⁻²⁹) / (4.0 × 10⁻¹⁴)

KE = (3637.44 × 10⁻²⁹) / (4.0 × 10⁻¹⁴)

KE = 909.36 × 10⁻¹⁵ J

KE = 9.09 × 10⁻¹³ J

Converting to MeV (1 eV = 1.6 × 10⁻¹⁹ J, 1 MeV = 1.6 × 10⁻¹³ J):

KE = (9.09 × 10⁻¹³) / (1.6 × 10⁻¹³) MeV

∴ KE ≈ 5.68 MeV (1 mark for formula + substitution; 1 mark for correct answer with unit)

(iii) The distance of closest approach is given by r₀ = (1/4πε₀) · (2Ze²) / KE, so r₀ is directly proportional to Z; since silver has a smaller atomic number (Z = 47) than gold (Z = 79), the distance of closest approach will decrease for the same kinetic energy. (1 mark)
Q28Short Answer3 marks

A science student reads that a hydrogen atom can exist in various excited states. In a hydrogen discharge tube, electrons transition from higher energy levels to lower ones, emitting photons. The student records the following two observed wavelengths in the emission spectrum: 486 nm (visible, blue-green) and 102.6 nm (ultraviolet).

(i) Identify the spectral series to which each wavelength belongs. Justify your answer.
(ii) For the transition that produces the 486 nm photon, use Bohr's energy formula to determine the initial quantum number n of the electron. (Given: R = 1.097 × 10⁷ m⁻¹, and for the Balmer series the final level is n = 2.)
(iii) Using Bohr's model, calculate the ratio of the radius of the orbit corresponding to the initial level (found in part ii) to the radius of the ground state orbit.
(iv) The student claims: 'If the intensity of light falling on a hydrogen atom is doubled, the atom will emit photons of higher energy.' Is this claim correct? Justify with reference to Bohr's model.

Show answer
(i) Identification of spectral series:

The wavelength 486 nm lies in the visible region of the electromagnetic spectrum. According to Bohr's model, transitions to the n = 2 level produce the Balmer series, which falls in the visible region.
∴ 486 nm belongs to the Balmer series.

The wavelength 102.6 nm lies in the ultraviolet region. Transitions to the n = 1 level produce the Lyman series, which lies in the UV region.
∴ 102.6 nm belongs to the Lyman series.

(1 mark — both identifications with correct justification)

─────────────────────────────────────
(ii) Finding the initial quantum number for the 486 nm (Balmer) transition:

By Bohr's model, the wavelength of emitted radiation is given by:
1/λ = R(1/n₁² − 1/n₂²)

where n₁ = 2 (Balmer series final level), n₂ is the initial level, R = 1.097 × 10⁷ m⁻¹.

Substituting:
1/(486 × 10⁻⁹) = 1.097 × 10⁷ × (1/4 − 1/n₂²)

→ 2.057 × 10⁶ = 1.097 × 10⁷ × (1/4 − 1/n₂²)

→ 1/4 − 1/n₂² = 2.057 × 10⁶ / 1.097 × 10⁷

→ 1/4 − 1/n₂² = 0.1875

→ 1/n₂² = 0.25 − 0.1875 = 0.0625

→ n₂² = 16

∴ n₂ = 4

The electron transitions from n = 4 to n = 2, emitting the 486 nm photon.

(1 mark — correct substitution and value n₂ = 4)

─────────────────────────────────────
(iii) Ratio of orbital radius (n = 4) to ground state radius (n = 1):

By Bohr's model, the radius of the n<super>th</super> orbit is:
rₙ = n² a₀
where a₀ = 0.529 Å is the Bohr radius (ground state, n = 1).

∴ r₄ = 4² × a₀ = 16 a₀
r₁ = 1² × a₀ = a₀

Ratio: r₄/r₁ = 16 a₀ / a₀

∴ r₄/r₁ = 16

The radius of the n = 4 orbit is 16 times the ground state orbit radius.

(1 mark — correct formula, substitution, and ratio)

─────────────────────────────────────
(iv) Evaluation of the student's claim:

The student's claim is INCORRECT.

According to Bohr's model, the energy of a photon emitted during a transition depends only on the difference between the quantised energy levels involved:

E = hν = E<sub>n₂</sub> − E<sub>n₁</sub> = −13.6(1/n₁² − 1/n₂²) eV

The energy levels of hydrogen are fixed and do not depend on the intensity of incident light. Increasing the intensity of incident light only increases the number of photons incident per unit time (i.e., the rate of excitation/absorption), but it does NOT change the energy of individual photons or the energy of the atomic levels.

Therefore, transitions still occur between the same fixed energy levels, and the emitted photons have exactly the same energy regardless of the intensity of the incident light.

∴ The student's claim is incorrect. Intensity affects the number of emitted photons, not their individual energy.

(1 mark — correct reasoning citing quantised energy levels and role of intensity)
Q29Short Answer3 marks

A science student studying atomic spectra notices that when hydrogen gas in a discharge tube is excited, it emits light of specific colours. She records that one particular spectral line in the visible region corresponds to an electron transition from the 4th orbit to the 2nd orbit.

(i) Name the spectral series to which this line belongs and state the region of the electromagnetic spectrum it lies in.

(ii) Using Bohr's formula Eₙ = −13.6/n² eV, calculate the energy of the emitted photon for this transition.

(iii) Calculate the wavelength of this emitted photon. (Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

(iv) If the same electron, instead of falling to n = 2, had fallen to n = 1, would the emitted photon have a longer or shorter wavelength? Justify your answer in one sentence.

Show answer
(i) The spectral series is the Balmer series (transitions to n = 2).
It lies in the visible region of the electromagnetic spectrum.
[1 mark]

(ii) By Bohr's formula, the energy of the nth orbit is Eₙ = −13.6/n² eV.

Energy of n = 4 orbit:
E₄ = −13.6 / 4² = −13.6 / 16 = −0.85 eV

Energy of n = 2 orbit:
E₂ = −13.6 / 2² = −13.6 / 4 = −3.40 eV

Energy of emitted photon:
ΔE = E₄ − E₂ = (−0.85) − (−3.40)
∴ ΔE = 2.55 eV
[1 mark]

(iii) Converting energy to joules:
E = 2.55 × 1.6 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J

Using the relation E = hc/λ:
λ = hc / E
λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (4.08 × 10⁻¹⁹)
λ = (19.89 × 10⁻²⁶) / (4.08 × 10⁻¹⁹)
∴ λ ≈ 4.87 × 10⁻⁷ m = 487 nm
[1 mark]

(iv) If the electron falls from n = 4 to n = 1, the energy difference ΔE is larger (since n = 1 has much lower energy than n = 2).

Since λ = hc/ΔE, a larger ΔE gives a smaller λ.
∴ The emitted photon would have a shorter wavelength (and would lie in the ultraviolet/Lyman series, not the visible region).
[1 mark]
Q30Short Answer3 marks

A hydrogen atom in its ground state absorbs a photon and gets excited to the n = 4 level. The excited electron then undergoes a cascade of transitions back to the ground state, emitting photons at each step.

(i) How many different spectral lines can be emitted in total from the n = 4 level as the electron returns to the ground state (considering ALL possible downward transitions)?

(ii) Which of these emitted photons belongs to the Balmer series? Write the transition and calculate the wavelength of this photon. (Given: R = 1.097 × 10⁷ m⁻¹)

(iii) The atom is now placed in a strong external electric field. State ONE way in which the Bohr model fails to explain the resulting spectral observations, and name the phenomenon responsible.

Show answer
CBSE MARKING SCHEME (4 marks)

─────────────────────────────────────
Part (i): Number of spectral lines [1 mark]
─────────────────────────────────────

By the formula for the number of spectral lines possible when an electron de-excites from level n:

Number of lines = n(n − 1) / 2

Here n = 4:

Number of lines = 4 × 3 / 2 = 6

∴ 6 different spectral lines can be emitted.

[The six transitions are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1]

─────────────────────────────────────
Part (ii): Balmer series photon — transition and wavelength [2 marks]
─────────────────────────────────────

The Balmer series corresponds to transitions ending at n₁ = 2.
From level n = 4, the Balmer series transition is: n = 4 → n = 2.

Using the Rydberg formula:

1/λ = R (1/n₁² − 1/n₂²)

where n₁ = 2, n₂ = 4, R = 1.097 × 10⁷ m⁻¹:

1/λ = 1.097 × 10⁷ × (1/4 − 1/16)

1/λ = 1.097 × 10⁷ × (4/16 − 1/16)

1/λ = 1.097 × 10⁷ × 3/16

1/λ = 1.097 × 10⁷ × 0.1875

1/λ = 2.057 × 10⁶ m⁻¹

∴ λ = 1 / (2.057 × 10⁶) ≈ 4.86 × 10⁻⁷ m = 486 nm

∴ The Balmer series photon corresponds to the 4 → 2 transition, with wavelength λ ≈ 486 nm (visible, blue-green — Hβ line).

─────────────────────────────────────
Part (iii): Failure of Bohr model in an external electric field [1 mark]
─────────────────────────────────────

The Bohr model assumes that electron orbits are perfectly circular and fixed; it cannot account for the splitting of spectral lines when an atom is placed in an external electric field.

Because the Bohr model does not incorporate the concept of orbital shapes (sublevels with different values of orbital quantum numbers l and m), it predicts only a single line for each transition and FAILS to explain the splitting of these lines into multiple closely spaced components.

∴ The phenomenon responsible is the STARK EFFECT (splitting of spectral lines in an external electric field), which requires quantum mechanical treatment (including orbital angular momentum quantum numbers) and CANNOT be explained by the Bohr model.

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