NCERT Solutions for Class 12 Physics Chapter 12: Atoms

Concept and context. Thomson’s model (1898) pictures the atom as a sphere of uniformly distributed positive charge (radius ≈ 10⁻¹⁰ m, the size of the atom) in which the electrons are embedded like plums in a pudding; the positive charge and the mass are spread over the whole atom. Rutherford’s nuclear model (1911), inferred from the Geiger–Marsden α-particle scattering experiment (about 1 α-particle in 8000 deflects by more than 90°), concentrates all the positive charge and almost all the mass in a tiny nucleus (≈ 10⁻¹⁵ m) with the electrons revolving around it at distances of the order of the atomic size (10⁻¹⁰ m). Each blank below is answered using these two pictures.

• Detail for (a): the radius of the nucleus (10⁻¹⁵ m) is about 10⁵ times smaller than the orbit radius; the “size of the atom” is the orbit size ≈ 10⁻¹⁰ m in the Rutherford model and the radius of the positive sphere ≈ 10⁻¹⁰ m in the Thomson model — equal.
• Detail for (b): in Thomson’s model the force on an electron displaced r from the centre of a uniformly positive sphere is proportional to r and directed towards the centre (like a spring): the equilibrium at the centre is stable and the electron can oscillate with a definite frequency. In Rutherford’s model, there is no position of zero force; the electron survives only because it is moving (centripetal acceleration).
• Detail for (c): an electron revolving around the nucleus is continuously accelerated; the radiation would take away its energy, so the orbit radius would shrink and the electron would spiral into the nucleus — and the emission would be a continuous spectrum, contradicting both the stability of atoms and the line spectra (NCERT Example 12.4: ν ≈ 6.6 × 10¹⁵ Hz initially).
• Detail for (d): Thomson’s model could not explain the large-angle scattering of α-particles, because a smeared charge cannot exert a strong enough force to turn an α-particle back; the non-uniform (nuclear) distribution of Rutherford’s model can.

Conclusion: (a) no different from; (b) Thomson’s model / Rutherford’s model; (c) Rutherford’s model; (d) Thomson’s model / Rutherford’s model; (e) both the models.

Concept and context. In the Geiger–Marsden experiment (NCERT Fig. 12.1) the large-angle scattering of α-particles (about 1 in 8000 by more than 90°) by gold atoms arises because the α-particle (mass 4 u, charge 2e) is repelled by a very massive, highly charged nucleus (gold: Z = 79, mass ≈ 197 u ≈ 50 times heavier than the α-particle) which hardly recoils. A hydrogen nucleus is a single proton (mass ≈ 1 u, charge +e), lighter than the incident α-particle. The nature of the scattering therefore changes completely.

1. Heavier projectile, lighter target: the mass of an α-particle (≈ 4 u) is about four times that of the proton (≈ 1 u). In an elastic collision with a nearly stationary particle that is lighter than the projectile, momentum conservation shows the projectile cannot be turned back or deflected through a large angle: the maximum deflection angle of the α-particle satisfies sin θ_max = m_p/m_α = 1/4, so θ_max ≈ 14.5°. It is like a bowling ball hitting a table-tennis ball — the heavy object ploughs on almost undisturbed while the light one is knocked away.
2. Weaker Coulomb force: the charge of a hydrogen nucleus (+e) is also small (compared with +79e of gold), so the repulsive force on the α-particle (F = (1/4πε₀)(2e × e)/r²) is much weaker and the closest-approach energy barrier much smaller: the α-particle (even at a few MeV) can approach very close to the proton and still lose only a small part of its momentum.
3. Expected result: no large-angle scattering (no back-scattering at all); almost all α-particles go straight through the thin solid hydrogen sheet with only small deflections (below about 14.5°), and the hydrogen nuclei (protons) are knocked out of the sheet in the forward direction — these recoil protons could be detected. The conclusions drawn from gold (a small massive positive nucleus) therefore could not have been drawn from a hydrogen target.

• Why gold was chosen: a heavy, high-Z, malleable metal that can be rolled into a very thin foil (2.1 × 10⁻⁷ m, a few thousand atoms thick) so that single scattering by the nucleus dominates.
• Also: energy conservation: the momentum transferred to a light proton is small, so the α-particle loses only a modest amount of energy.

Conclusion: with a solid hydrogen foil no large-angle scattering would be observed — the α-particles would pass through with only small deflections (at most ≈ 14.5°) while the lighter protons recoil forward — so the experiment would not reveal the nuclear structure as gold does.

Concept and context. According to Bohr’s third postulate, when an atom makes a transition from a higher energy state of energy E_i to a lower one of energy E_f, it emits a photon whose energy equals the difference between the energies of the two states: hν = E_i − E_f (NCERT Section 12.4). The frequency of the emitted radiation therefore follows directly from the energy gap: ν = (E_i − E_f)/h. Conversely, an atom can absorb a photon of exactly this frequency to move up.

Given: E_i − E_f = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J; h = 6.63 × 10⁻³⁴ J s.

1. Photon energy: hν = 3.68 × 10⁻¹⁹ J.
2. Frequency: ν = (3.68 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 5.55 × 10¹⁴ Hz.

ν = ΔE/h = 5.55 × 10¹⁴ Hz

• Check: λ = hc/ΔE = 1240 eV nm/2.3 eV = 539 nm; 3 × 10⁸/5.55 × 10¹⁴ = 5.4 × 10⁻⁷ m ✓.
• Meaning: energy gaps of a few eV between the levels of atoms produce visible light (1.8–3.1 eV); smaller gaps give infrared and larger gaps ultraviolet, as in the series of hydrogen.

Conclusion: the frequency of the radiation emitted is 5.55 × 10¹⁴ Hz (green light of wavelength about 540 nm).

Concept and context. In the Bohr (and Rutherford) model the electron of a hydrogen atom moves in a circular orbit of radius r under the attractive Coulomb force which provides the centripetal force: mv²/r = e²/(4πε₀r²). From this, the kinetic energy is K = ½mv² = e²/(8πε₀r), the electrostatic potential energy (zero at infinity) is U = −e²/(4πε₀r), so U = −2K and the total energy is E = K + U = −e²/(8πε₀r) = −K (NCERT Section 12.2.2, Eqs 12.3–12.4). The total energy is negative: the electron is bound to the nucleus. Hence K = −E and U = 2E.

Given: total energy of the ground state E = −13.6 eV.

1. Kinetic energy: K = −E = +13.6 eV (= 2.2 × 10⁻¹⁸ J).
2. Potential energy: U = 2E = −27.2 eV (= −4.4 × 10⁻¹⁸ J).
3. Check: K + U = 13.6 − 27.2 = −13.6 eV = E ✓.

K = +13.6 eV ; U = −27.2 eV

• Virial relation: for an inverse-square force law, 2K = −U (the “virial theorem”) — the result is general for all Bohr orbits: K_n = 13.6/n² eV, U_n = −27.2/n² eV.
• Meaning of E = −13.6 eV: 13.6 eV of energy must be supplied to the electron to free it (E = 0 at rest at infinity): the ionisation energy of hydrogen.

Conclusion: the kinetic energy of the electron is +13.6 eV and its potential energy is −27.2 eV.

Concept and context. By Bohr’s third postulate an atom can absorb a photon only if its energy exactly equals the difference between two stationary-state energies: hν = E_f − E_i. The energy of the n-th level of hydrogen is E_n = −13.6 eV/n²: E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV (NCERT Fig. 12.7). The photon needed to excite the atom from n = 1 to n = 4 has energy E₄ − E₁.

Given: n_i = 1, n_f = 4; E_n = −13.6/n² eV; h = 6.63 × 10⁻³⁴ J s; c = 3 × 10⁸ m s⁻¹.

1. Energy of the photon: ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV (= 12.75 × 1.6 × 10⁻¹⁹ = 2.04 × 10⁻¹⁸ J).
2. Frequency: ν = ΔE/h = (2.04 × 10⁻¹⁸)/(6.63 × 10⁻³⁴) = 3.08 × 10¹⁵ Hz.
3. Wavelength: λ = c/ν = (3 × 10⁸)/(3.08 × 10¹⁵) = 9.74 × 10⁻⁸ m = 97.4 nm (also 1240/12.75 = 97.3 nm).

ΔE = 12.75 eV ; ν = 3.08 × 10¹⁵ Hz ; λ ≈ 97 nm

• Check with the Rydberg formula: 1/λ = R(1/1² − 1/4²) = 1.097 × 10⁷ × 15/16 = 1.028 × 10⁷ m⁻¹ ⇒ λ = 97.3 nm ✓.
• After absorption the atom (in the excited state n = 4) may return to the ground state in one step (emitting the same 97.3 nm photon) or by cascading through intermediate levels (4 → 3 → 2 → 1, etc.), giving several spectral lines (Lyman, Balmer, Paschen series).

Conclusion: the photon has energy 12.75 eV, frequency 3.08 × 10¹⁵ Hz and wavelength ≈ 97 nm (ultraviolet).

Concept and context. In the Bohr model the electron revolves in a circular orbit satisfying (i) the centripetal force condition mv²/r = e²/(4πε₀r²) and (ii) the quantisation of angular momentum mvr = nh/2π. Together these give the radius r_n = n² a₀ (a₀ = 5.3 × 10⁻¹¹ m, the Bohr radius) and the speed v_n = e²/(2ε₀nh) = v₁/n where v₁ = 2.19 × 10⁶ m s⁻¹ (the speed in the first orbit ≈ c/137). The orbital period T_n = 2πr_n/v_n = 2π n² a₀/(v₁/n) = n³T₁ with T₁ = 2πa₀/v₁.

Given: v₁ = e²/(2ε₀h) = (1.6 × 10⁻¹⁹)²/(2 × 8.85 × 10⁻¹² × 6.63 × 10⁻³⁴) = 2.18 × 10⁶ m s⁻¹; r₁ = 5.3 × 10⁻¹¹ m.

(a) Speeds

• n = 1: v₁ = 2.18 × 10⁶ m s⁻¹ (≈ c/137).
• n = 2: v₂ = v₁/2 = 1.09 × 10⁶ m s⁻¹.
• n = 3: v₃ = v₁/3 = 7.27 × 10⁵ m s⁻¹.

(b) Orbital periods

1. n = 1: r₁ = 5.3 × 10⁻¹¹ m; T₁ = 2πr₁/v₁ = (2 × 3.14 × 5.3 × 10⁻¹¹)/(2.18 × 10⁶) = 1.52 × 10⁻¹⁶ s.
2. n = 2: r₂ = 4r₁ = 2.12 × 10⁻¹⁰ m; T₂ = 2πr₂/v₂ = (2 × 3.14 × 2.12 × 10⁻¹⁰)/(1.09 × 10⁶) = 1.22 × 10⁻¹⁵ s (= 8T₁).
3. n = 3: r₃ = 9r₁ = 4.77 × 10⁻¹⁰ m; T₃ = 2πr₃/v₃ = (2 × 3.14 × 4.77 × 10⁻¹⁰)/(7.27 × 10⁵) = 4.12 × 10⁻¹⁵ s (= 27T₁).

• Scaling laws: r_n ∝ n², v_n ∝ 1/n, T_n ∝ n³, E_n ∝ −1/n². The speed of the electron falls as it goes to higher orbits; the revolution frequency falls as n⁻³.
• Relativity check: v₁/c = 2.18 × 10⁶/3 × 10⁸ = 0.7 % — the non-relativistic Bohr treatment is a good approximation (fine-structure constant α = v₁/c ≈ 1/137).

Conclusion: (a) v = 2.18 × 10⁶, 1.09 × 10⁶ and 7.27 × 10⁵ m s⁻¹; (b) T = 1.52 × 10⁻¹⁶ s, 1.22 × 10⁻¹⁵ s and 4.12 × 10⁻¹⁵ s for n = 1, 2, 3 respectively.

Concept and context. In the Bohr model the radius of the n-th stationary orbit of hydrogen is r_n = n² r₁ where r₁ = a₀ = (h/2π)²(4πε₀)/(me²) = 5.3 × 10⁻¹¹ m is the Bohr radius. The radius thus increases as the square of the principal quantum number — a consequence of the quantisation of angular momentum (mvr = nh/2π) combined with the Coulomb force law.

1. n = 2: r₂ = 2² × r₁ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m (≈ 0.212 nm).
2. n = 3: r₃ = 3² × r₁ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m (≈ 0.477 nm).

r₂ = 4r₁ = 2.12 × 10⁻¹⁰ m ; r₃ = 9r₁ = 4.77 × 10⁻¹⁰ m

• Interpretation: higher orbits are much larger — the n = 3 orbit has a radius nine times that of the ground state; the atom swells as the electron is excited (the size ~10⁻¹⁰ m of the ground-state atom).
• Derivation: from mv²/r = e²/4πε₀r² and mvr = nh/2π: v = nh/(2πmr) ⇒ m (nh/2πmr)²/r = e²/(4πε₀r²) ⇒ r = n²h²ε₀/(πme²) ∝ n².

Conclusion: the radii of the second and third Bohr orbits are 2.12 × 10⁻¹⁰ m and 4.77 × 10⁻¹⁰ m.

Concept and context. Hydrogen atoms at room temperature are in the ground state (n = 1, E₁ = −13.6 eV). An incident electron can excite an atom by collision if its kinetic energy is at least equal to the energy needed to raise the atom to a higher level (the excess energy stays with the electron). The energy needed to excite from n = 1 to the n-th level is E_n − E₁ = 13.6(1 − 1/n²) eV: 10.2 eV (n = 2), 12.09 eV (n = 3), 12.75 eV (n = 4), … Because the bombarding electrons have only 12.5 eV, the atoms can be excited to levels up to n = 3 (12.09 eV ≤ 12.5 eV) but not to n = 4 (12.75 eV > 12.5 eV). The excited atoms then return to lower levels, emitting photons of wavelengths given by hν = E_i − E_f.

Step 1 — highest level reachable

• E₃ − E₁ = −1.51 − (−13.6) = 12.09 eV < 12.5 eV ✓ (n = 3 can be reached).
• E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV > 12.5 eV ✗ (n = 4 cannot be reached).

Step 2 — possible transitions from n = 3 (and n = 2)

1. 3 → 1 (Lyman series): ΔE = 12.09 eV; λ = hc/ΔE = 1240/12.09 = 102.6 nm (ultraviolet). Check with Rydberg: 1/λ = 1.097 × 10⁷ (1 − 1/9) = 9.75 × 10⁶ m⁻¹ ⇒ 102.6 nm.
2. 2 → 1 (Lyman series): ΔE = 10.2 eV; λ = 1240/10.2 = 121.6 nm (ultraviolet). Check: 1.097 × 10⁷ (1 − 1/4) = 8.23 × 10⁶ ⇒ 121.5 nm.
3. 3 → 2 (Balmer series): ΔE = 1.89 eV; λ = 1240/1.89 = 656 nm (red, Hα line). Check: 1.097 × 10⁷ (1/4 − 1/9) = 1.524 × 10⁶ ⇒ 656.3 nm.

• Answer in words: two lines of the Lyman series (102.5 nm and 121.5 nm) and the first line of the Balmer series (656.3 nm) are emitted — three lines in all.
• Why not more: reaching n = 4 or higher needs at least 12.75 eV, more than the 12.5 eV available; hence the Paschen, Brackett and Pfund series (which end on n = 3, 4, 5) and the higher lines of the Balmer series do not appear.
• Role of collision: in an inelastic collision the electron gives up exactly the excitation energy (12.09 eV or 10.2 eV) and keeps the rest of its energy.

Conclusion: the beam can excite hydrogen only up to n = 3, so the emitted wavelengths belong to the Lyman series (102.6 nm and 121.6 nm) and the Balmer series (first line, 656 nm).

Concept and context. Bohr’s second postulate states that the angular momentum of an electron in a stationary orbit is an integral multiple of h/2π: L = mvr = nh/2π (n = 1, 2, 3 …, the principal quantum number). Although the postulate was proposed for atomic electrons, the same quantisation applied (formally) to the Earth’s revolution around the Sun gives an enormous quantum number — which illustrates the correspondence principle: for macroscopic systems the quantum numbers are so large that the discreteness of the allowed values is utterly unobservable, and the classical (continuous) description holds.

Given: m = 6.0 × 10²⁴ kg; v = 3 × 10⁴ m s⁻¹; r = 1.5 × 10¹¹ m; h = 6.63 × 10⁻³⁴ J s.

1. Angular momentum of the Earth: L = mvr = (6.0 × 10²⁴)(3 × 10⁴)(1.5 × 10¹¹) = 2.7 × 10⁴⁰ kg m² s⁻¹.
2. Quantum number: n = L/(h/2π) = 2πL/h = (2 × 3.14 × 2.7 × 10⁴⁰)/(6.63 × 10⁻³⁴) = (1.70 × 10⁴¹)/(6.63 × 10⁻³⁴).
3. n ≈ 2.6 × 10⁷⁴.

n = 2π m v r / h ≈ 2.6 × 10⁷⁴

• Significance: successive allowed values (n and n + 1) differ by only one part in 10⁷⁴ — the “quantisation” of the Earth’s orbital angular momentum is completely undetectable; hence quantum effects are not seen in the planetary motion and Newtonian mechanics is perfectly adequate.
• Contrast: for the electron in hydrogen the quantum numbers are 1, 2, 3 …, with the discreteness of energy levels very evident (13.6 eV, 3.4 eV …).

Conclusion: the quantum number characterising the Earth’s revolution around the Sun is n ≈ 2.6 × 10⁷⁴ — so large that quantisation of its orbit is of no practical consequence.