NCERT Solutions Class 12 Physics Chapter 12 Atoms

What the question is asking. A side-by-side compare-and-contrast between the two earliest atomic models — J.J. Thomson's "plum-pudding" model (1904) and Ernest Rutherford's "nuclear" model (1911). For each blank, pick the model that fits.

Concept used — the two models in one sentence each.

• Thomson's model: the atom is a uniform sphere of positive charge of radius ∼ 10^-10 m, with electrons embedded in it like raisins in a pudding. Mass is spread throughout the whole sphere.
• Rutherford's model: almost all the mass and positive charge are concentrated in a tiny central nucleus of radius ∼ 10^-15 m; electrons orbit the nucleus in the otherwise-empty atomic volume of radius ∼ 10^-10 m.

Step 1 — (a) Atomic size. Both models predict the same outer atomic radius, ∼ 10^-10 m. Thomson's whole sphere is that size; Rutherford's orbiting-electron cloud is also that size. So Thomson's atomic size is no different from Rutherford's.

Step 2 — (b) Ground-state equilibrium. In Thomson's model the positive charge is spread out and the electrons sit at positions where the net electric force on them is zero — they are in stable equilibrium. In Rutherford's model the electron must orbit the nucleus (otherwise it falls in), so it always feels a net Coulomb attraction toward the nucleus. So: Thomson's model has stable equilibrium; Rutherford's model always has net force.

Step 3 — (c) Classical collapse. A classical orbiting charge accelerates, radiates electromagnetic energy, and spirals into the nucleus in ∼ 10^-8 s. This happens to Rutherford's model. Thomson's electrons just sit in equilibrium and do not radiate.

Step 4 — (d) Mass distribution. Thomson's mass is smeared evenly across the whole atom (nearly continuous). Rutherford's mass is overwhelmingly in the tiny nucleus (highly non-uniform). So continuous = Thomson's; non-uniform = Rutherford's.

Step 5 — (e) Positive charge holds most mass. In Thomson's pudding the positive sphere holds most of the mass (electrons are extremely light). In Rutherford's nucleus the positive nucleus also holds essentially all the mass. So the answer is both the models.

Final answer.

• (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

What the question is asking. Predict what would happen if the gold-foil target were replaced by a solid-hydrogen foil in the Geiger–Marsden α-scattering experiment.

Given.

• Original target: gold (Au), atomic number Z = 79, mass number A = 197.
• New target: hydrogen (H), Z = 1, A = 1 (the nucleus is a single proton).
• Projectile: α-particle Z = 2, A = 4.

Concept used — two effects matter.

(i) Coulomb repulsion strength. The closest distance of approach of an α to a nucleus of charge Ze is r_min = 14π₀ 2Ze²K, where K is the α's kinetic energy. The repulsive force is proportional to Z of the target nucleus, so a gold nucleus Z=79 repels an α about 79 times harder than a hydrogen nucleus does.

(ii) Mass of the target. A projectile can be back-scattered only if it strikes a target much heavier than itself. The gold nucleus A=197 is about 49 times heavier than the α A=4, so it acts as a near-fixed scattering centre. The hydrogen nucleus A=1 is roughly four times lighter than the α. In a head-on collision the α would barely slow down — it would just push the proton out of the way (like a bowling ball hitting a marble).

Step 1 — qualitative prediction. Two things change at once: (a) the Coulomb barrier is much smaller because Z=1, and (b) the target is much lighter than the projectile.

Step 2 — expected scattering pattern.

• Almost all α-particles will pass straight through with negligible deflection (because the Coulomb repulsion from a proton is small).
• There will be essentially no large-angle (or back-) scattering. A light proton cannot bounce a heavier α-particle straight back; conservation of momentum forbids it.
• Instead, the hydrogen protons would be kicked forward and ejected from the foil — but the α-detector at large angles would register almost nothing.

Step 3 — why this matters historically. The whole power of Rutherford's analysis depended on (i) a high-Z target so the deflection is dramatic, and (ii) a heavy target nucleus that does not recoil. Hydrogen fails both tests; the experiment would not have led to the discovery of the nucleus.

Final answer. If the gold foil were replaced by solid hydrogen, almost no α-particles would be scattered back. The angular distribution would be sharply peaked in the forward direction, with no observable large-angle or back-scattering. The experiment would not reveal the existence of an atomic nucleus.

Given.

• Energy gap between the two atomic levels, Δ E = 2.3 eV.
• Planck's constant, h = 6.62610^-34 J s.
• 1 eV = 1.610^-19 J.

Concept used — Bohr's frequency condition. When an atom drops from a higher energy level E₂ to a lower level E₁, the emitted photon carries away the entire energy difference: hν = E₂ - E₁ = Δ E. Solving for the frequency: ν = Δ Eh.

Step 1 — convert Δ E to joules. Δ E = 2.3 eV × 1.610^-19 J/eV = 3.6810^-19 J.

Step 2 — write the formula symbolically. ν = Δ Eh.

Step 3 — substitute the numbers. ν = 3.6810^-196.62610^-34 Hz.

Step 4 — divide. ν = 3.686.626× 10^-19+34 = 0.5554× 10¹⁵ Hz.

Step 5 — clean up to standard scientific notation. ν ≈ 5.55× 10¹⁴ Hz.

Step 6 — sanity-check the spectral region. Visible light spans roughly 410¹⁴ Hz (red) to 7.510¹⁴ Hz (violet). Our answer falls in the yellow-green part of the visible spectrum — exactly the kind of transition you would see in an atomic emission line.

Final answer. ν ≈ 5.55× 10¹⁴ Hz (visible, yellow-green).

Given.

• Total energy of the electron in the hydrogen ground state, E = -13.6 eV.
• The negative sign means the electron is bound — it would take +13.6 eV of external energy to remove the electron from the atom (the "ionisation energy").

Concept used — virial theorem for the inverse-square force. For a particle moving under a Coulomb (or gravitational) attractive 1/r² force, the time-averaged kinetic and potential energies satisfy K= -12U, E = K + U. Combining, E = K + U = -U2 + U = U2, ⇒ U = 2E, K = -E. This is the famous virial relation K = -E, U = 2E for hydrogen-like atoms.

Why is U negative? Convention: the potential energy of two point charges is taken to be zero when they are infinitely far apart. As the (positive) proton and (negative) electron come closer, work is done by the attractive force, so the system loses potential energy → U < 0. The proton–electron PE is U = -ke²/r.

Step 1 — find the kinetic energy. K = -E = -(-13.6 eV) = +13.6 eV.

Step 2 — find the potential energy. U = 2E = 2×(-13.6 eV) = -27.2 eV.

Step 3 — verify with the formula E = K + U. K + U = (+13.6) + (-27.2) = -13.6 eV = E.

Step 4 — sanity check. Kinetic energy is always positive K = 12mv² — yes, +13.6 eV is positive. Potential energy of an attractive bound system at finite r is negative — yes, -27.2 eV is negative. And |U| > K, as it must be for a bound state.

Final answer.

• Kinetic energy: K = +13.6 eV
• Potential energy: U = -27.2 eV

Given.

• Initial level: ground state, n₁ = 1, with E₁ = -13.6 eV.
• Final level: n₂ = 4, with E₄ = -13.6/16 = -0.85 eV.
• Planck constant h = 6.62610^-34 J s, speed of light c = 310⁸ m/s.

Concept used — Bohr energy levels and absorption. The allowed energies of the hydrogen electron are E_n = -13.6n² eV, n = 1,2,3,… When the atom absorbs a photon, the photon's energy equals the gap between the final and initial levels: hν = E_n2 - E_n1. Both E's are negative; E_n2 is the smaller (less negative) one. The difference is positive, as it should be for absorption.

Step 1 — compute the photon energy in eV. Δ E = E₄ - E₁ = (-0.85) - (-13.6) = 12.75 eV.

Step 2 — convert to joules. Δ E = 12.75× 1.610^-19 = 2.0410^-18 J.

Step 3 — frequency from hν = Δ E. ν = Δ Eh = 2.0410^-186.62610^-34 ≈ 3.08× 10¹⁵ Hz.

Step 4 — wavelength from λ = c/ν. λ = 310⁸3.0810¹⁵ ≈ 9.74× 10^-8 m = 97.4 nm.

Step 5 — identify the spectral region. Visible light is roughly 380 – 750 nm. Our wavelength ≈ 97 nm is much shorter — it lies in the ultraviolet (UV), specifically in the Lyman series all transitions ending at or starting from n=1 are in the UV.

Final answer.

• Photon energy: Δ E = 12.75 eV = 2.0410^-18 J.
• Frequency: ν ≈ 3.0810¹⁵ Hz.
• Wavelength: λ ≈ 97.4 nm (ultraviolet, Lyman series).

Given. Hydrogen atom, Bohr orbits n = 1, 2, 3. Constants: k = 910⁹, e = 1.610^-19 C, m_e = 9.1110^-31 kg, ₀ = 8.85410^-12, Bohr radius a₀ = 5.2910^-11 m.

Concept used — Bohr orbital speed and radius. The Bohr quantisation condition angular momentum is quantised in units of gives v_n = e²4π₀ ·1n = v₁n, v₁ ≈ 2.19× 10⁶ m/s. The orbital radius is r_n = n² a₀, a₀ = 5.2910^-11 m. The orbital period is the circumference divided by speed: T_n = 2π r_nv_n = 2π n² a₀v₁/n = 2π n³ a₀v₁.

Plain-English version. "The deeper the orbit (smaller n), the faster the electron moves and the shorter the period. Speed scales as 1/n; radius scales as n²; period scales as n³."

Step 1 — find v₁ carefully. v₁ = (910⁹)(1.610^-19)²(1.05510^-34) = 910⁹× 2.5610^-381.05510^-34. Numerator: 92.56 = 23.04; so num = 23.0410^-29. Divide: v₁ = 23.0410^-291.05510^-34 ≈ 21.810⁵ ≈ 2.1810⁶ m/s.

Step 2 — speeds in the three orbits. v₁ ≈ 2.1810⁶ m/s, v₂ = v₁2 ≈ 1.0910⁶ m/s, v₃ = v₁3 ≈ 7.2710⁵ m/s.

Step 3 — Bohr radii in the three orbits. r₁ = a₀ = 5.2910^-11 m, r₂ = 4a₀ = 2.1210^-10 m, r₃ = 9a₀ = 4.7610^-10 m.

Step 4 — compute the orbital periods. T₁ = 2π r₁v₁ = 2π(5.2910^-11)2.1810⁶ ≈ 3.3210^-102.1810⁶ ≈ 1.5210^-16 s. T₂ = 2π(2.1210^-10)1.0910⁶ ≈ 1.3310^-91.0910⁶ ≈ 1.2210^-15 s. T₃ = 2π(4.7610^-10)7.2710⁵ ≈ 2.9910^-97.2710⁵ ≈ 4.1110^-15 s.

Step 5 — check scaling. Since T_n ∝ n³: T₂/T₁ = 8 and T₃/T₁ = 27. Our numbers: 1.22/0.152 ≈ 8 , 4.11/0.152 ≈ 27 .

Final answer.

• Speeds: v₁ ≈ 2.1810⁶ m/s, v₂ ≈ 1.0910⁶ m/s, v₃ ≈ 7.2710⁵ m/s.
• Periods: T₁ ≈ 1.5210^-16 s, T₂ ≈ 1.2210^-15 s, T₃ ≈ 4.1110^-15 s.

Given.

• Radius of the innermost (Bohr-ground-state) orbit, r₁ = 5.310^-11 m this is the Bohr radius a₀.

Concept used — Bohr-orbit radius formula. In Bohr's quantised hydrogen, the allowed orbital radii are r_n = n² a₀, n = 1, 2, 3, … The radius scales as the square of the principal quantum number. So as you climb to higher orbits, the atom gets dramatically larger.

Step 1 — write the formula symbolically. r_n = n² r₁.

Step 2 — radius of n = 2. r₂ = 2² r₁ = 4× 5.310^-11 = 21.210^-11 = 2.1210^-10 m.

Step 3 — radius of n = 3. r₃ = 3² r₁ = 9× 5.310^-11 = 47.710^-11 = 4.7710^-10 m.

Step 4 — sanity check. A typical neutral atom has radius ∼ 10^-10 m, so r₂ = 2.1210^-10 and r₃ = 4.7710^-10 m sit nicely in the expected atomic-size range. Both are still much smaller than the wavelength of visible light ∼ 510^-7 m, which is why we cannot "see" atoms with ordinary microscopes.

Final answer.

• r₂ = 2.1210^-10 m
• r₃ = 4.7710^-10 m

Given.

• Kinetic energy of bombarding electrons, E = 12.5 eV.
• Hydrogen gas at room temperature — virtually all atoms are initially in the ground state, n = 1, with E₁ = -13.6 eV.

Concept used — atomic excitation by electron impact. An incoming electron can transfer energy to a hydrogen atom only in discrete amounts equal to the gaps between Bohr energy levels. The atom is initially in state n = 1. It can be excited to a higher state n₂ only if the bombarding electron has at least enough energy to bridge the gap Δ E_{1→ n2} = E_n2 - E₁ = 13.6(1 - 1n₂²) eV. Whatever energy is not absorbed leaves with the bombarding electron.

Step 1 — find the highest level reachable. Compute the required Δ E for the first few jumps:

• n=1 → 2: Δ E = 13.6(1 - 1/4) = 10.2 eV. (12.5 eV is enough.)
• n=1 → 3: Δ E = 13.6(1 - 1/9) = 12.09 eV. (12.5 eV is enough.)
• n=1 → 4: Δ E = 13.6(1 - 1/16) = 12.75 eV. (12.5 eV is NOT enough.)

So the highest level a 12.5 eV electron can excite the atom to is n₂ = 3.

Step 2 — list all possible downward transitions. Once excited to n = 3, the atom can de-excite to any lower level (directly or via intermediate steps), emitting a photon for each transition. The possible transitions are:

• n = 3 → 2 — Balmer series (visible).
• n = 3 → 1 — Lyman series (UV).
• n = 2 → 1 — Lyman series (UV).

Step 3 — compute the wavelengths. Use the Rydberg formula 1λ = R (1n₁² - 1n₂²), R = 1.09710⁷ m^-1.

(i) n = 3 → 2 Balmer-α, H-α: 1λ = 1.09710⁷ (14 - 19) = 1.09710⁷×536 ≈ 1.52410⁶ m^-1. λ ≈ 6.5610^-7 m = 656 nm (red, visible).

(ii) n = 3 → 1 Lyman-β: 1λ = 1.09710⁷ (1 - 19) = 1.09710⁷×89 ≈ 9.7510⁶ m^-1. λ ≈ 1.0310^-7 m = 103 nm (UV).

(iii) n = 2 → 1 Lyman-α: 1λ = 1.09710⁷ (1 - 14) = 1.09710⁷×34 ≈ 8.2310⁶ m^-1. λ ≈ 1.2210^-7 m = 122 nm (UV).

Step 4 — group by spectral series. The 656 nm line is in the Balmer series transitions ending at n = 2. The 103 nm and 122 nm lines are in the Lyman series transitions ending at n = 1.

Final answer. Three wavelengths will be emitted:

• ₃₂ ≈ 656 nm — Balmer-α, red (visible).
• ₃₁ ≈ 103 nm — Lyman-β, UV.
• ₂₁ ≈ 122 nm — Lyman-α, UV.

So both the Lyman series (UV) and the first line of the Balmer series (visible) are emitted.

Given.

• Orbital radius of Earth, r = 1.510¹¹ m.
• Orbital speed, v = 310⁴ m/s.
• Mass of Earth, m = 6.010²⁴ kg.
• Reduced Planck constant, = 1.05510^-34 J s.

Concept used — Bohr's angular-momentum quantisation. Bohr's postulate (originally for atomic electrons) says the orbital angular momentum can take only discrete values: L = m v r = n , n = 1, 2, 3, … This is the only ingredient we need. We are asked to find the value of n that fits Earth's actual orbit.

Why pretend planets obey Bohr? Strictly, Bohr's rule applies to atomic-scale bound systems. Here we extrapolate it to macroscopic motion just to find the equivalent n — and to discover how absurdly large it must be.

Step 1 — write the formula. n = m v r.

Step 2 — compute the numerator (the angular momentum). L = m v r = (6.010²⁴)(310⁴)(1.510¹¹). Multiply step-by-step: 6.0 × 3 × 1.5 = 27.0; 10²⁴10⁴10¹¹ = 10³⁹. So L = 2710³⁹ = 2.710⁴⁰ kg m²/s.

Step 3 — divide by . n = 2.710⁴⁰1.05510^-34 = 2.71.05510⁴⁰⁺³⁴ ≈ 2.5610⁷⁴.

Step 4 — interpret. The quantum number for Earth's orbit is roughly n ∼ 2.610⁷⁴ — an absolutely enormous number. The spacing between successive Bohr orbits at this n is so unimaginably small that the orbit looks perfectly continuous on every observable scale. This is exactly Bohr's correspondence principle in action: for very large quantum numbers, quantum predictions blend smoothly into classical mechanics.

Final answer. n ≈ 2.5610⁷⁴.