Atoms

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Exercise 12.1Bohr's Model and Hydrogen Spectrum

Choose the correct alternative from the clues given at the end of each statement: (a) The size of the atom in Thomson's model is .......... the atomic size in Rutherford's model. (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (c) A classical atom based on .......... is doomed to collapse. (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (e) The positively charged part of the atom possesses most of the mass in .......... (Clues: Thomson's model/Rutherford's model, both the models, none of the models, Thomson's model, Rutherford's model)

Solution

(a) The size of the atom in Thomson's model is the same as the atomic size in Rutherford's model. (Both models predict atomic size of approximately 10⁻¹⁰ m) (b) In the ground state of Thomson's model, electrons are in stable equilibrium, while in Rutherford's model, electrons always experience a net force. (In Thomson's model, electrons sit at fixed equilibrium positions; in Rutherford's, they orbit and experience centripetal force) (c) A classical atom based on Rutherford's model is doomed to collapse. (Orbital electrons radiate energy and spiral inward in classical theory) (d) An atom has a nearly continuous mass distribution in Thomson's model but has a highly non-uniform mass distribution in Rutherford's model. (e) The positively charged part of the atom possesses most of the mass in Rutherford's model.

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. What results do you expect?

Solution

Hydrogen nucleus (proton) is much lighter than an alpha particle: m_proton = 1.67 × 10⁻²⁷ kg vs m_alpha = 6.64 × 10⁻²⁷ kg When a heavier particle (alpha) collides with a lighter particle (proton): • The lighter proton would be knocked forward • The alpha particle would deflect very slightly • Large-angle scattering would NOT be observed (unlike with gold) Since the hydrogen nucleus is much lighter, it cannot send the alpha particle back. Most alpha particles would pass through with small deflections. Conclusion: The Rutherford experiment cannot be replicated with hydrogen — we would NOT observe large-angle scattering, and thus could not infer the existence of a tiny, massive nucleus from hydrogen alone.

What is the shortest wavelength present in the Paschen series of spectral lines?

Solution

Paschen series: transitions from higher levels to n = 3 Rydberg formula: 1/λ = R_H(1/n₁² – 1/n₂²) Shortest wavelength = series limit (n₂ → ∞, n₁ = 3): 1/λ_min = R_H(1/3² – 0) = R_H/9 R_H = 1.097 × 10⁷ m⁻¹ λ_min = 9/R_H = 9/(1.097 × 10⁷) = 8.204 × 10⁻⁷ m ≈ 820 nm The shortest wavelength in the Paschen series is ≈ 820 nm (infrared region).

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Solution

ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J Using: ΔE = hf f = ΔE/h = 3.68 × 10⁻¹⁹/6.626 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz The frequency of emitted radiation is 5.55 × 10¹⁴ Hz (visible light — green-yellow).

The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Solution

Total energy E = –13.6 eV For a circular orbit (Bohr model): Kinetic energy K = –E = +13.6 eV (positive) Potential energy U = 2E = –27.2 eV Verification: E = K + U = 13.6 + (–27.2) = –13.6 eV ✓ (In atomic physics: K = –E and U = 2E, since the Coulomb potential energy is negative and K = –½U for circular orbits)

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Solution

Energy of n = 1 level: E₁ = –13.6 eV Energy of n = 4 level: E₄ = –13.6/4² = –13.6/16 = –0.85 eV Energy of photon: ΔE = E₄ – E₁ = –0.85 – (–13.6) = 12.75 eV = 12.75 × 1.6 × 10⁻¹⁹ = 2.04 × 10⁻¹⁸ J Frequency: f = ΔE/h = 2.04 × 10⁻¹⁸/6.626 × 10⁻³⁴ = 3.08 × 10¹⁵ Hz Wavelength: λ = c/f = 3 × 10⁸/3.08 × 10¹⁵ = 9.74 × 10⁻⁸ m ≈ 97.4 nm (UV region)

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n = 2 and n = 3 orbits?

Solution

Bohr radius: rₙ = n²r₁ r₁ = 5.3 × 10⁻¹¹ m (given) Radius of n = 2 orbit: r₂ = 2² × r₁ = 4 × 5.3 × 10⁻¹¹ = 21.2 × 10⁻¹¹ m ≈ 2.12 Å Radius of n = 3 orbit: r₃ = 3² × r₁ = 9 × 5.3 × 10⁻¹¹ = 47.7 × 10⁻¹¹ m ≈ 4.77 Å

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution

Energy levels of hydrogen: E₁ = –13.6 eV, E₂ = –3.4 eV, E₃ = –1.51 eV, E₄ = –0.85 eV Energy needed to reach: n=2: 13.6 – 3.4 = 10.2 eV n=3: 13.6 – 1.51 = 12.09 eV n=4: 13.6 – 0.85 = 12.75 eV > 12.5 eV (not accessible) With 12.5 eV electrons, hydrogen atoms can be excited up to n = 3. Possible transitions and series: • n=3 → n=1: Lyman series (UV) • n=2 → n=1: Lyman series (UV) • n=3 → n=2: Balmer series (visible) Lyman series and Balmer series wavelengths will be emitted. Wavelengths: • n=2→1: λ = 121.6 nm (Lyman α) • n=3→1: λ = 102.6 nm (Lyman β) • n=3→2: λ = 656 nm (Balmer α, visible red)

In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6 × 10²⁴ kg.)

Solution

Bohr's quantisation condition: mvr = nh/2π m = 6 × 10²⁴ kg, v = 3 × 10⁴ m/s, r = 1.5 × 10¹¹ m n = 2πmvr/h = (2π × 6 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹)/(6.626 × 10⁻³⁴) = (2π × 2.7 × 10⁴⁰)/(6.626 × 10⁻³⁴) = (6.283 × 2.7 × 10⁴⁰)/(6.626 × 10⁻³⁴) = (16.96 × 10⁴⁰)/(6.626 × 10⁻³⁴) = 2.56 × 10⁷⁴ n ≈ 2.56 × 10⁷⁴ This enormous quantum number shows why quantum effects are completely negligible for macroscopic objects like the Earth.

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