Atoms — Exam Practice

Answer Negative energy indicates the electron is bound to the nucleus — energy must be supplied to free it entirely (to E = 0, corresponding to n → ∞). Positive total energy would instead describe a free, unbound electron.

Answer Setting Coulomb attraction as the centripetal force: ke²/r² = mv²/r → v² = ke²/(mr).

Bohr's quantization condition: mvr = nh/2π → v = nh/(2πmr).

Substituting and solving for r: r_n = n²h²/(4π²mke²) — which simplifies to r_n = n²a₀, where a₀ ≈ 0.529 Å is the Bohr radius.

Answer (a) Total energy = kinetic + potential energy: E = ½mv² − ke²/r. Using v² = ke²/(mr) (from the force balance) and r_n = n²a₀ (from Q2), substitution and simplification gives:
E_n = −13.6/n² eV.

(b) 1/λ = R(1/2² − 1/3²) = 1.097×10⁷ × (0.25 − 0.111) ≈ 1.524×10⁶ m⁻¹.
λ ≈ 1/(1.524×10⁶) ≈ 656 nm — the well-known red Hα line.

Solution The unexpectedly large-angle scattering of alpha particles could only be explained by a tiny, dense, positively charged nucleus at the atom's centre.

Solution L = nh/2π, n = 1, 2, 3, ... — Bohr's second postulate, later justified by de Broglie's standing-wave condition.

Solution E₁ = −13.6 eV, so exactly 13.6 eV must be supplied to free the electron entirely (raise it to E = 0).

Solution Transitions to n=1 involve the largest energy gaps, giving the shortest wavelengths of hydrogen's series — squarely in the ultraviolet.

Solution r₃ = n²a₀ = 9 × 0.529 Å = 4.76 Å.

Solution An orbit is stable only if it fits a whole number of de Broglie wavelengths around its circumference — this standing-wave condition, combined with λ = h/mv, directly reproduces Bohr's angular momentum quantization.

Solution E₄ = −13.6/16 = −0.85 eV. E₂ = −13.6/4 = −3.4 eV.

Energy released: E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV.