(a) Calculate the zero point energy for a particle in an infinite potential well for the following cases : (i) a 100 g ball…

(a) Zero-point energy in an infinite well

For an infinite well of width L, Eₙ = n²h²/(8mL²), so the zero-point energy is E₁ = h²/(8mL²), with h = 6.626×10⁻³⁴ J s and h² = 4.390×10⁻⁶⁷ J² s². (i) Ball: m = 0.100 kg, L = 5 m. E₁ = 4.390×10⁻⁶⁷/(8 × 0.100 × 25) = 2.20×10⁻⁶⁸ J = 1.4×10⁻⁴⁹ eV. (ii) Oxygen atom: m = 16 × 1.6605×10⁻²⁷ = 2.657×10⁻²⁶ kg, L = 2×10⁻¹⁰ m. E₁ = 4.390×10⁻⁶⁷/(8 × 2.657×10⁻²⁶ × 4×10⁻²⁰) = 5.16×10⁻²³ J = 3.2×10⁻⁴ eV. (iii) Electron: m = 9.109×10⁻³¹ kg, L = 10⁻¹⁰ m. E₁ = 4.390×10⁻⁶⁷/(8 × 9.109×10⁻³¹ × 10⁻²⁰) = 6.02×10⁻¹⁸ J = 37.6 eV. Comment: E₁ ∝ 1/(mL²) is the energy forced by confinement through the uncertainty principle (Δp ≳ ħ/L). For the ball it is about 10⁻⁶⁸ J, against thermal energy kT ≈ 4×10⁻²¹ J at room temperature, so it is unobservable. For atomic confinement it reaches eV scale and governs atomic structure.

(b) Charged oscillator in a uniform electric field

H = p²/2m + ½mω²X² − qEX. Complete the square with x₀ = qE/(mω²): ½mω²(X − x₀)² = ½mω²X² − qEX + q²E²/(2mω²), so H = p²/2m + ½mω²(X − x₀)² − q²E²/(2mω²). With the new coordinate ξ = X − x₀ (p is unchanged, so [ξ, p] = iħ), this is a standard oscillator of frequency ω plus a constant. The spectrum is Eₙ = (n + ½)ħω − q²E²/(2mω²), n = 0, 1, 2, … where the nth excited state has quantum number n and n = 0 is the ground state. The wave functions are the oscillator functions shifted by x₀: ψₙ(x) = (1/√(2ⁿn!))(mω/πħ)¹/4 exp[−mω(x − x₀)²/2ħ] Hₙ(√(mω/ħ)(x − x₀)). The field shifts the equilibrium to x₀ and lowers every level by q²E²/(2mω²), which has the dimensions of energy. The level spacing stays ħω.

(c) Minimum depth for two s-wave bound states

Take V = −V₀ for r < a and 0 for r > a, with l = 0 and u(r) = rR(r). The radial equation is −(ħ²/2m)u″ + Vu = Eu with E < 0. Inside: u = A sin kr, k² = 2m(V₀ + E)/ħ² (u(0) = 0). Outside: u = B e^(−κr), κ² = −2mE/ħ². Continuity of u′/u at r = a gives k cot ka = −κ, i.e. ξ cot ξ = −η with ξ = ka and η = κa. A new bound state appears as E → 0⁻, when η → 0, so ξ cot ξ = 0 and ξ = π/2, 3π/2, … The first bound state appears at k₀a = π/2 and the second at k₀a = 3π/2, with k₀² = 2mV₀/ħ². Two s-states need k₀a > 3π/2, so the minimum depth is V₀,min = 9π²ħ²/(8ma²). Since l = 0, there is no centrifugal barrier.

(d) Stern-Gerlach deflection of hydrogen atoms

Ground-state hydrogen has no orbital moment; the spin gives μ_z = ±μ_B. The force is F = μ_B (dB/dz) = 9.274×10⁻²⁴ × 10 = 9.27×10⁻²³ N. Time in the magnet: t = L/v. Take the thermal speed from ½mv² = (3/2)kT, so v² = 3kT/m. Deflection at exit: y = ½(F/m)t² = μ_B(dB/dz)L²/(2mv²) = μ_B(dB/dz)L²/(6kT). y = (9.274×10⁻²⁴ × 10 × 1)/(6 × 1.381×10⁻²³ × 400) = 9.274×10⁻²³/3.314×10⁻²⁰ = 2.8×10⁻³ m. The two spin states deflect in opposite directions, so the beam splits into two spots, each 2.8 mm from the axis. The mass cancels, so the atomic mass is not needed.

(e) Precession rate and torque on an l = 3 electron

The orbital magnetic moment is μ = −(e/2mₑ)L, so the magnetic field makes L precess at the Larmor angular frequency ω_L = eB/(2mₑ) = μ_B B/ħ = (9.274×10⁻²⁴ × 0.1)/(1.055×10⁻³⁴) = 8.8×10⁹ rad/s, that is ν = ω_L/2π = 1.4×10⁹ Hz. This rate does not depend on the angle. For l = 3, μ = μ_B√(l(l + 1)) = √12 μ_B = 3.21×10⁻²³ J/T. Torque: τ = μB sinθ = 3.21×10⁻²³ × 0.1 × sin30° = 1.6×10⁻²⁴ N m (equivalently √3 μ_B B), directed along μ × B and perpendicular to both.