(a) A particle of rest mass m₀ has a kinetic energy K, show that its de Broglie wavelength is given by λ =…

(a) Using the relativistic energy–momentum relation, E² = p²c² + m₀²c⁴. The total energy is E = K + m₀c². Hence p²c² = (K + m₀c²)² − m₀²c⁴ = K² + 2Km₀c² = K(K + 2m₀c²). Therefore p = √[K(K + 2m₀c²)]/c. By de Broglie’s relation, λ = h/p, so λ = hc/√[K(K + 2m₀c²)]. This is valid for a free relativistic particle of rest mass m₀.

For an electron, m₀c² = 0.511 MeV and K = 2 MeV. Then K + 2m₀c² = 2 + 1.022 = 3.022 MeV, K(K + 2m₀c²) = 2 × 3.022 = 6.044 MeV². Thus √[K(K + 2m₀c²)] = √6.044 MeV = 2.458 MeV = 2.458 × 10⁶ eV. Using hc = 1240 eV nm, λ = 1240 eV nm / (2.458 × 10⁶ eV) = 5.04 × 10⁻⁴ nm. Hence λ ≈ 0.504 pm = 5.04 × 10⁻¹³ m.

If K ≪ m₀c², then K + 2m₀c² ≈ 2m₀c². Therefore λ ≈ hc/√[2Km₀c²] = h/√(2m₀K). So the non-relativistic de Broglie formula is recovered, with condition K ≪ m₀c².

(b) For a simple harmonic oscillator, the normal state is the ground state n = 0. Its energy is E₀ = ½ ħω. The classical turning points ±A satisfy ½ mω²A² = ½ ħω, so A = √(ħ/(mω)). The ground-state wavefunction is ψ₀(x) = (mω/(πħ))^¼ exp(−mωx²/(2ħ)). Let α = mω/ħ. The probability inside the classical limits is P_in = ∫ from −A to A of |ψ₀|² dx = √(α/π) ∫ from −A to A of exp(−αx²) dx. Put u = √α x. Since A = 1/√α, the limits become u = −1 to +1. Hence P_in = (1/√π) ∫ from −1 to 1 of exp(−u²) du = (2/√π) ∫ from 0 to 1 of exp(−u²) du = erf(1). Using erf(1) = 0.8427007929, P_in = erf(1) ≈ 0.8427 = 84.27%. Therefore the probability outside the classical limits is P_out = 1 − erf(1) = erfc(1) = 0.1572992071 ≈ 0.1573. Thus P_out ≈ 15.73% ≈ 16%, as required.

(c) The normal Zeeman effect occurs when the resultant spin of the atom is zero, S = 0. The magnetic moment is due only to orbital motion: μ = −(μ_B/ħ)L, where μ_B is the Bohr magneton. In a constant magnetic field B along the z-axis, the interaction energy is ΔE = −μ·B = μ_B B mₗ. Thus a level with orbital quantum number l splits into 2l + 1 equally spaced sublevels. Electric dipole selection rules allow Δmₗ = 0, ±1, so a single spectral line ν₀ splits into three components: ν₀ − μ_B B/h, ν₀, ν₀ + μ_B B/h. The central component is π-polarized, and the two side components are σ⁺ and σ⁻ polarized. This is the normal Zeeman triplet.

The anomalous Zeeman effect occurs when S ≠ 0 and spin-orbit coupling is present. The total magnetic moment now involves both orbital and spin contributions. In the weak-field Zeeman regime, spin-orbit coupling dominates and J is a good quantum number. The energy shift is ΔE = gⱼ μ_B B Mⱼ, where Mⱼ = −J, …, +J, and the Landé g-factor is gⱼ = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)]. Because gⱼ differs from 1 and depends on L, S, J, the splitting is no longer a simple triplet. A fine-structure level splits into 2J + 1 components, and transitions obey ΔMⱼ = 0, ±1, producing many lines. This more complex pattern is called the anomalous Zeeman effect.

In hydrogen, the non-relativistic energy depends only on the principal quantum number n, so states with different l, mₗ, and mₛ are degenerate. A magnetic field along z reduces the spherical symmetry to axial symmetry. The perturbation H′ = (μ_B/ħ)(Lz + 2Sz)B depends on the magnetic quantum numbers. Therefore states with different mₗ or mⱼ acquire different energies: ΔE = μ_B B(mₗ + 2mₛ) in the simple spin-orbit-free description, or ΔE = gⱼ μ_B B Mⱼ when fine structure is included. This removes the m-degeneracy of the hydrogen levels. For example, an l = 1 level splits into mₗ = −1, 0, +1 components. Since the hydrogen electron has spin ½, its weak-field splitting is generally anomalous rather than a pure normal triplet. In strong fields, the Paschen–Back effect can occur, where L and S decouple and the shift becomes approximately μ_B B(mₗ + 2mₛ).