(a) Set up the Schrodinger’s wave equation for one dimensional potential barrier and obtain the probability of tunneling. (20…

(a) A particle of mass m and energy E impinges on a one‑dimensional rectangular barrier of height V₀ and width a, where the potential is V(x) = 0 for x < 0 and x > a, and V(x) = V₀ for 0 < x < a; we consider the tunnelling case E < V₀. The barrier, with its three regions I (x < 0), II (0 < x < a) and III (x > a), is sketched below.

The time‑independent Schrödinger equation is −(ℏ²/2m)(d²ψ/dx²) + V(x)ψ = Eψ. Defining the wave numbers k = √(2mE)/ℏ and κ = √(2m(V₀ − E))/ℏ, the region‑wise solutions are: Region I, ψ_I = A e^(ikx) + B e^(−ikx), the sum of an incident and a reflected plane wave; Region II, ψ_II = C e^(κx) + D e^(−κx), an evanescent (exponentially varying) wave since E < V₀; and Region III, ψ_III = F e^(ikx), a transmitted wave only, with no left‑moving component. At the two interfaces x = 0 and x = a both ψ and its derivative dψ/dx must be continuous, giving four linear equations among A, B, C, D, F. Eliminating the interior amplitudes C and D and solving for the transmission amplitude F/A, the transmission (tunnelling) probability is T = |F/A|² = [1 + V₀² sinh²(κa) / (4E(V₀ − E))]⁻¹. In the opaque (high or wide barrier) limit κa ≫ 1, sinh(κa) ≈ ½ e^(κa), so T ≈ [16E(V₀ − E)/V₀²] e^(−2κa): the tunnelling probability decays exponentially with the barrier width, the defining feature of quantum tunnelling.

(b) The required identity follows directly from the virial theorem for the Coulomb potential. The hydrogen Hamiltonian is H = T + V with V(r) = −e²/(4πε₀ r) ∝ 1/r. For a potential V = k rⁿ the virial theorem gives 2⟨T⟩ = n⟨V⟩; this follows because in a stationary state d/dt ⟨r·p⟩ = 0, which yields 2⟨T⟩ = ⟨r·∇V⟩ = n⟨V⟩. For the Coulomb case n = −1, so 2⟨T⟩ = −⟨V⟩, i.e. ⟨T⟩ = −½⟨V⟩. The energy of the *n*th stationary state is the sum of the expectation values, Eₙ = ⟨T⟩ + ⟨V⟩. Substituting the virial relation, Eₙ = −½⟨V⟩ + ⟨V⟩ = ½⟨V⟩, and since ⟨T⟩ = −½⟨V⟩ we also have Eₙ = −⟨T⟩. Hence the rigorous result is Eₙ = ½⟨V⟩ = −⟨T⟩. As an explicit check, for the ground state ⟨1/r⟩ = 1/a₀, so ⟨V⟩ = −e²/(4πε₀ a₀) = −27.2 eV = 2E₁, while E₁ = −13.6 eV = −⟨T⟩, in agreement. The relation Eₙ = ⟨V⟩ as printed is therefore off by a factor ½ (a misprint). The correct statement, derived above, is Eₙ = ½⟨V⟩ = −⟨T⟩.

(c)(i) For a fixed principal quantum number n, the orbital quantum number l takes the n values 0, 1, …, *n*−1. For each l the magnetic quantum number mₗ takes the 2*l*+1 values −*l*, …, +*l*, so the number of orbital states is Σₗ₌₀ⁿ⁻¹(2l+1) = 1 + 3 + 5 + … + (2n−1), the sum of the first n odd integers, which equals *n*². So there are *n*² spatial (orbital) states for a given n (e.g. n = 2 gives 1 + 3 = 4). Including the two spin projections the degeneracy becomes 2*n*².

(c)(ii) The term symbol is written ²ˢ⁺¹L_J. Reading off ⁴D₅/₂: the letter D gives L = 2; the superscript 4 = 2S+1 gives S = 3/2; the subscript gives J = 5/2. The spin multiplicity 2S+1 = 4 fixes S = 3/2, and S = 3/2 needs at least three electrons with parallel spins; three non‑equivalent electrons can also give L = 2 (a ⁴D term), so the minimum number of electrons involved is three. A possible configuration is 2p²3p, as in the 2p²(³P)3p ⁴D term of nitrogen (an even number of electrons, such as 3d⁴, can only give integer S and so never a quartet).