(a) Sketch the dc load line for the circuit shown. (10 marks) (b) A solid contains a dilute concentration of Nd³⁺ ions, each of…

(a) Applying KVL to the collector-emitter loop of the circuit: V_CC = I_C R_C + V_CE, so I_C = (V_CC − V_CE)/R_C = (20 − V_CE)/10 mA, with V_CE in volts.

Two end points fix the line:

• Cutoff (I_C = 0): V_CE = V_CC = 20 V, the point (20 V, 0).
• Saturation (V_CE = 0): I_C = 20 V/10 kΩ = 2 mA, the point (0, 2 mA).

The dc load line is the straight line joining these points on the I_C (vertical) versus V_CE (horizontal) axes, with slope −1/R_C = −0.1 mA/V. The Q-point is where this line meets the output characteristic for I_B ≈ (20 − V_BE)/R_B; since R_B is not given, Q cannot be fixed numerically.

(b) Nd³⁺ is 4f³. By Hund's rules, S = 3/2, L = 3 + 2 + 1 = 6, and since the shell is less than half full, J = L − S = 9/2.

g_J = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)] = 1 + [99/4 + 15/4 − 42]/(99/2) = 1 − 3/11 = 8/11.

Effective moment: p_eff = g_J√[J(J+1)] = (8/11)√(99/4) ≈ 3.62, so p_eff² = 144/11.

The ions are dilute and non-interacting, so Curie's law applies: χ = μ₀ n p_eff² μ_B²/(3k_BT).

With n = 10²⁵ m⁻³, T = 1 K, μ₀ = 4π × 10⁻⁷ H/m, μ_B = 9.274 × 10⁻²⁴ J/T and k_B = 1.381 × 10⁻²³ J/K, χ = (4π × 10⁻⁷ × 10²⁵ × 144/11 × (9.274 × 10⁻²⁴)²)/(3 × 1.381 × 10⁻²³) ≈ 3.4 × 10⁻⁴ (dimensionless SI volume susceptibility).

In cgs units this is χ_SI/4π ≈ 2.7 × 10⁻⁵. The result is valid in weak fields, where μ_B B ≪ k_BT.

(c) Internal conversion: an excited nucleus transfers its de-excitation energy ΔE directly to a bound atomic electron, usually in the K shell, through the electromagnetic interaction between nucleus and electron, instead of emitting a gamma photon. The ejected electron has a discrete energy E_e = ΔE − B_e, where B_e is its binding energy. The vacancy is then filled, emitting characteristic X-rays or Auger electrons. It is not beta decay, and no gamma photon is emitted in the process.

Internal conversion coefficient: α = N_e/N_γ = λ_e/λ_γ, the ratio of conversion electrons to gamma photons for the same transition. Shell-wise, α = α_K + α_L + α_M + …, and the fraction of decays by conversion is α/(1 + α).

It becomes important when:

• The transition energy is low, since gamma emission is then slow but conversion remains possible if ΔE > B_e.
• Z is high, because the K-shell electron density at the nucleus rises roughly as Z³.
• The multipolarity is high or the spin change large, since gamma emission is strongly retarded.
• The transition is 0⁺ → 0⁺ (E0), where single-photon emission is forbidden and conversion is the dominant mode.