(a) An n-p-n transistor with β = 49 is used in common-emitter amplifier mode with Vcc = 10V and RL = 2 kΩ. If a 100 kΩ resistor…

(a) Let the feedback resistor be R_f = 100 kΩ, collector load R_L = 2 kΩ, Vcc = 10 V, β = 49. The emitter is grounded and VBE = 0, so the base voltage is V_B = 0.

Let I_C be the quiescent collector current and I_B the base current. The current through the collector-to-base feedback resistor is the base current: I_B = (V_C − V_B)/R_f = V_C/R_f.

At the collector node, the current through R_L is I_C + I_B, because I_B leaves the collector node towards the base. Hence V_C = Vcc − (I_C + I_B)R_L.

Using the transistor relation I_C = β I_B, this becomes V_C = Vcc − (β I_B + I_B)R_L = Vcc − (β + 1)I_B R_L. But V_C = I_B R_f, so I_B R_f = Vcc − (β + 1)I_B R_L => I_B[R_f + (β + 1)R_L] = Vcc => I_B = Vcc/[R_f + (β + 1)R_L].

Therefore I_C = β I_B = β Vcc/[R_f + (β + 1)R_L].

Substituting the values, with Vcc in volts, resistances in kΩ and current in mA: I_C = 49 × 10/[100 + (49 + 1) × 2] mA = 490/[100 + 100] mA = 490/200 mA = 2.45 mA.

Check the active region: V_C = I_B R_f = (I_C/β)R_f = (2.45/49) × 100 V = 5 V. Thus V_CE = V_C − 0 = 5 V > 0, so the transistor is in the active region.

Final: I_C ≈ 2.45 mA.

(b) For scandium in the metallic state, the single 3d electron has orbital quantum number L = 2 and spin S = 1/2. Hund’s third rule applies: for a shell less than half filled, the ground-state total angular momentum is J = |L − S| = 2 − 1/2 = 3/2.

So the ground term is ²D_3/2, giving J = 3/2.

The Landé splitting factor is g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)].

Here J(J + 1) = (3/2)(5/2) = 15/4, S(S + 1) = (1/2)(3/2) = 3/4, L(L + 1) = 2 × 3 = 6.

Thus g_J = 1 + [15/4 + 3/4 − 6]/[2 × 15/4] = 1 + [18/4 − 24/4]/(15/2) = 1 + [−6/4]/(15/2) = 1 + [−3/2]/[15/2] = 1 − 1/5 = 4/5 = 0.8.

In a magnetic field B along z, the Zeeman energy is E = g_J μ_B B m_J, where m_J = −J, −J + 1, …, +J. The lowest energy level has m_J = −J = −3/2.

Therefore E_min = g_J μ_B B (−3/2) = (4/5)(−3/2) μ_B B = −(6/5) μ_B B.

With B = 0.5 T and μ_B = 9.274 × 10⁻²⁴ J/T, μ_B B = 9.274 × 10⁻²⁴ × 0.5 = 4.637 × 10⁻²⁴ J.

Hence E_min = −(6/5) × 4.637 × 10⁻²⁴ J = −5.564 × 10⁻²⁴ J ≈ −5.56 × 10⁻²⁴ J.

In electron-volts, E_min ≈ −5.56 × 10⁻²⁴ / 1.602 × 10⁻¹⁹ eV ≈ −3.47 × 10⁻⁵ eV.

Final: J = 3/2, g = 4/5 = 0.8, and the lowest Zeeman level energy is ≈ −5.56 × 10⁻²⁴ J = −3.47 × 10⁻⁵ eV.

(c) For an n-channel FET, pinch-off occurs when the depletion regions extending from the two gate junctions just meet at the centre of the channel. Let the full channel width be d = 6 × 10⁻⁴ cm = 6 × 10⁻⁶ m. The half-width is a = d/2 = 3 × 10⁻⁶ m.

For an abrupt p⁺–n junction, the depletion width into the n-side is W = sqrt[2 ε_s (V + V_bi)/(q N_D)]. Ignoring the built-in potential V_bi, or defining the internal pinch-off voltage, pinch-off occurs when W = a. Thus V_p = q N_D a²/(2 ε_s).

Since a = d/2, V_p = q N_D (d/2)²/(2 ε_s) = q N_D d²/(8 ε_s).

Given κ = 12 for silicon, ε_s = κ ε0 = 12 × 8.854 × 10⁻¹² F/m = 1.0625 × 10⁻¹⁰ F/m.

Also, N_D = 10¹⁵ cm⁻³ = 10²¹ m⁻³, q = 1.602 × 10⁻¹⁹ C, d = 6 × 10⁻⁶ m.

Therefore V_p = [1.602 × 10⁻¹⁹ × 10²¹ × (6 × 10⁻⁶)²]/[8 × 1.0625 × 10⁻¹⁰].

Compute the numerator: 1.602 × 10⁻¹⁹ × 10²¹ = 160.2 C/m³, and (6 × 10⁻⁶)² = 36 × 10⁻¹² m². So 160.2 × 36 × 10⁻¹² = 5.767 × 10⁻⁹.

The denominator is 8 × 1.0625 × 10⁻¹⁰ = 8.500 × 10⁻¹⁰.

Hence V_p = 5.767 × 10⁻⁹ / 8.500 × 10⁻¹⁰ V ≈ 6.79 V.

Final: V_p ≈ 6.79 V. This uses the given 6 × 10⁻⁴ cm as the full channel width. If the question instead means the half-width, the value would be four times larger, about 27.1 V.