(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) Collector-base feedback bias: R_f = 100 kΩ, R_L = 2 kΩ, V_CC = 10 V, β = 49, V_BE = 0, so V_B = 0 and V_C = I_B R_f.

The current through R_L is I_C + I_B, so V_CC = (I_C + I_B)R_L + I_B R_f. With I_C = βI_B, V_CC = I_B[(β + 1)R_L + R_f] I_B = 10 V/[50 × 2 kΩ + 100 kΩ] = 10 V/200 kΩ = 50 µA.

Therefore I_C = βI_B = 49 × 50 µA = 2.45 mA.

Check: V_C = V_CE = I_B R_f = 50 µA × 100 kΩ = 5 V, which is positive, so the transistor is in the active region. (Equivalently, V_CC − (I_C + I_B)R_L = 10 − 2.5 × 2 = 5 V.)

(b) Sc has one 3d electron: L = 2, S = 1/2. The shell is less than half full, so Hund's third rule gives J = |L − S| = 3/2, the ground term ²D₃/₂.

Landé factor: g_J = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)] = 1 + [15/4 + 3/4 − 6]/[2 × 15/4] = 1 − (3/2)/(15/2) = 4/5 = 0.8.

In a field B the energy is E = g_J μ_B B m_J, with m_J = −3/2, …, +3/2. The lowest level is m_J = −3/2: E_min = −(3/2)(4/5)μ_B B = −(6/5)μ_B B.

With μ_B = 9.274 × 10⁻²⁴ J/T and B = 0.5 T, μ_B B = 4.637 × 10⁻²⁴ J, so E_min ≈ −5.56 × 10⁻²⁴ J ≈ −3.47 × 10⁻⁵ eV.

This treats the 3d electron as an atomic (localised) moment, as the question intends.

(c) Pinch-off occurs when the depletion layers from the two gates meet at the channel centre. With full channel width d = 6 × 10⁻⁴ cm, the half-width is a = 3 × 10⁻⁶ m.

For an abrupt junction with the gate-channel built-in potential neglected, W = √(2ε V/(qN_D)). Setting W = a, V_p = qN_D a²/(2ε), with ε = 12 × 8.854 × 10⁻¹² = 1.0625 × 10⁻¹⁰ F/m, N_D = 10²¹ m⁻³.

V_p = (1.602 × 10⁻¹⁹ × 10²¹ × 9 × 10⁻¹²)/(2 × 1.0625 × 10⁻¹⁰) = 1.442 × 10⁻⁹/2.125 × 10⁻¹⁰ ≈ 6.8 V.

If 6 × 10⁻⁴ cm is instead taken as the half-width a, the value is four times larger, about 27 V.