Electronic configuration of ⁶C

⁶C = 1s², 2s² 2p²

The electronic configuration of ¹⁴Si

¹⁴Si = 1s², 2s² 2p⁶, 3s² 3p²

As they are away from Nucleus, so effect of nucleus is low for Si even for Sn and Pb are almost mettalic.

Here, R_C = 2 k$$\Omega $$ = 2 × 10³ $$\Omega $$

V₀ = 2 V, R_B = 1 k$$\Omega $$ = 1 × 10³ $$\Omega $$, $$\beta $$ = 100

The output voltage, across the load R_C

V₀ = I_CR_C = 2

The collector current (I_C)

I_C = $${2 \over {2 \times {{10}^3}}}$$ = 10^-3 A = 1 mA

Current gain($$\beta $$) = $${{{I_C}} \over {{I_B}}}$$ = 100

$$ \Rightarrow $$ I_B = $${{{I_C}} \over {100}}$$ = $${{{{10}^{ - 3}}} \over {100}}$$ = 10^-5 A

Input voltage, V_i

= I_BR_B = (10^–5 A) (1 × 10³ $$\Omega $$) = 10^–2 V = 10 mV

Here D₁ is in forward bias and D₂ is in reverse bias so, D₁ will conduct and D₂ will not conduct.
So, the current supplied by the battery is

I = $${5 \over {10}}$$ = 0.5 A

(n_i)² = n_e × n_h

(1.5 × 10¹⁶)² = n_e (4.5 × 10²²)

So n_e = 5 × 10⁹

Now n_h = 4.5 × 10²²

$$ \Rightarrow $$ n_h $$ >> $$ n_e

Hence, semiconductor is p-type

and n_e = 5 × 10⁹ m^–3