Question

Match List-I with List-II. Assume that the Si intrinsic semiconductor has \(5 \times 10^{28}\) atoms/m\(^3\). (Given: the intrinsic concentration of electrons in the semiconductor is \(10^{16}\ \text{m}^{-3}\)).

	LIST-I		LIST-II
A	When it is doped with 2 ppm concentration of a pentavalent atom, \(n_e\) is	I	\(4.0 \times 10^{8}\ \text{m}^{-3}\)
B	When it is doped with 2 ppm concentration of a pentavalent atom, \(n_h\) is	II	\(2.5 \times 10^{23}\ \text{m}^{-3}\)
C	When it is doped with 5 ppm concentration of a pentavalent atom, \(n_e\) is	III	\(10^{9}\ \text{m}^{-3}\)
D	When it is doped with 5 ppm concentration of a pentavalent atom, \(n_h\) is	IV	\(10^{23}\ \text{m}^{-3}\)

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-II, B-III, C-IV, D-I
• C. A-IV, B-III, C-II, D-I
• D. A-IV, B-II, C-III, D-I

Hint:

Find \(n_e\) from the ppm dopant count, then use \(n_h = n_i^2/n_e\).

Answer 1

The Correct Option is C

Step 1: Understand the concept:
When a pentavalent atom is added to silicon, each dopant atom gives one free electron. So the electron concentration \(n_e\) is almost equal to the dopant concentration. The hole concentration then follows from the mass action law \(n_e n_h = n_i^2\).

Step 2: Convert ppm to a number density:
1 ppm means 1 dopant atom in \(10^6\) atoms. The number of Si atoms is \(5 \times 10^{28}\) per cubic metre.
For 2 ppm: \[ N_D = 2 \times 10^{-6} \times 5 \times 10^{28} = 10^{23}\ \text{m}^{-3} \] For 5 ppm: \[ N_D = 5 \times 10^{-6} \times 5 \times 10^{28} = 2.5 \times 10^{23}\ \text{m}^{-3} \]

Step 3: Find n_e for A and C:
Doping adds far more electrons than the intrinsic \(10^{16}\), so \(n_e \approx N_D\).
A (2 ppm): \(n_e = 10^{23}\ \text{m}^{-3}\), which is IV.
C (5 ppm): \(n_e = 2.5 \times 10^{23}\ \text{m}^{-3}\), which is II.

Step 4: Find n_h for B and D:
Use \(n_h = \dfrac{n_i^2}{n_e}\) with \(n_i^2 = 10^{32}\).
B (2 ppm): \[ n_h = \frac{10^{32}}{10^{23}} = 10^{9}\ \text{m}^{-3} \] which is III.
D (5 ppm): \[ n_h = \frac{10^{32}}{2.5 \times 10^{23}} = 4.0 \times 10^{8}\ \text{m}^{-3} \] which is I.

Step 5: Match and check the options:
So A-IV, B-III, C-II, D-I.
Option (1) and (2) assign the wrong values to A. Option (4) gives B as II, but \(n_h\) cannot be as high as \(2.5 \times 10^{23}\) in an n-type sample. Only option (3) matches.

Final Answer:
The correct matching is A-IV, B-III, C-II, D-I, which is option (3). \[ \boxed{\text{A-IV, B-III, C-II, D-I}} \]