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: Plan:
We will find the donor density from ppm, treat electrons as equal to donors, then use $n_e n_h = n_i^2$ for holes. Then we compare with the list.

Step 2: Donor densities:
One ppm of $5 \times 10^{28}$ is $5 \times 10^{22}$ per cubic metre.
So 2 ppm gives $10 \times 10^{22} = 10^{23}$, and 5 ppm gives $25 \times 10^{22} = 2.5 \times 10^{23}$.

Step 3: Electrons:
Each donor releases one electron, and this number is much bigger than $n_i = 10^{16}$. So $n_e$ equals the donor density.
A is $10^{23}$, which is entry IV.
C is $2.5 \times 10^{23}$, which is entry II.

Step 4: Holes:
Minority hole density is $n_h = n_i^2 / n_e$.
For B: $n_h = 10^{32} / 10^{23} = 10^{9}$, entry III.
For D: $n_h = 10^{32} / (2.5 \times 10^{23}) = 0.4 \times 10^{9} = 4.0 \times 10^{8}$, entry I.

Step 5: Read off the answer:
The pairs are A-IV, B-III, C-II, D-I. Scanning the options, the third one has exactly this set.

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