Q: A silicon sample has n_i = 1.5 \times 10^{16}\text{ m}^{-3} and is doped so that n_e = 4.5 \times 10^{22}\text{ m}^{-3}. The hole concentration is:

5 \times 10^9\text{ m}^{-3}. n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} = \dfrac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^9\text{ m}^{-3}.

Question 1

In an intrinsic semiconductor, the numbers of free electrons and holes are:

Accepted Answer

Equal. Every electron promoted to the conduction band leaves behind exactly one hole, so n_e = n_h = n_i.

Question 2

An extrinsic semiconductor is one that has been:

Accepted Answer

Deliberately doped with a suitable impurity. Doping in a proportion as small as one part in a million transforms the conductivity.

Question 3

An n-type semiconductor is produced by doping with a:

Accepted Answer

Pentavalent impurity such as phosphorus or arsenic. The fifth valence electron is not needed for bonding and becomes a free conduction electron.

Question 4

A p-type semiconductor is produced by doping with a:

Accepted Answer

Trivalent impurity such as boron or indium. With only three valence electrons, the impurity leaves a vacancy in one bond — a hole.

Question 5

In an n-type semiconductor, the majority and minority carriers are respectively:

Accepted Answer

Electrons and holes. Donor atoms supply plenty of electrons while holes arise only from thermal generation.

Question 6

In a p-type semiconductor, the majority and minority carriers are respectively:

Accepted Answer

Holes and electrons. Acceptor atoms create an abundance of holes, leaving electrons as the scarce minority carriers.

Question 7

A doped semiconductor, whether n-type or p-type, is:

Accepted Answer

Electrically neutral overall. Each impurity atom brings its own nucleus and electrons, so the total charge balances exactly.

Question 8

In an intrinsic semiconductor at absolute zero, the Fermi level lies:

Accepted Answer

At the middle of the forbidden energy gap. Equal electron and hole populations place the Fermi level midway between the two bands.

Question 9

In an n-type semiconductor at room temperature, the concentration of majority carriers is determined mainly by the:

Accepted Answer

Concentration of donor impurity atoms. Essentially all donor atoms are ionised at room temperature, so n_e \approx N_D.

Question 10

A silicon sample has n_i = 1.5 	imes 10^{16}	ext{ m}^{-3} and is doped so that n_e = 4.5 	imes 10^{22}	ext{ m}^{-3}. The hole concentration is:

Accepted Answer

5 	imes 10^9	ext{ m}^{-3}. n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{22}} = \dfrac{2.25 	imes 10^{32}}{4.5 	imes 10^{22}} = 5 	imes 10^9	ext{ m}^{-3}.

Question 11

A semiconductor has n_i = 10^{16}	ext{ m}^{-3} and is doped to give n_e = 10^{22}	ext{ m}^{-3}. The hole concentration is:

Accepted Answer

10^{10}	ext{ m}^{-3}. n_h = \dfrac{n_i^2}{n_e} = \dfrac{10^{32}}{10^{22}} = 10^{10}	ext{ m}^{-3}.

Question 12

A semiconductor is doped with 10^{21} donor atoms per cubic metre. The majority carrier concentration is approximately:

Accepted Answer

10^{21}	ext{ m}^{-3} of electrons. At room temperature virtually every donor is ionised, so the electron concentration matches the doping.

Question 13

Germanium doped with phosphorus becomes:

Accepted Answer

An n-type semiconductor. Phosphorus is pentavalent, so it donates electrons.

Question 14

Silicon doped with boron becomes:

Accepted Answer

A p-type semiconductor. Boron is trivalent, so it accepts electrons and creates holes.

Question 15

In a doped semiconductor at a given temperature, if the electron concentration increases, the hole concentration:

Accepted Answer

Decreases, since n_en_h = n_i^2 is fixed. The mass-action law keeps the product constant, so the two concentrations move in opposite directions.

Question 16

In an intrinsic semiconductor the relation between electron, hole and intrinsic concentrations is:

Accepted Answer

n_e = n_h = n_i. Thermal generation always produces electrons and holes in pairs.

Question 17

A semiconductor has n_i = 2 	imes 10^{16}	ext{ m}^{-3} and is doped to give n_e = 8 	imes 10^{22}	ext{ m}^{-3}. The hole concentration is:

Accepted Answer

5 	imes 10^9	ext{ m}^{-3}. n_h = \dfrac{(2 	imes 10^{16})^2}{8 	imes 10^{22}} = \dfrac{4 	imes 10^{32}}{8 	imes 10^{22}} = 5 	imes 10^9	ext{ m}^{-3}.

Question 18

The numbers of valence electrons in silicon, phosphorus and boron are respectively:

Accepted Answer

4, 5 and 3. This is precisely why phosphorus donates an electron and boron accepts one.

Question 19

Typical doping levels in a semiconductor are of the order of one impurity atom in:

Accepted Answer

10^6 host atoms. Even this tiny proportion raises the conductivity by many orders of magnitude.

Question 20

The conductivity of an extrinsic semiconductor compared with that of the intrinsic material is:

Accepted Answer

Very much greater. Doping supplies carriers far more numerous than those produced by thermal generation alone.

Intrinsic & Extrinsic Semiconductors — MH-CET Physics MCQs with Solutions

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