Topic 1 : Semiconductor Material

1 . On the basis of electrical conductivity, which one of the following material has the smallest resistivity? ⇒  (NEET 2023 Manipur)

A. Germanium

B. Silver

C. Glass

D. Silicon

The Correct Answer is Option (B)

Electrical conductivity is a measure of a material's ability to conduct an electric current. The higher the conductivity, the lower the resistivity. So, if we are looking for the material with the smallest resistivity, we are looking for the material with the highest electrical conductivity. Among the options given, silver (Option B) is known to have the highest electrical conductivity and thus, the smallest resistivity. Germanium (Option A) and silicon (Option D) are semiconductors, meaning they have moderate conductivity that can be manipulated, while glass (Option C) is generally a poor conductor, or an insulator. Therefore, silver is the correct answer.

2 . A p-type extrinsic semiconductor is obtained when Germanium is doped with: ⇒  (NEET 2023 Manipur)

A. Antimony

B. Phosphorous

C. Arsenic

D. Boron

The Correct Answer is (D)

For p type semiconductor trivalent impurity added

3. As the temperature increases, the electrical resistance ⇒  (NEET 2022 Phase 1)

A. Increases for both conductors and semiconductors

B. Decreases for both conductors and semiconductors

C. Increases for conductors but decreases for semiconductors

D. Decreases for conductors but increases for semiconductors

The Correct Answer is (C)

As the temperature increases the resistivity of the conductor increases hence the electrical resistance increases. However for semiconductor the resistivity decreases with the temperature. Hence electrical resistance of semiconductor decreases.

4. The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them. ⇒  (NEET 2021)

A. No current will flow in p-type, current will only flow in n-type

B. Current in n-type = current in p-type

C. current in p-type > current in n-type

D. current in n-type > current in p-type

The Correct Answer is (D)

In N type semiconductor majority charge carriers are e and P type semiconductor majority charge carriers are holes.

I = neAVd = neA ( μ E)

As μ e > μ h Ie > Ih

5. For a p-type semiconductor, which of the following statements is true? ⇒  (NEET 2019)

A. Holes are the majority carriers and trivalent atoms are the dopants.

B. Holes are the majority carriers and pentavalent atoms are the dopants.

C. Electrons are the majority carriers and pentavalent atoms are the dopants

D. Electrons are the majority carriers and trivalent atoms are the dopants.

The Correct Answer is (A)

In p-type semiconductor, an intrinsic semiconductor is doped with trivalent impurities, that creates deficiencies of valence electrons called holes which are majority charge carriers.

6. In a n-type semiconductor, which of the following statement is true. ⇒  ( NEET 2013)

A. Holes are minority carries and pentavalent atoms are dopants.

B. Holes are majority carries and trivalent atoms are dopants.

C. Electrons are majority carries and trivalent atoms are dopants.

D. Electrons are minority carriers and pentavalent atoms are dopants.

The Correct Answer is (A)

In n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers and pentavalent atoms are dopants.

7. C and Si both have same lattice structure; having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because ⇒  (AIPMT 2012 Prelims)

A. In case of C the valence band is not completely filled at absolute zero temperature.

B. In case of C the conduction band is partly filled even at absolute zero temperature.

C. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.

D. The four bonding electrons in the case of C lie in the third orbit , whereas for Si they lie in the fourth orbit.

The Correct Answer is (C)

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.

8. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 1016 m 3. Doping by indium increases nh to 4.5 × 1022 m 3. The doped semiconductor is of ⇒  (AIPMT 2011 Mains)

A. p-type having electron concentration ne = 5 × 109 m 3

B. n-type with electron concentration ne = 5 × 1022 m 3

C. p-type with electron concentration ne = 2.5 × 1010 m 3

D. n-type with electron concentration ne = 2.5 × 1023 m 3

The Correct Answer is (A)

(ni)2 = ne × nh

(1.5 × 1016)2 = ne (4.5 × 1022)

So ne = 5 × 109

Now nh = 4.5 × 1022

nh >> ne

Hence, semiconductor is p-type

and ne = 5 × 109 m–3

9. If a small amount of antimony is added to germanium crystal ⇒  (AIPMT 2011 Prelims)

A. it becomes a p-type semiconductor

B. the antimony becomes an acceptor atom

C. there will be more free electrons than holes in the semiconductor

D. its resistance is increased

The Correct Answer is (C)

If a small amount of antimony is added to germanium crystal, crystal becomes n-type semiconductor. Hence, there will be more free electrons than holes.

10. Which one of the following statement is false ?

A. Pure Si doped with trivalent impurities gives a p-type semiconductor.

B. Majority carries in a n-type semiconductor are holes.

C. Minority carries in a p-type semiconductor are electrons.

D. The resistance of intrinisic semiconductor decreases with increase of temperature. ⇒  (AIPMT 2010 Prelims)

The Correct Answer is (B)

In a n-type semiconductors, electrons are majority carriers and holes are minority carriers.

11. Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point ? ⇒  (AIPMT 2010 Prelims)

A. metallic bonding

B. van der Waal's bonding

C. ionic bonding

D. covalent bonding

The Correct Answer is (A)

In case of metal, conductivity decreases with increase in temperature and metal has high melting point.

12. Sodium has body centred packing. Distance between two nearest atoms is 3.7 A ⇒  (AIPMT 2009 )

A. 4.3 A

B. 3.0 A

C. 8.6 A

D. 6.8 A

The Correct Answer is ()

Distance between nearest atoms in body centred cubic lattice (bcc),

d = 3 2 a

Given d = 3.7 A

a = 3.7 × 2 3 = 4.3 A

13. If the lattice parameter for a crystalline structure is 3.6 A , then the atomic radius in fcc crystal is ⇒  (AIPMT 2008)

A. 2.92 A

B. 1.27 A

C. 1.81 A

D. 2.10 A

The Correct Answer is (B)

The atomic radius in a f.c.c. crystal is = a 2 2

Atomic radius = 3.6 2 2 = 1.27 A

14. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is ⇒  (AIPMT 2007)

AIPMT 2007 Physics - Semiconductor Electronics Question 54 English

A. an insulator

B. a metal

C. an n-type semiconductor

D. a p-type semiconductor.

The Correct Answer is (D)

For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top Ev of the valence band. With very small supply of energy an electron from the valence band can jump to the level EC and ionise acceptor negatively.

15. For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct ? ⇒  (AIPMT 2007)

A. a b c  and   α = β = γ = 90o

B. a = b = c  and   α β γ = 90o

C. a = b = c  and   α = β = γ = 90o

D. a b c  and  ; α β   γ = 90o

The Correct Answer is (C)

In a cubic crystal structure
a = b = c  and   α = β = γ = 90o

16 Choose the only false statement from the following. ⇒  (AIPMT 2005)

A. In conductors the valence and conduction bands overlap.

B. Substances with energy gap of the order of 10 eV are insulators.

C. The resistivity of a semiconductor increases with increase in temperature.

D. The conductivity of a semiconductor increases with increase in temperature.

The Correct Answer is Option (C)

Option (a) is correct as in conductor conduction and valance band overlap and the conduction band is partially filled

Option (b) is correct as insulators have energy gap of 5 - 10 eV.

Option (c) is incorrect as resistivity decreases with increase in temperature.

Option (d) is correct as with increase in temperature, more and more electrons jump to conduction band, hence conductivity increases.

17. Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by (Eg)C, (Eg)si and (Eg)Ge respectively. Which one of the following relationships is true in their case? ⇒  (AIPMT 2005)

A. (Eg)C > (Eg)Si

B. (Eg)C < (Eg)Si

C. (Eg)C = (Eg)Si

D. (Eg)C < (Eg)Ge.

The Correct Answer is Option (A)

Band gap of carbon is 5.5 eV while that of silicon is 1.1 eV

(Eg)C > (Eg)Si

18. Copper has face centered cubic (fcc) lattice with interatomic spacing equal to 2.54 A . The value of lattice constant for this lattice is ⇒  (AIPMT 2005)

A. 2.54 A

B. 3.59 A

C. 1.27 A

D. 5.08 A

The Correct Answer is Option (B)

Lattice constant for (f.c.c.)

= a = interatomic spacing × 2 = 3.59 A

19. In semiconductors at a room temperature ⇒  (AIPMT 2004)

A. the valence band is partially empty and the conduction band is partially filled

B. the valence band is completely filled and the conduction band is partially filled

C. the valence band is completely filled

D. the conduction band is completely empty

The Correct Answer is Option (A)

In semiconductors at room temperature the electrons get enough energy so that they are able to over come the forbidden gap. Thus at room temperature the valence band is partially empty and conduction band is partially filled. Conduction band in semiconductor is completely empty only at 0 K.

20. Number of atom per unit cell in B.C.C. ⇒  (AIPMT 2002)

A. 9

B. 4

C. 2

D. 1

The Correct Answer is Option (C)

In body-centred cubic (b.c.c.) lattice there are eight atoms at the corners of the cube and one at the centre.

Therefore number of atom per unit cell

= 1 8 × 8 + 1 = 2

21. The cations and anions are arranged in alternate form in ⇒  (AIPMT 2000)

A. metallic crystal

B. ionic crystal

C. covalent crystal

D. semi-conductor crystal.

The Correct Answer is Option (B)

In an Ionic crystal, the cations and anions are arranged in alternate form.