Semiconductor Devices Test

Result

Semiconductor Devices Test - 82

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Weekly Quiz Competition

Question 1
4 / -1

In good conductors of electricity, the type of bondingthat exists is

A
Ionic

B
Vander Waals

C
Covalent

D
Metallic
Question 2
4 / -1

Bonding in a germanium crystal (semi-conductor) is

A
Metallic

B
Ionic

C
Vander Waal's type

D
Covalent
Question 3
4 / -1

For a cubic crystal structure which one of the following relations indicating the cell characteristics in correct?

A
a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°

B
a ≠ b ≠ c and α = β = γ = 90°

C
a = b = c and α ≠ β ≠ γ ≠ 90°

D
a = b = c and α = β = γ = 90°

Solution
Question 4
4 / -1

A material has N atom in its crystal structure which is a hexagonal close packed. Then, the number of electronicstates in a band is

A
N

B
2N

C
4N

D
6N
Question 5
4 / -1

A P–type semiconductor can be obtained by adding

A
Arsenic to pure silicon

B
Gallium to pure silicon

C
Antimony to pure germanium

D
Phosphorous to pure germanium
Question 6
4 / -1

The valence of an impurity added to germanium crystalin order to convert it into a P–type semi-conductor is

A
6

B
5

C
4

D
3
Question 7
4 / -1

In P–type semiconductor, there is

A
An excess of one electron

B
Absence of one electron

C
A missing atom

D
A donor level
Question 8
4 / -1

The valency of the impurity atom that is to be added to germanium crystal so as to make it a N–typesemiconductor, is

A
6

B
5

C
4

D
3
Question 9
4 / -1

Silicon is a semiconductor. If a small amount of as is added to it, then its electrical conductivity

A
Decreases

B
Increases

C
Remains unchanged

D
Becomes zero
Question 10
4 / -1

To obtain P–type Si semiconductor, we need to dopepure Si with

A
Aluminum

B
Phosphorous

C
Oxygen

D
Germanium

User

Question Analysis

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Re-Attempt Weekly Quiz Competition

Semiconductor Devices Test - 82

Result

Semiconductor Devices Test - 82

Score

-

Rank

-

SHARING IS CARING

Question 1

Ionic

Vander Waals

Covalent

Metallic

Question 2

Metallic

Ionic

Vander Waal's type

Covalent

Question 3

a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°

a ≠ b ≠ c and α = β = γ = 90°

a = b = c and α ≠ β ≠ γ ≠ 90°

a = b = c and α = β = γ = 90°

Solution

Question 4

N

2N

4N

6N

Question 5

Arsenic to pure silicon

Gallium to pure silicon

Antimony to pure germanium

Phosphorous to pure germanium

Question 6

6

5

4

3

Question 7

An excess of one electron

Absence of one electron

A missing atom

A donor level

Question 8

6

5

4

3

Question 9

Decreases

Increases

Remains unchanged

Becomes zero

Question 10

Aluminum

Phosphorous

Oxygen

Germanium

User

Question Analysis

My Perfomance

Score

-

Rank

-

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