MCQ Question for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

MCQs MCQs Class 12

Refer to MCQ Class 12 Semiconductor Electronics: Materials, Devices and Simple Circuits provided below which is an important chapter in Class 12 Physics. Students should go through the MCQs questions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits with answers given below so that they are able to understand the complete topic properly. It’s important to understand the entire chapter by reading Class 12 Physics Notes also. Also, refer to MCQ Questions for Class 12 Physics for all chapters.

MCQ on Semiconductor Electronics: Materials, Devices and Simple Circuits Class 12 Physics PDF with Answers

All multiple choice questions with solutions provided below have been developed based on the latest syllabus and examination pattern issued for class 12 by CBSE and NCERT. As Semiconductor Electronics: Materials, Devices and Simple Circuits is a very important and scoring chapter in Physics Class 12, therefore, the students should carefully learn the questions and answers given below which will help them to get better scores in upcoming examinations for class 12th.

Question. When using a triode, as an amplifier, the electrons are emitted by
(a) grid and collected by cathode only
(b) cathode and collected by the anode only
(c) anode and collected by cathode only
(d) anode and collected by the grid and by cathode.

Answer

B

Question. For amplification by a triode, the signal to be amplified is given to
(a) the cathode
(b) the grid
(c) the glass envelope
(d) the anode 

Answer

B

Question. For an electronic valve, the plate current I and plate voltage V in the space charge limited region are related as
(a) I is proportional to V 3/2
(b) I is proportional to V 2/3
(c) I is proportional to V
(d) I is proportional to V 2 

Answer

A

Question. When a triode is used as an amplifier the phase difference between the input signal voltage and the output is
(a) 0
(b) p
(c) p/2
(d) p/4.

Answer

B

Question. Choose the only false statement from the following.
(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. 

Answer

C

Question. 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?
(a) (Eg)C > (Eg)Si
(b) (Eg)C < (Eg)Si
(c) (Eg)C = (Eg)Si
(d) (Eg)C < (Eg)Ge 

Answer

A

Question. In semiconductors at a room temperature
(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.

Answer

A

Question. C and Si both have same lattice structure; having 4 bonding electrons in each. However, C is insulator whereas Si is intrinsic semiconductor. This is because
(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. 

Answer

C

Question. At absolute zero, Si acts as
(a) non metal
(b) metal
(c) insulator
(d) none of these.

Answer

C

Question. For a p-type semiconductor, which of the following statements is true?
(a) Electrons are the majority carriers and pentavalent atoms are the dopants.
(b) Electrons are the majority carriers and trivalent atoms are the dopants.
(c) Holes are the majority carriers and trivalent atoms are the dopants.
(d) Holes are the majority carriers and pentavalent atoms are the dopants. 

Answer

C

Question. In a n-type semiconductor, which of the following statement is true?
(a) Holes are minority carriers and pentavalent atoms are dopants.
(b) Holes are majority carriers and trivalent atoms are dopants.
(c) Electrons are majority carriers and trivalent atoms are dopants.
(d) Electrons are minority carriers and pentavalent atoms are dopants. 

Answer

A

Question. If a small amount of antimony is added to germanium crystal
(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.

Answer

C

Question. In a p type semiconductor, the majority carriers of current are
(a) protons
(b) electrons
(c) holes
(d) neutrons 

Answer

C

Question. Which of the following, when added as an impurity into the silicon produces n type semiconductor?
(a) B
(b) Al
(c) P
(d) Mg 

Answer

C

Question. To obtain a p-type germanium semiconductor, it must be doped with
(a) indium
(b) phosphorus
(c) arsenic
(d) antimony. 

Answer

A

Question. When arsenic is added as an impurity to silicon, the resulting material is
(a) n-type conductor
(b) n-type semiconductor
(c) p-type semiconductor
(d) none of these. 

Answer

B

Question. When n type semiconductor is heated
(a) number of electrons increases while that of holes decreases
(b) number of holes increases while that of electrons decreases
(c) number of electrons and holes remain same
(d) number of electrons and holes increases equally.

Answer

D

Question. The increase in the width of the depletion region in a p-n junction diode is due to
(a) forward bias only
(b) reverse bias only
(c) both forward bias and reverse bias
(d) increase in forward current 

Answer

B

Question. The barrier potential of a p-n junction depends on
(1) type of semiconductor material
(2) amount of doping
(3) temperature
Which one of the following is correct?
(a) (1) and (2) only
(b) (2) only
(c) (2) and (3) only
(d) (1), (2) and (3)

Answer

D

Question. In an unbiased p-n junction, holes diffuse from the p-region to n-region because of
(a) he attraction of free electrons of n-region
(b) the higher hole concentration in p-region than that in n-region
(c) the higher concentration of electrons in the n-region than that in the p-region
(d) the potential difference across the p-n junction.

Answer

B

Question. In a p-n junction
(a) high potential at n side and low potential at p side
(b) high potential at p side and low potential at n side
(c) p and n both are at same potential
(d) undetermined. 

Answer

A

Question. Depletion layer consists of
(a) mobile ions
(b) protons
(c) electrons
(d) immobile ions 

Answer

D

Question. The depletion layer in the p-n junction region is caused by
(a) drift of holes
(b) diffusion of charge carriers
(c) migration of impurity ions
(d) drift of electrons.

Answer

B

Question. In a p-n junction diode, change in temperature due to heating
(a) affects only reverse resistance
(b) affects only forward resistance
(c) does not affect resistance of p-n junction
(d) affects the overall V – I characteristics of p-n junction.

Answer

D

Question. In forward biasing of the p-n junction
(a) the positive terminal of the battery is connected to p-side and the depletion region becomes thick.
(b) the positive terminal of the battery is connected to n-side and the depletion region becomes thin.
(c) the positive terminal of the battery is connected to n-side and the depletion region becomes thick.
(d) the positive terminal of the battery is connected to p-side and the depletion region becomes thin.

Answer

D

Question. Application of a forward bias to a p-n junction
(a) widens the depletion zone
(b) increases the potential difference across the depletion zone
(c) increases the number of donors on the n side
(d) decreases the electric field in the depletion zone.

Answer

D

Question. Reverse bias applied to a junction diode
(a) lowers the potential barrier
(b) raises the potential barrier
(c) increases the majority carrier current
(d) increases the minority carrier current 

Answer

B

Question. Barrier potential of a p-n junction diode does not depend on
(a) diode design
(b) temperature
(c) forward bias
(d) doping density

Answer

A

Question. In forward bias, the width of potential barrier in a
p-n junction diode
(a) remains constant
(b) decreases
(c) increases
(d) first (a) then (b)

Answer

B

Question. In a junction diode, the holes are due to
(a) extra electrons
(b) neutrons
(c) protons
(d) missing of electrons

Answer

D

Question. The cause of the potential barrier in a p-n junction diode is
(a) depletion of negative charges near the junction
(b) concentration of positive charges near the junction
(c) depletion of positive charges near the junction
(d) concentration of positive and negative charges near the junction. 

Answer

D

Question. A semiconducting device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be
(a) a p-type semiconductor
(b) an intrinsic semiconductor
(c) a p-n junction
(d) an n-type semiconductor.

Answer

C

Question. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz 

Answer

D

Question. A p-n junction diode can be used as
(a) condenser
(b) regulator
(c) amplifier
(d) rectifier

Answer

D

Question. An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to
(a) 10.4 × 10–26 m
(b) 654 nm
(c) 654 Å
(d) 654 × 10–11 m

Answer

B

Question. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
(a) 4000 nm
(b) 6000 nm
(c) 4000 Å
(d) 6000 Å 

Answer

C

Question. A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
(a) 1 × 1014 Hz
(b) 20 × 1014 Hz
(c) 10 × 1014 Hz
(d) 5 × 1014 Hz 

Answer

D

Question. Zener diode is used for
(a) amplification
(b) rectification
(c) stabilisation
(d) producing oscillations in an oscillator. 

Answer

C

Question. In a p-n junction photo cell, the value of the photoelectromotive force produced by monochromatic light is proportional to
(a) the barrier voltage at the p-n junction
(b) the intensity of the light falling on the cell
(c) the frequency of the light falling on the cell
(d) the voltage applied at the p-n junction.

Answer

B

Question. The device that can act as a complete electronic circuit is
(a) junction diode
(b) integrated circuit
(c) junction transistor
(d) zener diode.

Answer

B

Question. For transistor action, which of the following statements is correct?
(a) Base, emitter and collector regions should have same doping concentrations.
(b) Base, emitter and collector regions should have same size.
(c) Both emitter junction as well as the collector junction are forward biased.
(d) The base region must be very thin and lightly doped.

Answer

D

Question. Metallic solids are always opaque because
(a) they reflect all the incident light.
(b) they scatter all the incident light.
(c) the incident light is readily absorbed by the free electrons in a metal.
(d) the energy band traps the incident.

Answer

C

Question. The manifestation of band structure in solids is due to
(a) Bohr’s correspondence principle
(b) Pauli’s exclusion principle
(c) Heisenberg’s uncertainty principle
(d) Boltzmann’s law

Answer

B

Question. In a p-type semiconductor the acceptor level is situated 60 meV above the valence band. The maximum wavelength of light required to produce a hole will be
(a) 0.207 × 10–5 m
(b) 2.07 × 10–5 m
(c) 20.7 × 10–5 m
(d) 2075 × 10–5 m

Answer

B

Question. The band gap in germanium and silicon in ev respectively is
(a) 1.1, 0
(b) 0, 1.1
(c) 1.1, 0.7
(d) 0.7, 1.1

Answer

D

Question. In a npn transistor 1010 electrons enter the emitter in 10–6 s. 4% of the electrons are lost in the base. The current transfer ratio will be
(a) 0.98
(b) 0.97
(c) 0.96
(d) 0.94

Answer

C

Question. A diode having potential difference 0.5 V across its junction which does not depend on current, is connected in series with resistance of 20Ω across source. If 0.1 A current passes through resistance then what is the voltage of the source?
(a) 1.5 V
(b) 2.0 V
(c) 2.5 V
(d) 5 V

Answer

C

Question. Assuming that the silicon diode having resistance of 20 Ω , the current through the diode is (knee voltage 0.7 V)

(a) 0 mA
(b) 10 mA
(c) 6.5 mA
(d) 13.5 mA

Answer

C

Question. Which one is the weakest type of bonding in solids ?
(a) Ionic
(b) Covalent
(c) Metallic
(d) Vander Wall’s

Answer

D

Question. The transistor are usually made of
(a) metal oxides with high temperature coefficient of resistivity
(b) metals with high temperature coefficient of resistivity
(c) metals with low temperature coefficient of resistivity
(d) semiconducting materials having low temperature coefficient of resistivity

Answer

A

Assertion-Reason Questions

In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.

Question. Assertion (A): The electrical conductivity of n-type semiconductor is higher than that of p-type semiconductor at a given temperature and voltage applied.
Reason (R): The mobility of electron is higher than that of hole.

Answer

A

Question. Assertion (A): A p-n junction with reverse bias can be used as a photo-diode to measure light intensity.
Reason (R): In a reverse bias condition, the current is small but it is more sensitive to change in incident light intensity.

Answer

A

Question. Assertion (A): The forbidden energy gap between the valence and conduction bands is greater in silicon than in germanium.
Reason (R): Thermal energy produces fewer minority carriers in silicon than in germanium.

Answer

B

Question. Assertion (A): When the temperature of a semiconductor is increased, then its resistance decreases.
Reason (R): The energy gap between valence and conduction bands is very small for semiconductors.

Answer

A

Question. Assertion (A): A p-type semiconductor has negative charge on it.
Reason (R): p-type impurity atom has positive charge carrier (electrons) in it.

Answer

D

Question. Assertion (A): A p-n junction diode can be used even at ultra high frequencies.
Reason (R): Capacitive reactance of p-n junction diode increases as frequency increases.

Answer

C

Question. Assertion (A): The energy gap between the valence band and conduction band is greater in silicon than in germanium.
Reason (R): Thermal energy produces fewer minority carriers in silicon than in germanium.

Answer

A

Question. Assertion (A): The temperature coefficient of resistance is positive for metals and negative for p-type semiconductors.
Reason (R): The effective charge carriers in metals are negatively charged electrons, whereas in p-type semiconductors, they are positively charged.

Answer

A

Question. Assertion (A): The colour of light emitted by a LED depends on as reverse biasing.
Reason (R): The reverse biasing of p-n junction will lower the width of depletion layer.

Answer

D

Question. Assertion (A): Diamond behaves such as an insulator.
Reason (R): There is a large energy gap between valence band and conduction bond of diamond.

Answer

A

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