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Q1 In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.Ans: The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.
Q2 Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Ans: The statement is (d) true.
It is because in a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.
Q3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E g ) C , (E g ) Si and (E g ) Ge . Which of the following statements is true?
(a) (E g ) Si < (E g ) Ge < (E g ) C
(b) (E g ) C < (E g ) Ge > (E g ) Si
(c) (E g ) C > (E g ) Si > (E g ) Ge
(d) (E g ) C = (E g ) Si = (E g ) Ge
Ans: The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given three elements.
Hence answer is (c) is correct.
Q4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Ans: In an unbiased p-n junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration.
Thus answer (c) is correct.
Q5 When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Ans: When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.
Hence, The correct statement is (c).
Q6 For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Ans: The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased.
Q7 For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Ans: The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
Hence, the correct statement is (c).
Q8 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Ans: Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴ Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴ Output frequency = 2 × 50 = 100 Hz
Q9 For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Ans: Collector resistance, R C = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, R B = 1 kΩ = 1000 Ω
Input signal voltage = V i
Base current = I B
We have the amplification relation as: V/Vi = β Rc/Rb
Voltage amplification, Vi = V Rb / β Rc
= 2x1000 / 100x2000 = 0.01 V
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation: RB = Vi / IB
= 0.01 / 1000 = 10 μA
Therefore, the base current of the amplifier is 10 μA.
Q10 Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
Ans: Given that,
Voltage gain of the first amplifier, V 1 = 10
Voltage gain of the second amplifier, V 2 = 20
Voltage of input signal, V i = 0.01 V
Voltage of output AC signal = V o
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages,
i.e., V = V 1 × V 2 = 10 × 20 = 200
It can be calculated by the relation: V = Vo/Vi
V 0 = V × V i = 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
Q11 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Ans: Given that,
Energy band gap of the given photodiode, E g = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10 −9 m
The energy of a signal is given by the relation: E = hc/λ
Where, h = Planck’s constant = 6.626 × 10 −34 Js
c = Speed of light = 3 × 10 8 m/s
E = 6.626 x 10-34 x 3 x 108 / 6000 x 10-9 = 3.313 x 10-20 J
But 1.6 × 10 −19 J = 1 eV
E = 3.313 × 10 −20 J
∴E = 3.313 × 10 −20 J = 3.313 x 10-20 / 1.6 x 10-19 = 0.207 eV
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Q12 The number of silicon atoms per m 3 is 5 × 10 28 . This is doped simultaneously with 5 × 10 22 atoms per m 3 of Arsenic and 5 × 10 20 per m 3 atoms of Indium. Calculate the number of electrons and holes. Given that n i = 1.5 × 10 16 m −3 . Is the material n-type or p-type?
Ans: Number of silicon atoms, N = 5 × 10 28 atoms/m 3
Number of arsenic atoms, n As = 5 × 10 22 atoms/m 3
Number of indium atoms, n In = 5 × 10 20 atoms/m 3
Number of thermally-generated electrons, n i = 1.5 × 10 16 electrons/m 3
Number of electrons, n e = 5 × 10 22 − 1.5 × 10 16 ≈ 4.99 × 10 22
Number of holes = n h
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as: n e n h = n i2
Therefore, nh = n i2 / ne = (1.5x1016)2 / 4.99 x 1022 ≈ 4.51 x 109
Therefore, the number of electrons is approximately 4.99 × 10 22 and the number of holes is about 4.51 × 10 9 . Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.
Semiconductor Electronics Question Answers: NCERT Class 12 Physics
Welcome to the Chapter 14 - Semiconductor Electronics, Class 12 Physics - NCERT Solutions page. Here, we provide detailed question answers for Chapter 14 - Semiconductor Electronics.The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics and excel in their exams. By going through these Semiconductor Electronics question answers, you can strengthen your foundation and improve your performance in Class 12 Physics. Whether you're revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
This fourteenth and second last chapter has eleven topics and makes you aware of the differentiation between extrinsic and intrinsic semiconductors, energy bands and their I-V characteristics, diodes, and types of diodes and their characteristics and uses of diodes, junction transistor and characteristics of a transistor and application. Here you also get knowledge about the concept of analog and digital signals, logic gates. Solving the question at the end of the chapter will tell you how well you have learned them with NCERT Solutions for Class 12 Physics.
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Exercise 1
Key Features of NCERT Class 12 Physics Chapter 'Semiconductor Electronics' question answers :
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Popular Questions of Class 12 Physics
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A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
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A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
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A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
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A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
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A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Recently Viewed Questions of Class 12 Physics
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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
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(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
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(a) Identify an equipotential surface of the system.
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An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
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Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E:
(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates?
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(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10 −10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.
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You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
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A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
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