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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.
Chapter 14 Semiconductor Electronics
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.
Download pdf of NCERT Solutions for Class Physics Chapter 14 Semiconductor Electronics
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Exercise 1
Popular Questions of Class 12 Physics
- Q:-
What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
- Q:-
An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
- Q:-
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?
- Q:-
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?
- Q:-
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?
- Q:- A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
- Q:-
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.
- Q:-
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.)
- Q:-
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?
- Q:-
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
- Q:-
Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
- Q:-
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
- Q:-
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
- Q:-
The radius of the innermost electron orbit of a hydrogen atom is 5.3 ×10 −11 m. What are the radii of the n = 2 and n =3 orbits?
- Q:-
Answer the following questions:
(a) The earths magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earths core is known to contain iron. Yet geologists do not regard this as a source of the earths magnetism. Why?
(c) The charged currents in the outer conducting regions of the earths core are thought to be responsible for earths magnetism. What might be the battery (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earths field in such distant past?
(e) The earths field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f ) Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.] - Q:-
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure). Show
that the capacitance of a spherical capacitor is given by
where r1 and r2 are the radii of outer and inner spheres, respectively.
- Q:-
Two charges -q and +q are located at points (0, 0, - a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
- Q:-
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
- Q:-
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
- Q:-
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
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