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(c) 2

At absolute zero, Si acts as an insulator because it has no free electrons in the conduction band.

(c) 2

N = N_B + \(\frac{N_C}{8}\) = 1 + \(\frac{8}{8}\) = 2

Correct answer is (a) 4

N = \(\frac{N_c}{8}\) + \(\frac{N_F}{2}\) = \(\frac{8}{8}\) + \(\frac{6}{2}\) = 4.

(b) α = β = 90° , γ ≠ 90° 

For a monoclinic crystal, α ≠ β ≠ c and α = β = 90°≠ γ.

Correct answer is (c) 0° – 180°

1. Different colours are obtained by changing the concentration of arsenic and phosphors in Gallium Arsenide Phosphide.

2. LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc.

3. LEDs have the following advantages over conventional incandescent low power lamps:

• Low operational voltage and less power.
• Fast action and no warm-up time required. 
• The bandwidth of emitted light is 100A° to 500A° or in other words it is nearly (but not exactly) monochromatic.
• Long life and ruggedness Fast on-o switching capability.

(b) a vacancy created when an electron leaves a covalent bond.

(a) phosphorous

As phosphorous is pentavalent, it produces ntype semiconductor when added to silicon.

(c) Holes are minority carriers and pentavalent atoms are the dopants.

Semiconductor – P-type, n-type.

The answer is (C) Holes are minority carriers and pentavalent atoms are the dopants.

An n-type semiconductor is obtained by doping a semiconductor with a pentavalent impurity. The impurity so added produces free electrons. Therefore, in an n-type semiconductor, the electrons are majority carriers and holes are minority carriers and pentavalent atoms are the dopants.

(c) (E_g)_c > (E_g)_Si > (E_g)_Ge

(b) The base region must be very thin are lightly doped,

(c) The emitter junction is forward biase and collector junction is reverse biased.

The answer is (C) (Eg)_c > (Eg)_Si > (Eg)_Ge

Carbon is insulator and hence the value of energy band gap is maximum for it. Of the other two semiconductors, the value of energy band gap for germanium is lesser.

50 Hz, 100 Hz

The answer is (C) Hole concentration in p-region is more as compared to n-region.

In an unbiased p-n junction, hole concentration in p-region is more as compared to n-region.

Given Input frequency = 50 Hz

Output frequency

For Half wave rectifier = 50 Hz 

For full wave rectifier = 50 × 2 = 100 Hz.

The answer is (C) Lowers the potential barrier.

When a p-n junction is forward biased, the applied voltage opposes the barrier voltage across the junction. As a result, the potential barrier gets lowered.

reverse bias 

(c) is low at high and low frequencies a constant in the middle frequency large.

(c) lowers the potential barrier.

1. Collector 

2. Emitter 

3. EB junction 

4. CB junction

(a) Assertion is correct, but reason is incorrect

1. Conductor – Graphite, Ni

2. Insulator – Calcite

3. Semiconductor – Ga,As.

(b) if either or both inputs are 1

From the V-I characteristics a junction diode, it is clear that it allows the current to pass only when it is forward biased. So when an alternatively voltage is applied across the diode, current flows only during that part of the cycle when it is forward biased.

(a) electric field is zero

When a p – n junction is reverse biased, the width of the depletion layer becomes large and so the electric field (E = v/d) becomes very small, nearly zero.

(b) decreases

If the forward voltage in a diode is increased, the width of depletion region decreases.

The space charge region at the p-n junction which consists only of immobile ions and is depleted of mobile charge carriers is called depletion region. 

The diffusion of majority charge carriers i.e., free electrons and holes across the p–n junction causes the depletion region. 

Yes. It is different since in a periodic crystal lattice each bound electron is influenced by many neighbouring atoms.

(d) i_e > i_c

As i_e = i_c+ i_b

∴ i_i > i_c

(a) ions

The potential barrier in the depletion layer is due to the presence of immobile ions.

The answer is (A) 0.

(b) electrons move from base to collector

(c) electrons move from base to collector

When n-p-n transistor is used an amplifier, the majority carrier electrons move from base to collector.

(a) 49

β = \(\frac{I_C}{I_B}\) = \(\frac{5.488}{0.112}\) = 49.

(a) emitter

Emitter of a transistor is heavily doped so as to act as source of majority charge carriers.

The energy band formed due to the valence orbitals is called valence band and that formed due to the unoccupied orbitals is called the conduction band. The energy gap between the valence band and the conduction band is called forbidden energy gap.

(a) 0

 In a common – base amplifier, the input and output voltages are in the same phase.

(a) a little below the conduction band

For a heavily doped n-type semiconductor, the fermi level lies slightly below the bottom of the conduction band.

Consider a solid in crystalline form. Inside the crystal each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges. Because of this, each electron will have a different energy level. These different energy levels with continuous energy variation form what are called energy bands. The energy band which includes the energy levels of the valence electrons is called the valence band. The energy band above the valence band is called the conduction band. With no external energy, all the valence electrons will reside in the valence band. 

If there is some energy gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons are available in the conduction band. This makes the material an insulator. But some of the electrons from the valence band may gain external energy to cross the gap between the conduction band and the valence band. Then these electrons will move into the conduction band. 

At the same time they will create vacant energy levels in the valence band where other valence electrons can move. Thus the process creates the possibility of conduction due to electrons in conduction band as well as due to vacancies in the valence band. 

The gap between the top of the valence band and bottom of the conduction band is called the energy band gap (Energy gap E_g). It may be large, small, or zero, depending upon the material. 

(a) a metal

In metals, the valence band and conduction band may overlap at low temperature.

In a common emitter amplifier, the input and output voltages are 180° out of phase or in, opposite phases. The reason for this can be seen from the fact that as the input voltage rises, so the current increases through the base circuit.

(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction

(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice versa.