#### Page No 509:

#### Question 14.1:

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.

#### Answer:

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.

#### Question 14.2:

Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

#### Answer:

The correct statement
is **(d). **

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.

#### Question 14.3:

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}

#### Answer:

The correct statement
is **(c)**.

Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.

The energy band gap of
these elements are related as: (*E*_{g})_{C} > (*E*_{g})_{Si} > (*E*_{g})_{Ge}

#### Question 14.4:

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.

#### Answer:

The correct statement
is **(c)**.

The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

#### Page No 510:

#### Question 14.5:

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.

#### Answer:

The correct statement
is **(c)**.

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.

#### Question 14.6:

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.

#### Answer:

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.

#### Question 14.7:

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.

#### Answer:

The correct statement
is **(c)**.

The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.

#### Question 14.8:

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.

#### Answer:

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

#### Question 14.9:

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.

#### Answer:

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:

Voltage amplification

Therefore, the input signal voltage of the amplifier is 0.01 V.

Base resistance is given by the relation:

Therefore, the base current of the amplifier is 10 μA.

#### Question 14.10:

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.

#### Answer:

Voltage gain of the
first amplifier, *V*_{1} = 10

Voltage gain of the
second amplifier, *V*_{2} = 20

Input signal voltage, *V*_{i} = 0.01 V

Output AC signal
voltage = *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

We have the relation:

*V*_{0} = *V* × *V*_{i}

= 200 × 0.01 = 2 V

Therefore, the output AC signal of the given amplifier is 2 V.

#### Question 14.11:

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

#### Answer:

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* =

Where,

*h* = Planck’s constant

= 6.626 × 10^{−34} Js

c = Speed of light

= 3 × 10^{8} m/s

*E*

= 3.313 × 10^{−20} J

But 1.6 × 10^{−19} J = 1 eV

∴*E* = 3.313 × 10^{−20} J

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.

#### Question 14.12:

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?

#### Answer:

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*_{i}^{2}

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.

#### Question 14.13:

In an intrinsic
semiconductor the energy gap *E*_{g}is
1.2 eV. Its hole mobility is much smaller than electron mobility and
independent of temperature. What is the ratio between conductivity at
600K and that at 300K? Assume that the temperature dependence of
intrinsic carrier concentration *n*_{i}is
given by

where *n*_{0 }is
a constant.

#### Answer:

Energy gap of the given
intrinsic semiconductor, *E*_{g} = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

Where,

*k*_{B} =
Boltzmann constant = 8.62 × 10^{−5} eV/K

T = Temperature

*n*_{0} =
Constant

Initial temperature, *T*_{1} = 300 K

The intrinsic carrier-concentration at this temperature can be written as:

… (1)

Final temperature, *T*_{2} = 600 K

The intrinsic carrier-concentration at this temperature can be written as:

… (2)

The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

Therefore, the ratio
between the conductivities is 1.09 × 10^{5}.

#### Page No 511:

#### Question 14.14:

In a p-n junction diode, the current I can be expressed as

where *I*_{0} is called the reverse saturation current, *V *is the voltage
across the diode and is positive for forward bias and negative for
reverse bias, and *I *is the current through the diode, *k*_{B}is the Boltzmann constant (8.6×10^{−5} eV/K)
and T is the absolute temperature. If for a given diode *I*_{0} = 5 × 10^{−12 }A and T = 300 K, then

**(a) ** What will
be the forward current at a forward voltage of 0.6 V?

**(b) ** What will be the increase in the current if the voltage
across the diode is increased to 0.7 V?

**(c) ** What is the
dynamic resistance?

**(d) ** What will
be the current if reverse bias voltage changes from 1 V to 2 V?

#### Answer:

In a p-n junction diode, the expression for current is given as:

Where,

*I*_{0} =
Reverse saturation current = 5 × 10^{−12} A

*T* = Absolute
temperature = 300 K

*k*_{B} =
Boltzmann constant = 8.6 × 10^{−5} eV/K = 1.376 ×
10^{−23} J K^{−1}

V = Voltage across the diode

**(a)** Forward voltage, *V* = 0.6 V

∴Current, *I*

Therefore, the forward current is about 0.0256 A.

**(b)** For forward voltage, *V*^{’} = 0.7 V,
we can write:

Hence,
the increase in current, Δ*I *= *I*^{‘} − *I*

= 1.257 − 0.0256 = 1.23 A

**(c)** Dynamic resistance

**(d)** If the reverse bias voltage changes from 1 V to 2 V, then
the current (*I*) will almost remain equal to *I*_{0} in both cases. Therefore, the dynamic resistance in the reverse bias
will be infinite.

#### Question 14.15:

You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

#### Answer:

**(a) ***A* and *B* are the inputs and *Y* is the
output of the given circuit. The left half of the given figure acts
as the NOR Gate, while the right half acts as the NOT Gate. This is
shown in the following figure.

Hence, the output of the NOR Gate =

This
will be the input for the NOT Gate.
Its output will be = *A* + *B *

∴*Y* = *A* + *B*

Hence, this circuit functions as an OR Gate.

**(b) ***A* and *B* are the inputs and *Y* is the
output of the given circuit. It can be observed from the following
figure that the inputs of the right half NOR Gate are the outputs of
the two NOT Gates.

Hence, the output of the given circuit can be written as:

Hence, this circuit functions as an AND Gate.

#### Question 14.16:

Write the truth table for a NAND gate connected as given in Fig. 14.45.

Hence identify the exact logic operation carried out by this circuit.

#### Answer:

*A* acts as the
two inputs of the NAND gate and *Y* is the output, as shown in
the following figure.

Hence, the output can be written as:

The truth table for
equation (*i*) can be drawn as:

**A****Y**0

1

1

0

This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

#### Question 14.17:

You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

#### Answer:

In both the given circuits, *A* and *B* are the inputs and *Y* is the output.

**(a)** The
output of the left NAND gate will be,
as shown in the following figure.

Hence, the output of the combination of the two NAND gates is given as:

Hence, this circuit functions as an AND gate.

**(b)** is
the output of the upper left of the NAND gate and is
the output of the lower half of the NAND gate, as shown in the
following figure.

Hence, the output of the combination of the NAND gates will be given as:

Hence, this circuit functions as an OR gate.

#### Question 14.18:

Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

#### Answer:

*A* and *B* are the inputs of the given circuit. The output of the first NOR gate
is.
It can be observed from the following figure that the inputs of the
second NOR gate become the out put of the first one.

Hence, the output of the combination is given as:

The truth table for this operation is given as:

**A****B****Y****(=****A****+****B****)**0

0

0

0

1

1

1

0

1

1

1

1

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

#### Page No 512:

#### Question 14.19:

Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

#### Answer:

**(a) ***A* acts as the two inputs of the NOR gate and *Y* is the output, as shown in the following figure. Hence, the output of
the circuit is.

The truth table for the same is given as:

**A****Y**0

1

1

0

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

**(b) ***A* and *B* are the inputs and *Y* is the
output of the given circuit. By using the result obtained in solution **(a)**, we can infer that the outputs of the first two NOR gates
areas
shown in the following figure.

are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:

The truth table for the same can be written as:

**A****B****Y****(=****A****⋅****B****)**0

0

0

0

1

0

1

0

0

1

1

1

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

**NCERT Solutions for Class 12 Physics Chapters**

**Physics Part I**

- Chapter 1 – Electric Charges And Fields
- Chapter 2 – Electrostatic Potential And Capacitance
- Chapter 3 – Current Electricity
- Chapter 4 – Moving Charges And Magnetism
- Chapter 5 – Magnetism And Matter
- Chapter 6 – Electromagnetic Induction
- Chapter 7 – Alternating Current
- Chapter 8 – Electromagnetic Waves

**Physics Part II**

- Chapter 1 – Ray Optics And Optical Instruments
- Chapter 2 – Wave Optics
- Chapter 3 – Dual Nature Of Radiation And Matter
- Chapter 4 – Atoms
- Chapter 5 – Nuclei
- Chapter 6 – Semiconductor Electronics: Materials Devices and Simple Circuits
- Chapter 7 – Communication Systems

**NCERT Solutions for Class 12-science:**