According to the question,
 
D₂ is in reverse bias, so it acts as open circuit
\(\begin{aligned} R_{\mathrm{eq}} & =2+1=3 \Omega \end{aligned}\)
\(\begin{aligned} I & =\frac{V}{R_{\mathrm{eq}}}=\frac{6}{3}=2 \mathrm{~A} \end{aligned}\)

The given figure in question common emitter (CE) configuration of an n-p-n transistor. The base-emitter junction is forward biased and collector-base junction is reverse biased.
As, the base resistance R_B decreases, the input circuit will become more forward biased thus decreasing the base current (I_B ) (i.e. less number electron-hole recombination in base due to reduction in area of base) and increasing the emitter current (I_E ). This will increase the collector current (I_C) as I_E = I_B + I_C .
When I_C increase which flows through the lamp, the voltage across the bulb will also increase, thus making the lamp brighter and as the voltmeter is connected in parallel with the lamp, the reading in the voltmeter will also increase.

The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

In a transistor, the base region brings an interaction between emitter and collector. The base is kept thin and slightly doped so that, when emitter of transistor is forward biased, the less number of electron-hole combination takes place in base region and majority of charge carriers going from emitter towards base, pass over to collector. As a result of it, the collector current becomes quite good in comparison to base current.

A common emitter transistor amplifier is preferred when we require better current gain and power gain. But common base transistor amplifier is preferred when we require better voltage gain without any phase change of signal voltage

When a forward bias is applied to a p-n junction, the size of the depletion layer decreases. The resistance of the junction becomes low. The movement of the majority carriers takes place across the junction, resulting current, known as forward current which increases rapidly with increase in forward voltage.

(i) SSI stands for Small Scale Integration circuits.
(ii) VLSI stands for Very Large Scale Integration circuits.

(i) MSI stands for Medium Scale Integration circuits.
(ii) LSI stands for Large Scale Integration circuits.

No. Any slab, howsoever flat, will have roughness much larger than interatomic crystal spacing \((2\ to\ 3\overset { \circ }{ A } )\), so continuous contact at the atomic level is not possible.

Note that thermally generated electrons (ni ~10¹⁶m^–3) are negligibly small as compared to those produced by doping.
Therefore, ne \(\approx\) ND
Since n_e n_h = \(n_{i}^{2}\) , The number of holes
n_h = (2.25 x 10³² ) / (5 x10²²)
= ~ 4.5 x 10⁹ m^–3

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.

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

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.

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\)  Output frequency = 2 x 50 = 100 Hz.

Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
(a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
\(r_{f b}=\Delta V / \Delta I=0.1 \mathrm{~V} / 10 \mathrm{~mA}=10 \ \Omega\)
(b) From the curve at V = –10 V, I = –1 \(\mu\) A,
Therefore,
\(r_{r b}=10 \mathrm{~V} / 1 \mu \mathrm{A}=1.0 \times 10^{7} \ \Omega\) 

(i) Dynamic output resistance is given as,
\({ R }_{ e }={ \left( \frac { \Delta { V }_{ CE } }{ \Delta { I }_{ C } } \right) }_{ { I }_{ B }=constant }=\frac { 12-8 }{ (3.6-3.4)\times { 10 }^{ -3 } } \)
\(\frac { 4 }{ 0.2\times { 10 }^{ -3 } } =20\kappa \Omega \)
(ii) DC current gain,
\(\beta _{ DC }=\frac { { I }_{ C } }{ { I }_{ B } } =\frac { 3.5mA }{ 30\mu A } =\frac { 3.5\times { 10 }^{ -3 } }{ 30\times { 10 }^{ -6 } } =\frac { 350 }{ 3 } =116.67\)
(iii) AC current gain,
\(\beta _{ AC }=\frac { \triangle I_{ C } }{ \triangle { I }_{ B } } =\frac { \left( 4.7-3.5 \right) mA }{ \left( 40-30 \right) \mu A } =\frac { 1.2\times { 10 }^{ -3 } }{ 10\times { 10 }^{ -6 } } =120\)

As, Concentration of Si atom \(=5\times { 10 }^{ 28 }\) /m³
The doping of indium is 1 atom 10⁶ atoms of Si. But indium has three valence electrons and each doped indium atom creates one hole in Si crystal. Hence, it acts as an acceptor atom.
Concentration of acceptor atoms,
\({ n }_{ h }=5\times { 10 }^{ 28 }\times { 10 }^{ -6 }=5\times { 10 }^{ 22 }/{ m }^{ 3 }\)
Intrinsic carrier concentration,
\({ n }_{ i }=1.5\times { 10 }^{ 16 }/{ m }^{ 3 }\)
Hole concentration is increased
\(=\frac { { n }_{ h } }{ { n }_{ i } } =\frac { 5\times { 10 }^{ 22 } }{ 1.5\times { 10 }^{ 16 } } =3.33\times { 10 }^{ 6 }\)
New electron concentration,
\({ n }_{ e }=\frac { { n }_{ i }^{ 2 } }{ { n }_{ b } } =\frac { { \left( 1.5\times { 10 }^{ 16 } \right) }^{ 2 } }{ 5\times { 10 }^{ 22 } } =0.45\times { 10 }^{ 10 }/{ m }^{ 3 }\)
Electron concentration has been reduced
\(=\frac { { n }_{ i } }{ { n }_{ e } } =\frac { 1.5\times { 10 }^{ 16 } }{ 0.45\times { 10 }^{ 10 } } =3.33\times { 10 }^{ 6 }\)/m³
This means that the hole concentration has been increased over its intrinsic concentration by the same amount with which the electron concentration has been decreased.
The conductivity of doped silicon is given by
\(\sigma =e\left( { n }_{ e }H_{ e }+{ n }_{ h }H_{ h } \right) \)
\(\\ =\quad 1.6\times { 10 }^{ -19 }\left( 0.45\times { 10 }^{ 10 }\times 0.135+5\times { 10 }^{ 22 }\times 0.048 \right)\)
= 384 S/m
Resistivity,  \(\rho =\frac { 1 }{ \sigma } =\frac { 1 }{ 384 } =0.0026\Omega -m\)
Conductivity of pure Si crystal,
\(\sigma ={ en }_{ i }\left( { H }_{ e }+{ H }_{ h } \right) \)
\(\\ =1.6\times { 10 }^{ -19 }\times 1.5\times { 10 }^{ 16 }\left( 0.135+0.048 \right) \)
\(\\ =0.4392\times { 10 }^{ -3 }S/m\)
Resistivity, \(\rho =\frac { 1 }{ \sigma } =\frac { 1 }{ 0.4392\times { 10 }^{ -3 } } =2276.\Omega -m\)
Thus, we see that the conductivity of doped Si becomes much greater than its intrinsic conductivity and the resistivity has become much smaller than the intrinsic resistivity.

LASER stands for light amplification  by stimulated emission of Radiation.LED stands for light emitting diode.
(i) The frequency of light emitted by an LED is related to the band gap of the semiconductor used in LED i.e., a type of material used in marketing the LED.
(ii) The intensity of light emitted by LED depends upon the doping level of the semiconductor used.

Here,\(\beta =59,{ I }_{ e }=6.0\quad mA\)
\(\beta =\frac { { I }_{ c } }{ { I }_{ b } } =\frac { { I }_{ e }-{ I }_{ b } }{ { I }_{ b } } =\frac { { I }_{ e } }{ { I }_{ b } } =-1\)
\( \therefore \ 59=\frac { 6 }{ { I }_{ b } } -1 \ or \ \frac { 6 }{ { I }_{ b } } =60 \ or \ { I }_{ b }=\frac { 6 }{ 60 } =0.1 \ mA\)
\( { I }_{ c }={ I }_{ e }-{ I }_{ b }=6.0-0.1=5.9\ mA\)

(i) Since on doping a semiconductor, the hole concentration ( \({ n }_{ p }=4\times10^{ 10 }m^{ -3 })\) becomes greater than electron concentration \(({ n }_{ e }=2\times10^{ 8 }m^{ -3 })\)so p-type semiconductor is formed on doping.
(ii) Here
,\(n_{ 1 }=2\times10^{ 8 }m^{ -3 };\)
\( { n }_{ p }=4times10^{ 10 }m^{ -3 };{ n }_{ e }=?\)
\( { n }_{ e }=\frac { { n }_{ i } }{ n_{ p } } =\frac { (2\times10^{ 8 })^{ 2 } }{ 4\times10^{ 10 } } =10^{ 6 }m^{ -3 }\)
(iii) Energy gap decreases with doping due to formation of acceptor energy levels just above the valence band in the energy gap region.

(c) Carbon is an insulator because all of its valence electrons are used up in covalent bonding, while silicon is a semiconductor because some of its valence electrons are not involved in bonding, creating a small number of free electrons in the conduction band. This slight difference in electronic configuration results in a significant difference in their electrical properties, despite having the same lattice structure.