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.

Because in n-type semiconductor, electrons (having negative charge) are majority carriers which are responsible for the conduction

In an energy band, the highest energy level occupied by electron at 0K is called fermi level and its energy is called fermi -energy.

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.

(i) 

(ii) When the input voltage is negative, then it would be forward bias for diode. So, waveform would pass through the diode and we get no voltage on output. If input voltage is positive, then diode is in reverse bias and behaves as a open circuit. Now, input wave form is passed to the output terminals.

The property of a p-n junction that makes it suitable for rectifying alternating voltages is its ability to act as a diode, allowing current flow in only one direction. This property is a result of the fornnations of a depletion region at the interface between the p-type and n-type semi-conductor which acts as a barrier for the flow of current in one direction making the p-n junction a useful device for rectifying AC signals.
For the difference between half wave and a full wave rectifier, 

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,
\(r_{r b}=10 \mathrm{~V} / 1 \mu \mathrm{A}=1.0 \times 10^{7} \ \Omega\) 

\({ \left( 10111.10 \right) }_{ 2 }\)

\({ V }_{ c }={ V }_{ s }+{ V }_{ F }={ V }_{ s }+\frac { { V }_{ 0 } }{ 100 } \)
or
\({ V }_{ c }=0.2+\frac { -10 }{ 100 } \)
\(=0.2-0.1=0.1\)
\({ A }_{ 0 }=\frac { { V }_{ 0 } }{ { V }_{ c } } =\frac { -10 }{ 0.1 } =\frac { -10 }{ 0.1 } =-100\)
\({ A }_{ 0 }=\frac { -10 }{ 0.2 } =-50\)

\(\beta =\frac { { I }_{ c } }{ { I }_{ b } } =\frac { 13.5 }{ 150\times 10^{ -3 } } =90\)
And \({ I }_{ e }={ I }_{ b }+{ I }_{ c }=13.5+150\times { 10 }^{ -3 }\)
\(=13.5+0.15\)
or   \({ I }_{ e }=13.65 \ mA\)

Given \(a\)= 0.95
So 
\(\beta =\frac { a }{ 1-a } =\frac { 0.95 }{ 1-0.95 }\)
\(=\frac { 95 }{ 5 } =19\).

(i) New semiconductor must be n-type, because the electron concentration increases.
(ii) Given, 
\({ n }_{ i }=6\times 10^{ 8 }/{ m }^{ 3 }\)
\(\\ { n }_{ e }=9\times { 10 }^{ 12 }/{ m }^{ 3 }\)
\(\\ { n }_{ e }{ n }_{ h }={ n }_{ i }^{ 2 }\)
\(\\ \Rightarrow { n }_{ h }=\frac { { n }_{ i }^{ 2 } }{ { n }_{ e } } =\frac { (6\times 10^{ 8 })^{ 2 } }{ 9\times { 10 }^{ 12 } } \)
\(\\ =\frac { 36\times { 10 }^{ 16 } }{ 9\times 10^{ 12 } } \)
\(\\ =4\times 10^{ 4 }/{ m }^{ 3 }\)

(a) Split the gate,
  
The truth table for the given circuit is as shown below:

Here, for given A and B as inputs, C is the output of OR gate and input of NOT gate 1, D is the output of NOT gate 1 and input of NOT gate 2, then Y is finally output.

This is same as OR gate. So, this circuit acts as OR gate.  

Energy band gap of the given photodiode, E_g = 2.8 eV
Wavelength, λ = 6000 nm = 6000 x 10⁻⁹ m
The energy of a signal is given by the relation:
\(E=\frac{h c}{\lambda}\)
Where
h = Planck’s constant
= 6.626 x 10⁻³⁴ Js
c = Speed of light
= 3 x 10⁸ m/s
\(E=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-9}}\)
= 3.313 x 10⁻²⁰ J
But 1.6 x 10⁻¹⁹ J = 1 eV
∴ E = 3.313 x 10⁻²⁰ J
\(=3.313 \times \frac{10^{-20}}{1.6 \times 10^{-19}}=0.207 \mathrm{eV}\)
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.

For each atom doped with arsenic, one free electron is received. Similarly, for each atom doped of indium, a vacancy is created. So, number of free electrons introduced by pentavalent impurity is
\(N_{\mathrm{As}}=5 \times 10^{22} \mathrm{~m}^{-3}\)
The number of holes introduced by trivalent impurity added is
\(N_{\mathrm{I}}=5 \times 10^{20} \mathrm{~m}^{-3}\)
So, net number of electrons added is
n_e = N_As - N₁
\(\begin{aligned} =5 \times 10^{22}-5 \times 10^{20} \end{aligned}\)
\(\begin{aligned} =4.95 \times 10^{22} \mathrm{~m}^{-3} \end{aligned}\)
 We know that, \(n_e n_h=n_i^2\)
So,  \(n_h=\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}\)
\(=4.54 \times 10^9 \mathrm{~m}^{-3}\)
As, n_e > n_h (number of holes). So, the material is n-type semiconductor.

(i) (d): \(E=\frac{V}{l}=\frac{2}{0.1}=20 \mathrm{~V} / \mathrm{m} ;\)
\(A=1.0 \mathrm{~cm}^{2}=1.0 \times 10^{-4} \mathrm{~m}^{2}\)
\(\begin{array}{l} v_{e}=\mu_{e} E=0.14 \times 20=2.8 \mathrm{~m} \mathrm{~s}^{-1} \\ I_{e}=n_{e} A e v_{e} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 2.8 \\ =6.72 \times 10^{-7} \mathrm{~A} \end{array}\)
(ii) (c): In a pure semiconductor,
\(n_{e}=n_{h}=1.5 \times 10^{16} \mathrm{~m}^{-3}\)
\(\begin{array}{l} v_{h}=\mu_{h} \times E=0.05 \times 20=1.0 \mathrm{~ms}^{-1} \\ I_{h}=n_{h} \mathrm{Aev}_{h} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 1.0 \\ =2.4 \times 10^{-7} \mathrm{~A} \end{array}\)
(iii) (a): \(\sigma=e n_{e} m_{e}\)
\(\text { or } n_{e}=\frac{\sigma}{e \mu_{e}}=\frac{6.4 \times 10^{2}}{\left(1.6 \times 10^{-19}\right) \times 0.14}\)
\(=3.14 \times 10^{22} \approx 3 \times 10^{22} \mathrm{~m}^{-3}\)
(iv) (c): \(n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}=5 \times 10^{9} \mathrm{~m}^{-3}\)
(v) (c): In case of a p-n junction diode, width of the depletion region decreases as the forward bias voltage increases.

(i) (a) :The rms value of the output voltage at the load resistance \(V_{\mathrm{rms}}=\frac{V_{0}}{\sqrt{2}}\)
(ii) (d)
(iii) (a)
(iv) (c): The given circuit works as a half wave rectifier. In this circuit, we will get current through R when p-n junction is forward biased and no current when p-n junction is reverse biased. Thus the current (I) through resistor (R) will be shown in option (c).
(v) (c)