Semiconductors and Electronic Devices MCQ Questions & Answers in Modern Physics | Physics

Pauli’s exclusion principle.

$$R = \frac{{\Delta V}}{{\Delta I}} = \frac{{2.1 - 2}}{{\left( {800 - 400} \right) \times {{10}^{ - 3}}}} = \frac{1}{4} = 0.25\,\Omega $$

In the circuit emitter is common.

Neighbour distance of a body centred cubic cell $$d = \frac{{\sqrt 3 }}{2}a,$$   where $$a$$ is the lattice parameter.
$$\eqalign{ & \Rightarrow 3.7 = \frac{{\sqrt 3 a}}{2} \cr & {\text{or}}\,\,a = \frac{{2 \times 3.7}}{{\sqrt 3 }} = 4.3\,\mathop {\text{A}}\limits^ \circ \cr} $$

In a photoconductive cell, when monochromatic light is incident on the transparent metallic film, a force produced called the photo electromotive force, stimulates the emission of an electric current where photovoltaic action creates a potential difference between two points.
The magnitude of this current depends upon the intensity of incident light. Hence, photo electromotive force produced by monochromatic light is proportional to the intensity of light falling on the cell.

Output of upper AND gate $$ = \overline A B$$
Output of lower AND gate $$ = A\overline B $$
Thus, output of OR gate $$ = \overline A B + A\overline B $$
This is Boolean expression for XOR gate.

We know that in forward biasing of $$p$$ - $$n$$ junction the current is of the order of milliampere while in reverse biasing the current is of the order of microampere (negligible). Thus, device is a $$p$$ - $$n$$ junction.

Specific resistance is resistivity which is given by $$\rho = \frac{m}{{n{e^2}\tau }}$$
where $$n$$ = no. of free electrons per unit volume
and $$\tau $$ = average relaxation time
For a conductor with rise in temperature $$n$$ increases and $$\tau $$ decreases. But decrease in $$\tau $$ is more dominant than increase in $$n$$ resulting an increase in the value of $$\rho .$$
For a semiconductor with rise in temperature, $$n$$ increases and $$\tau $$ decreases. But the increase in $$n$$ is more dominant than decrease in $$\tau $$ resulting in a decrease in the value of $$\rho .$$

Many electronic devices require a source of energy at a specific frequency which may range from a few $$Hz$$  to several $$MHz.$$  This is achieved by an electronic device called an oscillator. An oscillator can produce waves from small $$\left( {20\,Hz} \right)$$  to extremely high frequencies $$\left( { > 100\,MHz} \right).$$

Voltage gain $$ = \frac{{{A_V}}}{{1 + \beta {A_V}}}$$
$$\eqalign{ & 10 = \frac{{{A_V}}}{{1 + \frac{9}{{100}}{A_V}}} \cr & {A_V} = \frac{{10}}{{0.1}} = 100 \cr} $$