Semiconductor Electronics: Materials, Devices and Simple Circuits MCQs for NEET — Physics Questions with Answers

Majority charge carriers are electrons and minority charge carrier are holes in the given specimen of semiconductor, so the semiconductor is n-type

(c) The n-type semi-conductor can be produced by doping an impurity atom of valence 5 i.e., pentavalent atoms. i.e., phosphorus.

When a small amount of antimony is added to

germanium crystal, the crystal becomes n-type

semiconductor because antimoney is a 

pentavalent substrate. It excess free electrons.

In forward biasing of p-n junction the positive terminal of the battery is connected to p-side and the depletion region becomes thin.

 $$\begin{gathered} {\mathrm{n}}_{\mathrm{i}}^2={\mathrm{n}}_{\mathrm{e}}{\mathrm{n}}_{\mathrm{h}} \\ {\mathrm{n}}_{\mathrm{e}}=\frac{({\mathrm{n}}_{\mathrm{i}}{)}^2}{{\mathrm{n}}_{\mathrm{h}}} \\ {\mathrm{n}}_{\mathrm{e}}=\frac{(1.5\times {10}^{16}{)}^2}{(4.5\times {10}^{22})} \\ {\mathrm{n}}_{\mathrm{e}}=5\times {10}^9 {\mathrm{m}}^{-3} \end{gathered}$$

 So, ${\mathrm{n}}_{\mathrm{h}}\gg {\mathrm{n}}_{\mathrm{e}}$ semiconductor is p-type.

p-type semiconductor are obtained by adding a small amount of trivalent impurity to a pure sample of semiconductor(Ge).

Majority charge carriers-holes

Minority charge carries-electrons

n-type semiconductor are obtained by adding a small amount of pentavalent impurity to a pure sample of semiconductor (Ge).

Majority charge carries-electrons

The resistance of intrinsic semiconductors decreases with increase of temperature.

Integrated circuits are miniature electronic circuit produced within a single crystal of a semiconductors such as silicon. They contain a million or so transistors and resistors or capacitors. They are widely used in memory circuits, micro computers, pocket calculators and electronic watches on account of their low cost and bulk, reliability into specific regions of the semiconductor crystals.

Key Idea    Only signals having wavelength less than threshold wavelength will be detected.

 $$\begin{gathered} \mathrm{Energy} \mathrm{E}=\mathrm{hv}=\mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}} \end{gathered}$$

Substituting the value of h, c and E in the above equation 

                  $\mathrm{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.5\times 1.6\times {10}^{-19}}=5000 \stackrel{\circ }{\mathrm{A}}$

   As  $4000 \stackrel{\circ }{\mathrm{A}}<5000 \stackrel{\circ }{\mathrm{A}}$

 Signal of wavelength $4000 \stackrel{\circ }{\mathrm{A}}$ can be detected by the photodiode.

Neighbour distance of a body centred cubic cell $d=\frac{\sqrt{3}}{2}a, where a is the lattice parameter.$

$\Rightarrow 3.7=\frac{\sqrt{3}a}{2}$

or               $a=\frac{2\times 3.7}{\sqrt{3}}=4.3\stackrel{0}{A}$

Atomic radius is the distance between two nearest neighbours. For fcc crystal

Atomic radius = $\frac{1}{2\sqrt{2}}\times lattice parameter$

or               $r=\frac{a}{2\sqrt{2}}$

or             $r=\frac{3.6}{2\sqrt{2}} \stackrel{0}{A}=1.27 \stackrel{0}{A}$