NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics
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1. In an n-type silicon, which of the following statements 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.
Sol. (c) The n-type semiconductor, is obtained by doping the Ge or Si with pentavalent atoms. In n-type semiconductor, electrons are majority carriers and holes are minority carriers.
2. Which of the statements given in Q. 1 is true for p-type semiconductors?
Sol. (d) A p-type semiconductor is obtained by doping Ge or Si with trivalent atom. In p-semiconductor holes are majority carriers and electrons are minority carriers.
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 (Eg)G, (Eg)Si and (Eg) Ge. Which of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
Sol. (c) The energy band-gap is largest for carbon, less for silicon and least for germanium.
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 of the above
Sol. (c) In an unbiased p-n junction, the diffusion of charge takes place across the junction takes place from higher concentration to lower concentration. Therefore, hole concentration in p-region is more as compared to n-region.
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
Sol. (c) When a forward bias is applied across the p-n junction, the applied voltage opposes the barrier voltage. Due to this, the decreased potential barrier across the junction is decreased.
6. 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?
Sol. Half wave rectifier rectifies only the half of AC input i.e., it conducts once during an AC cycle while a full-wave rectifier rectifies both the half cycles of the AC input i.e. or we can say, conducts twice during a cycle.
∴The output frequency for half-wave is 50 Hz.
The output frequency of a full-wave rectifier is double i.e., 2 × 50 = 100 Hz.
7. A p-n photodiode is fabricated from a semiconductor with band-gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
$$\textbf{sol.}\space\text{Energy, E =}\frac{hc}{\lambda}=\frac{6.6×10^{\normalsize-34}×3×10^{8}}{6000×10^{\normalsize-9}×1.6×10^{\normalsize-19}}\text{eV}$$
= 2.06 eV
The band-gap is 2.8 eV and energy E is less than the band-gap (E < Eg), so it cannot detect the wavelength 6000 nm.
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