Semiconductors & Energy Bands — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Semiconductors & Energy Bands MCQs with step-by-step solutions (21 questions). Part of Semiconductor Devices. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Semiconductors & Energy Bands · easy · theory
The energy gap in a solid is the separation between the:
A. Top of the valence band and the bottom of the conduction band ✓ Correct
B. Two halves of the valence band
C. Nucleus and the innermost electron shell
D. Two halves of the conduction band
Solution: No electron states exist within this forbidden gap, so an electron must acquire at least $E_g$ to become free.
Q2 — Semiconductors & Energy Bands · easy · theory
In a conductor, the valence band and the conduction band:
A. Overlap each other ✓ Correct
B. Are separated by more than $3\text{ eV}$
C. Are both completely empty
D. Are separated by about $1\text{ eV}$
Solution: With overlapping bands, electrons can move to higher states under even the feeblest field.
Q3 — Semiconductors & Energy Bands · easy · theory
In an insulator, the forbidden energy gap is typically:
A. About $1\text{ eV}$
B. Zero
C. Greater than about $3\text{ eV}$ ✓ Correct
D. Negative
Solution: Diamond, for example, has a gap of about $6\text{ eV}$, far beyond the reach of thermal energy at room temperature.
Q4 — Semiconductors & Energy Bands · easy · theory
In a semiconductor, the forbidden energy gap is typically:
A. About $100\text{ eV}$
B. Of the order of $1\text{ eV}$ ✓ Correct
C. Exactly zero
D. Greater than $5\text{ eV}$
Solution: A gap of this size lets a useful number of electrons cross it by thermal excitation at ordinary temperatures.
Q5 — Semiconductors & Energy Bands · medium · theory
The energy gaps of silicon and germanium at room temperature are approximately:
A. $1.1\text{ eV}$ and $0.7\text{ eV}$ ✓ Correct
B. $0.1\text{ eV}$ and $0.2\text{ eV}$
C. $3.0\text{ eV}$ and $5.0\text{ eV}$
D. $0.7\text{ eV}$ and $1.1\text{ eV}$
Solution: Germanium has the narrower gap, so it generates more thermal carriers and has the larger leakage current.
Q6 — Semiconductors & Energy Bands · easy · theory
The electrical resistivity of a semiconductor lies:
A. Below that of a conductor
B. Above that of an insulator
C. Between that of a conductor and that of an insulator ✓ Correct
D. Exactly equal to that of a conductor
Solution: This intermediate behaviour, controllable by doping, is what makes semiconductors so useful.
Q7 — Semiconductors & Energy Bands · medium · theory
With rising temperature, the resistance of a semiconductor:
A. Becomes infinite
B. Decreases, because many more charge carriers are generated ✓ Correct
C. Increases, because of lattice vibrations
D. Remains unchanged
Solution: The rapid growth in carrier number outweighs the increased scattering, giving a negative temperature coefficient.
Q8 — Semiconductors & Energy Bands · medium · theory
At absolute zero, a pure semiconductor behaves as:
A. A perfect conductor
B. A perfect insulator ✓ Correct
C. A superconductor
D. A normal semiconductor
Solution: Every valence electron is locked in a covalent bond, leaving the conduction band completely empty.
Q9 — Semiconductors & Energy Bands · hard · numerical
A semiconductor has an energy gap of $1.1\text{ eV}$. The longest wavelength that can excite an electron across it is approximately ($hc = 1240\text{ eV}\cdot\text{nm}$):
A. $1127\text{ \AA}$
B. $1127\text{ nm}$ ✓ Correct
C. $2254\text{ nm}$
D. $564\text{ nm}$
Solution: $\lambda = \dfrac{hc}{E_g} = \dfrac{1240}{1.1} \approx 1127\text{ nm}$, which lies in the infrared.
Q10 — Semiconductors & Energy Bands · hard · numerical
Germanium has an energy gap of $0.7\text{ eV}$. The corresponding threshold wavelength is approximately:
A. $1127\text{ nm}$
B. $700\text{ nm}$
C. $868\text{ nm}$
D. $1771\text{ nm}$ ✓ Correct
Solution: $\lambda = \dfrac{1240}{0.7} \approx 1771\text{ nm}$.
Q11 — Semiconductors & Energy Bands · medium · numerical
A photon of wavelength $1240\text{ nm}$ can just excite an electron across the gap of a semiconductor. The energy gap is:
A. $1\text{ eV}$ ✓ Correct
B. $2\text{ eV}$
C. $1240\text{ eV}$
D. $0.5\text{ eV}$
Solution: $E_g = \dfrac{1240}{1240} = 1\text{ eV}$.
Q12 — Semiconductors & Energy Bands · medium · numerical
A material absorbs light of wavelength $620\text{ nm}$ across its band gap. The energy gap is:
A. $1\text{ eV}$
B. $4\text{ eV}$
C. $2\text{ eV}$ ✓ Correct
D. $0.5\text{ eV}$
Solution: $E_g = \dfrac{1240}{620} = 2\text{ eV}$.
Q13 — Semiconductors & Energy Bands · medium · numerical
A semiconductor has an energy gap of $2.5\text{ eV}$. The maximum wavelength that can excite an electron across it is:
A. $1240\text{ nm}$
B. $248\text{ nm}$
C. $620\text{ nm}$
D. $496\text{ nm}$ ✓ Correct
Solution: $\lambda = \dfrac{1240}{2.5} = 496\text{ nm}$.
Q14 — Semiconductors & Energy Bands · medium · numerical
An energy gap of $1.1\text{ eV}$ expressed in joule is approximately:
A. $6.9 \times 10^{-19}\text{ J}$
B. $1.76 \times 10^{-18}\text{ J}$
C. $1.1 \times 10^{-19}\text{ J}$
D. $1.76 \times 10^{-19}\text{ J}$ ✓ Correct
Solution: $1.1 \times 1.6 \times 10^{-19} = 1.76 \times 10^{-19}\text{ J}$.
Q15 — Semiconductors & Energy Bands · hard · numerical
An insulator has an energy gap of $6\text{ eV}$. The wavelength needed to excite an electron across it is approximately:
A. $620\text{ nm}$
B. $1240\text{ nm}$
C. $103\text{ nm}$
D. $207\text{ nm}$ ✓ Correct
Solution: $\lambda = \dfrac{1240}{6} \approx 207\text{ nm}$, in the deep ultraviolet — which is why insulators are transparent to visible light.
Q16 — Semiconductors & Energy Bands · easy · numerical
A semiconductor sample has resistivity $0.5\,\Omega\cdot\text{m}$. Its conductivity is:
A. $0.25\text{ S/m}$
B. $0.5\text{ S/m}$
C. $2\text{ S/m}$ ✓ Correct
D. $5\text{ S/m}$
Solution: $\sigma = \dfrac{1}{\rho} = \dfrac{1}{0.5} = 2\text{ S/m}$.
Q17 — Semiconductors & Energy Bands · medium · numerical
At room temperature, germanium has more thermally generated carriers than silicon because germanium has:
A. More valence electrons per atom
B. A larger energy gap
C. A higher melting point
D. A smaller energy gap ✓ Correct
Solution: With only $0.7\text{ eV}$ to cross rather than $1.1\text{ eV}$, far more electrons make the jump at a given temperature.
Q18 — Semiconductors & Energy Bands · hard · numerical
Gallium arsenide has an energy gap of $1.43\text{ eV}$. The corresponding wavelength is approximately:
A. $433\text{ nm}$
B. $1127\text{ nm}$
C. $620\text{ nm}$
D. $867\text{ nm}$ ✓ Correct
Solution: $\lambda = \dfrac{1240}{1.43} \approx 867\text{ nm}$.
Q19 — Semiconductors & Energy Bands · medium · numerical
Two semiconductors have energy gaps of $1.1\text{ eV}$ and $2.2\text{ eV}$. The ratio of their threshold wavelengths is:
A. $1 : 1$
B. $1 : 2$
C. $4 : 1$
D. $2 : 1$ ✓ Correct
Solution: $\lambda \propto \dfrac{1}{E_g}$, so the smaller gap corresponds to the longer wavelength.
Q20 — Semiconductors & Energy Bands · medium · numerical
The energy needed to move an electron across a gap of $1.1\text{ eV}$ is supplied by a photon of energy at least:
A. $1.76 \times 10^{-20}\text{ J}$
B. $1.1 \times 10^{-19}\text{ J}$
C. $1.76 \times 10^{-19}\text{ J}$ ✓ Correct
D. $3.52 \times 10^{-19}\text{ J}$
Solution: Converting: $1.1\text{ eV} = 1.76 \times 10^{-19}\text{ J}$.
Q21 — Semiconductors & Energy Bands · medium · numerical
A semiconductor begins to absorb at $900\text{ nm}$. Its energy gap is approximately:
A. $1.1\text{ eV}$
B. $0.72\text{ eV}$
C. $2.76\text{ eV}$
D. $1.38\text{ eV}$ ✓ Correct
Solution: $E_g = \dfrac{1240}{900} \approx 1.38\text{ eV}$.