JEE Physics Semiconductor Electronics MCQs Set 18

Multiple Choice Questions (MCQs) for JEE Physics: Semiconductor Electronics

Access targeted multiple-choice questions for Semiconductor Electronics designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Semiconductor Electronics MCQs for JEE Physics

View or download the dedicated Semiconductor Electronics MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: The forbidden energy gaps of a conductor, a semiconductor and an insulator are \( E_{G1},E_{G2},E_{G3} \) respectively. The correct relation is:
(a) \( E_{G1}\lt E_{G2}\lt E_{G3} \)
(b) \( E_{G1}=E_{G2}=E_{G3} \)
(c) \( E_{G1}\gt E_{G2}\gt E_{G3} \)
(d) \( E_{G1}\lt E_{G2}\gt E_{G3} \)
Show Answer & Explanation

Answer: (a) \( E_{G1}\lt E_{G2}\lt E_{G3} \)

Explanation:
1. For a conductor the bands overlap or the gap is zero, a semiconductor has a small gap (about 1 eV) and an insulator has a large gap (more than about 3 eV).
2. Hence the gaps increase in the order conductor, semiconductor, insulator.

Teacher's Note:
1. Typical values: Ge 0.7 eV, Si 1.1 eV, diamond about 5.4 eV.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Question 2: Fermi energy is the:
(a) minimum energy of the electrons in a metal at 0 K
(b) minimum energy of the electrons in a metal at 0 °C
(c) maximum energy of the electrons in a metal at 0 °C
(d) maximum energy of the electrons in a metal at 0 K
Show Answer & Explanation

Answer: (d) maximum energy of the electrons in a metal at 0 K

Explanation:
1. At absolute zero the electrons fill the allowed levels from the bottom up, and the highest filled level is called the Fermi level.

Teacher's Note:
1. The Fermi energy is defined at 0 K, not at 0 °C (273 K).
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Question 3: Which of the following statements is true?
(a) In a conductor the conduction band is completely empty
(b) In a semiconductor the conduction band is partially filled at low temperature
(c) In an insulator the conduction band is completely empty
(d) In an insulator the conduction band is completely filled with electrons
Show Answer & Explanation

Answer: (c) In an insulator the conduction band is completely empty

Explanation:
1. In an insulator the valence band is full and the conduction band is empty, separated by a wide gap, so no conduction occurs.

Teacher's Note:
1. In a conductor the bands overlap or the conduction band is partly filled.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Question 4: At 0 K the valence band of a semiconductor is:
(a) completely empty
(b) completely filled
(c) partially filled
(d) nothing can be said
Show Answer & Explanation

Answer: (b) completely filled

Explanation:
1. At 0 K no electron has enough energy to cross the gap, so the valence band is full and the conduction band empty.

Teacher's Note:
1. This is why a pure semiconductor is an insulator at 0 K.
2. Check units and orders of magnitude before choosing an option.

Question 5: The free-electron concentration n in the conduction band of an intrinsic semiconductor at temperature T (gap \( E_g \)) varies as:
(a) \( n=AT^{3/2}e^{-E_g/2kT} \)
(b) \( n=ATe^{-E_g/kT} \)
(c) \( n=AT^2e^{-E_g/kT} \)
(d) \( n=AT^2e^{-E_g/2kT} \)
Show Answer & Explanation

Answer: (a) \( n=AT^{3/2}e^{-E_g/2kT} \)

Explanation:
1. From the density of states and the Fermi distribution, \( n_i\propto T^{3/2}e^{-E_g/2kT} \).

Teacher's Note:
1. The factor 2 in the exponent arises because the Fermi level lies in the middle of the gap, so each carrier needs only half the gap on average.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Question 6: An electric field is applied to a semiconductor. If the temperature is raised, the number of charge carriers n and their average drift speed v change as:
(a) both n and v increase
(b) v increases but n decreases
(c) both n and v decrease
(d) n increases but v decreases
Show Answer & Explanation

Answer: (d) n increases but v decreases

Explanation:
1. More bonds break at higher temperature, so n increases.
2. The lattice vibrates more, so collisions are more frequent, the relaxation time falls and the drift speed per unit field falls.

Teacher's Note:
1. The rise in n dominates, so the resistance of a semiconductor falls overall.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Question 7: The electrical conductivity of pure germanium can be increased by which of the following?
(a) raising the temperature only
(b) doping with donor impurity only
(c) raising the temperature, doping with acceptors, doping with donors, or shining ultraviolet light on it (all of these)
(d) raising the temperature and shining ultraviolet light, but not doping
Show Answer & Explanation

Answer: (c) raising the temperature, doping with acceptors, doping with donors, or shining ultraviolet light on it (all of these)

Explanation:
1. Heat and ultraviolet light create extra electron-hole pairs, and either donor or acceptor impurity adds extra carriers.

Teacher's Note:
1. So every one of the four methods raises the conductivity.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Question 8: An N-type semiconductor is:
(a) negatively charged
(b) electrically neutral
(c) positively charged
(d) charged, with the sign depending on temperature
Show Answer & Explanation

Answer: (b) electrically neutral

Explanation:
1. Donor atoms are neutral; each extra electron is balanced by the positive donor ion that remains in the lattice.

Teacher's Note:
1. Doping changes the carrier type, not the net charge.
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Question 9: In an extrinsic semiconductor the numbers of holes \( n_p \) and electrons \( n_e \) satisfy:
(a) \( n_p\ne n_e \)
(b) \( n_p=n_e \)
(c) \( n_p=n_e=n_i \)
(d) \( n_p\text{ and }n_e\text{ are both zero} \)
Show Answer & Explanation

Answer: (a) \( n_p\ne n_e \)

Explanation:
1. Doping makes one carrier type the majority, so the numbers are unequal.

Teacher's Note:
1. The product \( n_en_p=n_i^2 \) still holds.
2. Check units and orders of magnitude before choosing an option.

Question 10: In a P-type semiconductor the acceptor level lies 57 meV above the valence band. The maximum wavelength of light that can produce a hole is:
(a) 57 Å
(b) 57 × 10⁻³ Å
(c) 11.61 × 10 Å
(d) 217100 Å
Show Answer & Explanation

Answer: (d) 217100 Å

Explanation:
1. \( \lambda=\frac{hc}{E}=\frac{12375\ \text{eV\,\AA}}{0.057\ \text{eV}}\approx2.17\times10^{5}\ \text{\AA} \).

Teacher's Note:
1. Use hc = 12375 eV Å (about 12400 eV Å) for quick conversions.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Question 11: A P-type silicon semiconductor is made by adding one atom of indium per \( 5\times10^{7} \) atoms of silicon. If silicon has \( 5\times10^{28} \) atoms per cubic metre, the number of acceptor atoms per cubic centimetre is:
(a) \( 5\times10^{15}\ \text{atoms/cm}^3 \)
(b) \( 2\times10^{30}\ \text{atoms/cm}^3 \)
(c) \( 1\times10^{15}\ \text{atoms/cm}^3 \)
(d) \( 2.5\times10^{36}\ \text{atoms/cm}^3 \)
Show Answer & Explanation

Answer: (c) \( 1\times10^{15}\ \text{atoms/cm}^3 \)

Explanation:
1. \( N_A=\frac{5\times10^{28}}{5\times10^{7}}=10^{21}\ \text{m}^{-3} \).
2. Convert to cm⁻³: divide by \( 10^{6} \), giving \( 10^{15}\ \text{cm}^{-3} \).

Teacher's Note:
1. Divide by 10⁶ to convert m⁻³ to cm⁻³.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Question 12: If an N-type semiconductor is heated:
(a) the number of electrons increases and that of holes decreases
(b) the numbers of both electrons and holes increase equally (by thermally generated pairs)
(c) the number of holes increases and that of electrons decreases
(d) the numbers of electrons and holes become equal
Show Answer & Explanation

Answer: (b) the numbers of both electrons and holes increase equally (by thermally generated pairs)

Explanation:
1. Heating creates equal numbers of new electrons and holes, so both increase by the same amount; electrons remain the majority carriers.

Teacher's Note:
1. Electrons and holes become equal only at very high temperature when the material becomes effectively intrinsic.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Question 13: What is the conductivity of pure silicon at 300 K if the electron-hole pairs per cm³ are \( 1.072\times10^{10} \), \( \mu_n=1350\ \text{cm}^2/\text{V s} \) and \( \mu_p=480\ \text{cm}^2/\text{V s} \)?
(a) \( 3.14\times10^{-6}\ \text{mho/cm} \)
(b) \( 3\times10^{6}\ \text{mho/cm} \)
(c) \( 10^{-6}\ \text{mho/cm} \)
(d) \( 10^{6}\ \text{mho/cm} \)
Show Answer & Explanation

Answer: (a) \( 3.14\times10^{-6}\ \text{mho/cm} \)

Explanation:
1. \( \sigma=n_ie(\mu_n+\mu_p)=1.072\times10^{10}\times1.6\times10^{-19}\times(1350+480) \)
2. \( \sigma\approx3.14\times10^{-6}\ \text{mho/cm} \).

Teacher's Note:
1. Both carriers contribute, so add the two mobilities.
2. Check units and orders of magnitude before choosing an option.

Question 14: The forbidden energy gap of germanium is 0.75 eV. The maximum wavelength of radiation that can produce an electron-hole pair in it is:
(a) 4200 Å
(b) 4700 Å
(c) 4000 Å
(d) 16500 Å
Show Answer & Explanation

Answer: (d) 16500 Å

Explanation:
1. \( \lambda_{max}=\frac{12375}{0.75}=16500\ \text{\AA} \)

Teacher's Note:
1. Photons with wavelength longer than this have energy below the gap and cannot excite an electron.
2. Compare with the corresponding statement for the other kind of semiconductor or junction to avoid mixing them up.

Question 15: The mobility of electrons in N-type germanium is 5000 cm²/V s and the conductivity is 5 mho/cm. If the effect of holes is negligible, the impurity concentration is:
(a) \( 9.25\times10^{14}\ \text{cm}^{-3} \)
(b) \( 6\times10^{13}\ \text{cm}^{-3} \)
(c) \( 6.25\times10^{15}\ \text{cm}^{-3} \)
(d) \( 9\times10^{13}\ \text{cm}^{-3} \)
Show Answer & Explanation

Answer: (c) \( 6.25\times10^{15}\ \text{cm}^{-3} \)

Explanation:
1. \( \sigma=ne\mu\Rightarrow n=\frac{5}{1.6\times10^{-19}\times5000}=6.25\times10^{15}\ \text{cm}^{-3} \)

Teacher's Note:
1. Keep mobility and conductivity in the same (cm) units.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Question 16: Pure silicon at 300 K has equal electron and hole concentrations of \( 1.5\times10^{16}\ \text{m}^{-3} \). Doping with indium raises the hole concentration to \( 4.5\times10^{22}\ \text{m}^{-3} \). The electron concentration in the doped silicon is:
(a) \( 6.0\times10^{6}\ \text{m}^{-3} \)
(b) \( 5.0\times10^{9}\ \text{m}^{-3} \)
(c) \( 7.0\times10^{3}\ \text{m}^{-3} \)
(d) \( 4.0\times10^{9}\ \text{m}^{-3} \)
Show Answer & Explanation

Answer: (b) \( 5.0\times10^{9}\ \text{m}^{-3} \)

Explanation:
1. \( n_en_h=n_i^2\Rightarrow n_e=\frac{(1.5\times10^{16})^2}{4.5\times10^{22}}=5.0\times10^{9}\ \text{m}^{-3} \)

Teacher's Note:
1. Indium is an acceptor, so holes become the majority and electrons the minority.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Question 17: The length of a germanium rod is 0.58 cm and its cross-sectional area is 1 mm². For germanium \( n_i=2.5\times10^{19}\ \text{m}^{-3} \), \( \mu_h=0.19\ \text{m}^2/\text{V s} \), \( \mu_e=0.39\ \text{m}^2/\text{V s} \). The resistance of the rod is:
(a) 2.5 kΩ
(b) 5.0 kΩ
(c) 7.5 kΩ
(d) 10.0 kΩ
Show Answer & Explanation

Answer: (a) 2.5 kΩ

Explanation:
1. \( \sigma=n_ie(\mu_e+\mu_h)=2.5\times10^{19}\times1.6\times10^{-19}\times0.58=2.32\ \text{S/m} \)
2. \( R=\frac{l}{\sigma A}=\frac{0.58\times10^{-2}}{2.32\times10^{-6}}\approx2.5\times10^{3}\ \Omega \)

Teacher's Note:
1. Convert all lengths to metres: 0.58 cm = 5.8 × 10⁻³ m and 1 mm² = 10⁻⁶ m².
2. A quick sketch of the band diagram or the circuit makes this type of question easy.

Question 18: The contributions of electrons and holes to the total current in a semiconductor are 3/4 and 1/4 respectively. If the drift speed of electrons is 5/2 times that of holes at this temperature, the ratio of the concentrations of electrons and holes is:
(a) 5 : 6
(b) 3 : 2
(c) 2 : 3
(d) 6 : 5
Show Answer & Explanation

Answer: (d) 6 : 5

Explanation:
1. \( I\propto nv\Rightarrow \frac{n_e}{n_h}=\frac{I_e}{I_h}\cdot\frac{v_h}{v_e}=\frac{3/4}{1/4}\times\frac{2}{5}=\frac{6}{5} \)

Teacher's Note:
1. Current is proportional to carrier density times drift speed (area and charge are the same).
2. Check units and orders of magnitude before choosing an option.

Question 19: The diffusion current in a P-N junction is greater than the drift current in magnitude if the junction is:
(a) reverse biased
(b) unbiased
(c) forward biased
(d) never, in any case
Show Answer & Explanation

Answer: (c) forward biased

Explanation:
1. Forward bias lowers the barrier, letting many majority carriers diffuse across, so diffusion current exceeds drift current.

Teacher's Note:
1. Unbiased, the two currents are equal and opposite; reverse biased, drift exceeds diffusion.
2. Questions of this kind are asked directly in JEE Main, so keep the key fact at your fingertips.

Question 20: The depletion region of an open-circuit P-N diode contains:
(a) free electrons
(b) unmasked immobile impurity ions
(c) free holes
(d) neutral impurity atoms
Show Answer & Explanation

Answer: (b) unmasked immobile impurity ions

Explanation:
1. Majority carriers diffuse across the junction and recombine, leaving behind uncovered (immobile) donor and acceptor ions.

Teacher's Note:
1. These ions set up the junction electric field.
2. Eliminate the options that contradict a basic property of the device, then check the remaining one.

Question 21: A hole diffusing from the p-side to the n-side of a P-N junction means that:
(a) a bond is broken on the n-side and the freed electron jumps to a broken bond on the p-side to complete it
(b) a bond is broken on the n-side and the freed electron jumps to the conduction band
(c) a conduction electron on the p-side jumps into a broken bond to complete it
(d) a bond is broken on the p-side and the freed electron jumps to a broken bond on the n-side
Show Answer & Explanation

Answer: (a) a bond is broken on the n-side and the freed electron jumps to a broken bond on the p-side to complete it

Explanation:
1. A hole moves in the direction opposite to the electron that fills it.

Teacher's Note:
1. So hole motion p to n corresponds to a valence electron on the n-side moving into a vacancy on the p-side.
2. Link the answer to the physical picture of carriers and junctions rather than memorising it as a bare fact.

Question 22: The depletion layer in a P-N junction is caused by:
(a) drift of holes
(b) migration of impurity ions
(c) drift of electrons
(d) diffusion of free charge carriers
Show Answer & Explanation

Answer: (d) diffusion of free charge carriers

Explanation:
1. Electrons diffuse from n to p and holes from p to n; the carriers recombine near the junction and leave a region without free carriers.

Teacher's Note:
1. Drift acts afterwards and opposes the diffusion.
2. Check units and orders of magnitude before choosing an option.

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Practice MCQs for JEE Physics Semiconductor Electronics

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