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Properties of Semiconductors

Practice Questions:

Question 1: easy

In case of a semiconductor, which of the following statement is wrong :

1. Doping increases conductivity

2. Temperature coefficient of resistance is negative

3. Resisitivity is in between that of a conductor and insulator

4. At absolute zero temperature, it behaves like a conductor

View Answer

The statement "At absolute zero temperature, it behaves like a conductor" is wrong for a semiconductor.

Explanation:

At absolute zero (0 K), intrinsic semiconductors (like silicon and germanium) do not have free charge carriers (electrons and holes) because all the electrons occupy the valence band, and there is no thermal energy available to promote them to the conduction band.
Therefore, intrinsic semiconductors behave like insulators at absolute zero, not conductors.

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Question 2: moderate

Intrinsic germanium and silicon at absolute zero temperature behave like :

1. Superconductor

2. Good semiconductor

3. Ideal insulator

4. Conductor

View Answer

At absolute zero temperature (0 K), intrinsic germanium and silicon behave like insulators due to the following reasons:

No Free Charge Carriers: At absolute zero, all the electrons in the material occupy the lowest energy states, and there are no thermally excited electrons available to conduct electricity. This means there are no free charge carriers (electrons or holes).
Wide Band Gap: Both germanium and silicon have a band gap (about 0.66 eV for germanium and 1.1 eV for silicon). At absolute zero, the thermal energy is insufficient to excite electrons across this band gap from the valence band to the conduction band.

As a result, intrinsic germanium and silicon cannot conduct electricity at absolute zero, behaving as insulators.

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Question 3: easy

A conducting wire of Copper and Germanium are cooled from room temperature to temperature 80K, then their resistance will :

1. increase

2. decrease

3. copper's increase and Germanium's decrease

4. copper's decrease and Germanium's increase

View Answer

When a conducting wire of Copper and Germanium is cooled from room temperature to 80K:

Copper (metal): Its resistance will decrease because metals have lower resistance at lower temperatures due to reduced electron scattering.
Germanium (semiconductor): Its resistance will increase because the number of charge carriers (electrons and holes) decreases at lower temperatures, leading to higher resistance.

Thus, Copper's resistance decreases, while Germanium's resistance increases when cooled to 80K

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Question 4: easy

Which statement is correct ?

1. N-type germanium is negatively charged and P-type germanium is positively charged

2. Both N-type and P-type germanium are neutral

3. N-type germanium is positively charged and P-type germanium is negatively charged

4. Both N-type and P-type germanium are negatively charged

View Answer

Both N-type and P-type germanium are electrically neutral because, while they have an imbalance of charge carriers (electrons or holes), the overall number of positive and negative charges remains equal.

- **N-type**: Donor atoms add extra electrons (negatively charged carriers), but the atoms themselves become positively charged ions. This ensures charge neutrality.

- **P-type**: Acceptor atoms create holes (positively charged carriers) by accepting electrons, but the atoms become negatively charged ions. This balances the charge.

In both cases, the total positive and negative charges cancel each other out, keeping the material electrically neutral.

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Question 5: easy

Wires P and Q have the same resistance at ordinary (room) temperature. When heated, resistance of P increases and that of Q decreases. We conclude that :

1. P and Q are conductors of different materials

2. P is N-type semiconductor and Q is P-type semiconductor

3. P is semiconductor and Q is conductor

4. P is conductor and Q is semiconductor

View Answer

Resistance of conductor increases with increasing temperature and resistance of semiconductor decreases with increasing temperature.

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Question 6: easy

The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be
0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.

1. \[2.16\times 10^{19}/m^{3}\]

2. \[4.32\times 10^{19}/m^{3}\]

3. \[10^{20}/m^{3}\]

4. \[10^{30}/m^{3}\]

View Answer

The resistivity ( $ρ$ ) of a pure (intrinsic) semiconductor is given by the formula:

$\rho = \frac{1}{q (n_i) (\mu_n + \mu_p)}$

$ρ = q ( n ^{i} ) ( μ ^{n} + μ ^{p} ) 1$

Where:

$\rho = 0.5 \, \Omega\text{m}$

$ρ = 0.5 Ω m$ (resistivity),
$q = 1.6 \times 10^{-19} \, \text{C}$

$q = 1.6 \times 1 0^{-19} C$ (charge of an electron),
$n_i$

$n_{i}$ = intrinsic carrier concentration (to be calculated),
$\mu_n = 0.39 \, \text{m}^2/\text{V-s}$

$μ_{n} = 0.39 m^{2} / V-s$ (electron mobility),
$\mu_p = 0.19 \, \text{m}^2/\text{V-s}$

$μ_{p} = 0.19 m^{2} / V-s$ (hole mobility).

Rearranging the formula to solve for

$n_i$

$n_{i}$ :

$n_i = \frac{1}{q \rho (\mu_n + \mu_p)}$

Substitute the values:

$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.39 + 0.19)}$

$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.58)}$

$n_{i} = \frac{1}{4.64 \times 1 0^{- 20}} = 2.16 \times 1 0^{19} m^{- 3}$

Thus, the intrinsic carrier concentration

$n_i$

$n_{i}$ is

$2.16 \times 10^{19} \, \text{m}^{-3}$

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Question 7: moderate

If the ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4 then what is the ratio of their drift velocities :

1. 5/8

2. 4/5

3. 5/4

4. 4/7

View Answer

The ratio of electron to hole currents is given by:

$\frac{I_n}{I_p} = \frac{7}{4}$

The ratio of electron to hole concentrations is:

$\frac{n}{p} = \frac{7}{5}$

Using the current formula

$I = q n v A$

$I = q n v A$ , the ratio of drift velocities (

$\frac{v_n}{v_p}$

$v ^{p} v ^{n}$ ) can be found as:

$\frac{I_{n}}{I_{p}} = \frac{n v_{n}}{p v_{p}}$

Substitute the given ratios:

$\frac{7}{4} = \frac{\frac{7}{5} \cdot v_n}{v_p}$

Simplifying:

$\frac{v_{n}}{v_{p}} = \frac{5}{4}$

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Question 8: moderate

GaAs (with a band gap = 1.5 eV) as an LED can emit :

1. Blue light

2. Green light

3. Infrared rays

4. Ultraviolet rays

View Answer

GaAs (Gallium Arsenide) with a band gap of 1.5 eV can emit light in the infrared region of the electromagnetic spectrum.

The energy of a photon emitted by an LED is related to its wavelength by the equation:

$E = \frac{hc}{\lambda}$

Solving we get

$λ= 826 nm which lies in infrared region.💬 Discuss this QuestionQuestion 9: easy Which one of the following statement is FALSE : 1. The resistance of intrinisic semiconductor decreases with increase of temperature. 2. Pure Si doped with trivalent impurities gives a p-type semiconductor. 3. Majority carriers in a n-type semiconductor are holes. 4. Minority carriers in a p-type semiconductor are electrons. View Answer Majority charge carrier in N-type semiconductor is electrons and in P-type semiconductor is holes 💬 Discuss this QuestionQuestion 10: easy Representation of N-type semiconductor is : 1. 2. 3. 4. View Answer In N type semiconductor number of electrons is more than number of holes 💬 Discuss this Question12 Next »document.addEventListener('DOMContentLoaded', function () {
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