Quick Summary
Concept Tested: Electrical Conductivity in Metals vs. Semiconductors | Chapter: Solid State | Difficulty: ★★★☆☆ | Time: 2 mins | Answer: D
Why: Metals conduct via free electrons; increasing temperature increases lattice vibrations, reducing electron mobility and conductivity. Semiconductors conduct via thermally excited carriers; increasing temperature exponentially increases carrier concentration and conductivity.
The Question
Consider the following statements regarding defects in solids and their properties:
(A) Density of crystals remains unaffected due to Frenkel defect.
(B) In bcc unit cell the void space is 32%.
(C) Density of crystals decreases due to Schottky defect.
(D) Electrical conductivity of semiconductors and metals increases with increase in temperature.
Quick Answer
Answer: (D)
The statement (D) is incorrect because the temperature dependence of conductivity is opposite in metals and semiconductors. In metals, conductivity decreases with temperature due to increased scattering of free electrons. In semiconductors, conductivity increases with temperature due to the exponential generation of charge carriers.
Why Other Options Are Incorrect
Option (A) Analysis: Frenkel defect involves the movement of an ion from its lattice site to an interstitial position, creating a vacancy-interstitial pair. The total mass and volume of the crystal remain constant because the atom is simply relocated. Therefore, density ($\rho = \frac{m}{V}$) remains unaffected. The statement is actually correct, so it cannot be the answer to a question asking for the incorrect statement.
Option (B) Analysis: For a Body-Centered Cubic (bcc) unit cell, the packing efficiency is calculated as follows:
Number of atoms per unit cell = 2
Volume of atoms = $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$
Edge length ($a$) = $\frac{4r}{\sqrt{3}}$
Volume of unit cell = $a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}$
Packing efficiency = $\frac{\text{Volume of atoms}}{\text{Volume of unit cell}} = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{\pi\sqrt{3}}{8} \approx 0.68$ or $68\%$
Since packing efficiency is $68\%$, the void space is $100\% – 68\% = 32\%$. This statement is correct.
Option (C) Analysis: Schottky defects occur due to the simultaneous vacancy formation of cations and anions. The removal of ions from lattice sites reduces the mass and volume of the crystal. Consequently, the density of the crystal decreases. This statement is correct.
Video Solution
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Understanding the Concept
Electrical conductivity ($\sigma$) depends on the number of charge carriers ($n$) and their mobility ($\mu$), as given by the equation:
$$\sigma = n e \mu$$
In Metals: The number of free electrons ($n$) is constant. However, as temperature increases, lattice vibrations (phonons) increase, causing more frequent collisions between electrons and ions. This reduces the mean free path ($\lambda$) of electrons, thereby decreasing mobility ($\mu$). Since $\sigma \propto \mu$, conductivity decreases.
In Semiconductors: The number of charge carriers is very low at room temperature but increases exponentially with temperature due to thermal excitation across the band gap. The increase in $n$ is so significant that it outweighs the slight decrease in mobility. Thus, conductivity increases.
Detailed Step-by-Step Solution
Step 1: Analyze the Mechanism of Conduction in Metals
Metals contain a sea of delocalized electrons (free electrons) that move freely through the lattice. Conductivity is determined by the ease with which these electrons drift under an electric field.
The mobility ($\mu$) of electrons is defined as the ratio of drift velocity to the electric field. It is inversely proportional to the collision frequency of electrons with the lattice ions. As temperature rises, lattice ions vibrate with higher amplitude, increasing the collision frequency and reducing the mean free path ($\lambda$).
From the relation $\mu = \frac{e\lambda}{m}$, if $\lambda$ decreases, $\mu$ decreases. Since conductivity $\sigma = ne\mu$, and $n$ is constant in metals, the conductivity of metals decreases with increasing temperature.
Step 2: Analyze the Mechanism of Conduction in Semiconductors
Semiconductors have a band gap ($E_g$) between the valence band and the conduction band. At absolute zero, they behave as insulators. As temperature increases, electrons gain thermal energy ($kT$) and are excited to the conduction band.
The number of charge carriers ($n$) in semiconductors follows an exponential relationship:
$$n \propto e^{-E_g / 2kT}$$
As $T$ increases, the exponent becomes less negative, causing $n$ to increase exponentially. Although the mobility decreases slightly due to scattering, the drastic increase in $n$ dominates, resulting in a net increase in conductivity with temperature.
Step 3: Evaluate Option (D)
Option (D) claims that the conductivity of both semiconductors and metals increases with temperature. Based on Step 1 and Step 2, we know this is false. Metals show a decreasing trend, while semiconductors show an increasing trend.
Final Answer
Correct Option: (D)
The statement is incorrect because the electrical conductivity of metals decreases with increasing temperature, whereas the electrical conductivity of semiconductors increases with increasing temperature.
Essential Formulas for This Topic
1. Conductivity in Metals: $\sigma = n e \mu$ (where $n$ is constant, $\mu \propto \frac{1}{T}$)
2. Conductivity in Semiconductors: $\sigma = n e \mu$ (where $n \propto e^{-E_g/2kT}$ dominates)
3. Mobility: $\mu = \frac{e\lambda}{m}$
4. Packing Efficiency (BCC): $68\%$ (Void space $32\%$)
5. Defect Density: $\rho_{defect} = \frac{\text{Number of defects}}{\text{Number of atoms}}$
Common Mistakes to Avoid
Mistake 1: Assuming “Higher Temperature = Higher Conductivity” for All Materials
Wrong Thinking: Since conductivity usually increases with temperature in daily life (e.g., resistors heat up), one might assume metals conduct better when hot.
Correct Approach: Always distinguish between metals and semiconductors/insulators. Metals lose conductivity due to scattering; semiconductors gain it due to carrier generation.
Mistake 2: Misidentifying the Effect of Defects on Density
Wrong Thinking: Frenkel defects or Schottky defects always change the density of the crystal.
Correct Approach: Frenkel defects do not change density because the atom is just moved to an interstitial site. Schottky defects reduce density because atoms leave the lattice entirely.
Mistake 3: Calculating Void Space Incorrectly
Wrong Thinking: Void space is simply $100\%$ minus the number of atoms.
Correct Approach: Void space is calculated as $100\%$ minus the packing efficiency. For BCC, packing efficiency is $68\%$, so void space is $32\%$.
Key Concept Summary
- Metal Conductivity: Dominated by electron mobility; decreases with temperature ($\sigma \downarrow$ as $T \uparrow$).
- Semiconductor Conductivity: Dominated by intrinsic carrier concentration; increases with temperature ($\sigma \uparrow$ as $T \uparrow$).
- Frenkel Defect: Vacancy + Interstitial pair; Density unchanged.
- Schottky Defect: Pair of vacancies; Density decreases.
Golden Rule: Metals and semiconductors exhibit opposite temperature dependence of electrical conductivity.
Frequently Asked Questions
Q: Why do metals have high conductivity while insulators do not?
A: Metals have a large number of free electrons (high $n$) and their valence and conduction bands overlap, allowing electrons to move easily. Insulators have a large band gap; electrons are bound to atoms and cannot move freely.
Q: Does temperature affect the number of free electrons in metals?
A: No, the number of free electrons ($n$) in a metal is essentially constant regardless of temperature. The change in conductivity is solely due to the change in mobility.
Q: What is the dominant factor for conductivity in semiconductors at high temperatures?
A: The dominant factor is the exponential increase in the number of charge carriers ($n$) due to thermal excitation across the band gap.
Q: If a crystal has a high density of Schottky defects, what happens to its molar mass?
A: The molar mass remains the same because the atoms are still present in the crystal, just missing from their lattice sites. However, the mass per unit volume (density) decreases.
Prerequisites to Solve This Question
- Understanding of Crystal Structure (Unit Cells, BCC, FCC).
- Knowledge of Crystal Defects (Frenkel, Schottky) and their impact on density.
- Basic Band Theory of Solids (Conduction band, Valence band, Band gap).
- Drift Velocity and Mean Free Path concepts.
After Solving This, You Can:
- ✔ Identify the correct trend of conductivity with temperature for different material classes.
- ✔ Calculate void space and packing efficiency for cubic unit cells.
- ✔ Analyze the impact of point defects on the physical properties of solids.
Study Tips for This Topic
For JEE Main, focus on the contrasting trends between metals and semiconductors. Often, questions ask you to identify which statement is false by combining a true fact about metals with a false fact about semiconductors (or vice versa). Memorize the packing efficiencies: BCC ($68\%$), FCC ($74\%$), SC ($52\%$) to quickly answer void space questions.
Difficulty Rating & Exam Frequency
Difficulty: Moderate (Requires understanding of concepts, not just memorization).
JEE Main Frequency: High. Concepts of Solid State and Thermodynamics of materials appear frequently.
JEE Advanced Frequency: Moderate. Often appears in the Chemistry section testing conceptual clarity.
Related Questions from Solid State
Written by Nishant Kumar Gupta
Founder of Padho Likho JEE & Senior Chemistry Educator — 12+ Years Experience, Ex-Faculty Allen/Aakash/Narayana.
Last Updated: July 2026
Question Source: JEE Main 2012 PYQ
Topic: Solid State