Question:

Metallic conductors and semiconductors are heated separately. What are the changes in conductivity?

Updated On: Apr 16, 2025
  • Increase, Increase
  • Decrease, Decrease
  • Increase, Decrease
  • Decrease, Increase
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The Correct Option is D

Approach Solution - 1

When metallic conductors and semiconductors are heated, their conductivity changes differently due to their distinct electronic structures and conduction mechanisms.

Metallic Conductors: 

In metallic conductors, conductivity is primarily determined by the density of free electrons and their mobility. When the temperature increases, the atoms in the metal lattice vibrate more vigorously. These increased vibrations cause more frequent collisions between the free electrons and the lattice ions. This effectively reduces the mean free path and, consequently, the mobility of the electrons. Since conductivity (\(\sigma\)) is directly proportional to the number density of free electrons (\(n\)), the elementary charge (\(e\)), and the electron mobility (\(\mu\)), and inversely proportional to the relaxation time (\(\tau\)), we can represent conductivity as:

\(\sigma = n e \mu \) or \(\sigma \propto \frac{1}{\tau}\)

As temperature increases, \(\tau\) decreases due to increased collisions, leading to a decrease in conductivity. The number density of free electrons in metals remains relatively constant with temperature changes.

Semiconductors:

In semiconductors, conductivity depends strongly on the number of charge carriers (electrons and holes) available for conduction. At low temperatures, the number of charge carriers is relatively small. However, as the temperature increases, more electrons gain enough thermal energy to jump from the valence band to the conduction band, creating more free electrons and holes. This process is known as intrinsic carrier generation.

The conductivity of a semiconductor increases exponentially with temperature. The relationship can be approximated by:

\(\sigma = \sigma_0 e^{-\frac{E_g}{2kT}}\)

Where:

\(\sigma_0\) is a constant,

\(E_g\) is the band gap energy,

\(k\) is the Boltzmann constant, and

\(T\) is the absolute temperature.

As temperature increases, the exponential term increases significantly, leading to a substantial increase in conductivity. While the mobility of carriers might decrease slightly with temperature, the increase in carrier concentration dominates the overall conductivity behavior.

Conclusion:

Therefore, when metallic conductors are heated, their conductivity decreases, and when semiconductors are heated, their conductivity increases.

Correct Answer: Decrease, Increase

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Approach Solution -2

Correct Answer:

Option 4: Decrease, Increase

Explanation:

Metallic Conductors:

In metals, conductivity is due to the movement of free electrons. As temperature increases, thermal vibrations of the lattice increase, scattering electrons and decreasing conductivity.

Semiconductors:

In semiconductors, conductivity is due to the movement of electrons and holes. As temperature increases, more electrons gain energy to move to the conduction band, increasing the number of charge carriers and thus increasing conductivity.

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Approach Solution -3

Conductivity in materials depends on the availability of free charge carriers and their mobility.

  1. For metallic conductors: Conductivity decreases with an increase in temperature. This is because as temperature rises, the lattice vibrations in the metal increase, leading to more frequent collisions between electrons and the lattice ions. This reduces the mobility of electrons, thereby decreasing conductivity.
  2. For semiconductors: Conductivity increases with an increase in temperature. In semiconductors, thermal energy excites more electrons from the valence band to the conduction band, increasing the number of free charge carriers (electrons and holes). This leads to higher conductivity at elevated temperatures.

Thus, the correct answer is that the conductivity of metals decreases while the conductivity of semiconductors increases when heated.

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