The temperature coefficient of electrical resistivity ($ \alpha $) for pure semiconductors is: · Electronics MDCAT Past Paper MCQ

Physics Electronics PMDC Conceptual Practice

PMDC Verified Question 327 of 494

The temperature coefficient of electrical resistivity ($ \alpha $) for pure semiconductors is:

Negative (resistivity decreases as temperature rises: $ \alpha < 0 $)

Positive (resistivity increases as temperature rises: $ \alpha > 0 $)

Strictly zero at all temperatures

Infinite at room temperature

Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:

Option A: Negative (resistivity decreases as temperature rises: $ \alpha < 0 $)

Concept:

Because higher temperatures break covalent bonds and exponentially increase free carrier concentration, semiconductor resistivity decreases with temperature, giving a negative temperature coefficient ($ \alpha < 0 $).

Formula:

$$\rho(T) = \rho_0 \exp\left(\frac{E_g}{2 k T}\right) \implies \alpha = \frac{1}{\rho} \frac{d\rho}{dT} = -\frac{E_g}{2 k T^2} < 0$$

Solution:

In metallic conductors, increased temperature causes lattice vibrations that increase resistance ($ \alpha > 0 $).

In semiconductors, thermal generation of new electron-hole pairs outweighs scattering effects, lowering resistance ($ \alpha < 0 $).

Why other options are incorrect:

Option B: A positive temperature coefficient ($ \alpha > 0 $) is characteristic of metallic conductors.
Option C: Zero temperature coefficient is approximated by specialized precision alloys like Manganin.
Option D: Semiconductor resistivity is finite at room temperature.

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