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dimaraw [331]
4 years ago
8

Which is true about the resistivities of a metal and a semiconductor as the temperature is increased?

Physics
2 answers:
allochka39001 [22]4 years ago
7 0

Answer:

The resistivity of metals decreases on increase in temperature whereas the resistivity of semiconductors increases with increase in temperature.

Explanation:

The majority charge carriers in metals are electrons which are explained by the electron-sea model. The electron sea model considers metal to have a sea of conducting electrons in between the nucleus arrangement.

The nucleus is bind firmly within the lattice structure and when the temperature has increased this accounts for the lattice vibrations as a result offering obstructions to the valence electrons which increases its bulk resistivity.

The variation in the resistivity of metal is given by:

\rho=\rho_0[1+\alpha_t(T-T_0)]

where:

\alpha_t=temperature coefficient of resistivity which is defined as the fractional change in resistivity per unit change in temperature.

\rm \rho\ and\ \rho_0\ are\ the\ resistivity\ at\ temperature\ T\ and\ T_0\ respectively

At temperatures 500 K above the room temperature we have the variation of resistivity of the metal in a linear manner.

  • For semiconductors, the value of coefficient of resistivity is negative which implies that the resistivity of semiconductors decreases as temperature increases.
MAVERICK [17]4 years ago
4 0

Answer:

Explanation:

According to the electrical conductivity, there are three types of materials:

Metals are the conductors which allows the electric current to flow through them. For example, iron , copper, silver, etc.

As the temperature of a metal increases the resistivity also increases.

Semiconductors are the material which are insulators at room temperature but they behaves like a conductor as the temperature increases.

The resistivity of a semiconductor decreases as the temperature increases.

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m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

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