Step 1: Understanding the Question:
The question asks about the effect of temperature on the electrical resistance and electrical conductivity of a semiconductor material.
Step 2: Key Formula or Approach:
In semiconductors, the electrical conductivity $\sigma$ is directly related to the concentration of charge carriers (electrons and holes), which increases with temperature due to thermal excitation across the energy bandgap. The relationship between conductivity and resistance $R$ is inversely proportional:
$$\sigma = \frac{1}{\rho} \propto \frac{1}{R}$$
Step 3: Detailed Explanation:
Unlike conductors, semiconductors have a negative temperature coefficient of resistance. At low temperatures, the valence band is completely full and the conduction band is nearly empty.
As the temperature increases, covalent bonds break due to thermal energy, causing more electrons to jump from the valence band to the conduction band. This drastically increases the concentration of free charge carriers.
Because more charge carriers are available to conduct current, the electrical conductivity ($\sigma$) of the semiconductor increases.
Since resistance ($R$) is inversely proportional to conductivity, the increase in conductivity means that the resistance of the semiconductor simultaneously decreases.
Therefore, as temperature increases, resistance decreases and electrical conductivity increases.
Step 4: Final Answer:
The resistance decreases and electric conductivity increases, which matches option (D).