Question

Temperature coefficient of resistivity ($\delta$) is given as $\alpha = \frac{1}{\delta}\frac{d\delta}{dT}$ where $\delta$ is resistivity of material at temperature $T$. Choose the correct option(s):

• A. $\delta = \delta_0 e^{\alpha(T-T_0)}$; $\delta_0$ is resistivity at temperature $T_0$.
• B. $\delta = \delta_0(1+\alpha(T-T_0))$ if $\alpha(T-T_0)<<1$.
• C. $\delta = \sqrt{\delta_0(1+\alpha(T-T_0))}$ if $\alpha(T-T_0)<<1$.
• D. With increase in temperature, number density of conduction electrons and average relaxation time decrease in a conductor and so its resistivity increases.

Answer 1

Answer: (A), (B)

(A) and (B)

Answer 2

The definition of the temperature coefficient of resistivity is $\alpha = \frac{1}{\delta}\frac{d\delta}{dT}$. Rearranging this gives $\frac{d\delta}{\delta} = \alpha dT$. Integrating both sides from $(T_0, \delta_0)$ to $(T, \delta)$ and assuming $\alpha$ is constant, we get $\int_{\delta_0}^{\delta} \frac{d\delta'}{\delta'} = \int_{T_0}^{T} \alpha dT'$, which leads to $\ln(\frac{\delta}{\delta_0}) = \alpha(T-T_0)$, and thus $\delta = \delta_0 e^{\alpha(T-T_0)}$. This confirms option (A).

For the condition $\alpha(T-T_0)<<1$, we can use the Taylor expansion of $e^x \approx 1+x$ for small $x$. Substituting $x = \alpha(T-T_0)$, we get $\delta \approx \delta_0(1+\alpha(T-T_0))$, confirming option (B).

Option (C) is incorrect because the derived relationship does not match the form $\sqrt{\delta_0(1+\alpha(T-T_0))}$.

Option (D) is incorrect for conductors. While resistivity increases with temperature due to a decrease in average relaxation time ($\tau$), the number density of conduction electrons ($n$) in conductors remains nearly constant with temperature. Resistivity is proportional to $1/\tau$. Thus, the statement that the number density of conduction electrons decreases is false.