Discussion
As per the Drude model of metals, the electrical resistance of a metallic wire of length \(L\) and cross-section area \(A\) is
(Consider \(\tau\) as the relaxation time, \(m\) as electron mass, \(n\) as carrier concentration and \(e\) as electronic charge)
Answer: A
The resistance is \(\frac{mL}{ne^{2}A\tau}\).
In the Drude model an electron accelerated by a field \(E\) gains drift velocity \(v_{d} = \frac{eE\tau}{m}\) between collisions, so the current density is
\[ \begin{aligned} J &= nev_{d} = \frac{ne^{2}\tau}{m}E \Rightarrow \sigma = \frac{ne^{2}\tau}{m} \\ R &= \frac{L}{\sigma A} = \frac{mL}{ne^{2}A\tau} \end{aligned} \]
The other options carry spurious factors of 2 or 4 that the Drude derivation does not produce: the resistance is simply the length over conductivity times area.
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