A potential difference of 1.0 V is applied across a metal wire of length 2 m. The free electrons in the wire drift with a speed of \(2.25\times10^{-3}\ \mathrm{m\,s^{-1}}\). The relaxation time of the electrons is (Take \(e/m=1.8\times10^{11}\ \mathrm{C\,kg^{-1}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.5\times10^{-14}\ \mathrm{s}\)
The relaxation time \( \tau \) is the average time between two collisions of an electron. Between collisions the field speeds the electron up, and on average this gives the drift speed \( v_d=eE\tau/m \).
First find the field inside the wire. It is the potential difference per unit length.
\( \begin{aligned} E &= \frac{V}{l} \\ &= \frac{1.0}{2} \\ &= 0.5\ \mathrm{V\,m^{-1}} \end{aligned} \)
Now take \( \tau \) from the drift speed relation.
\( \begin{aligned} \tau &= \frac{v_d}{(e/m)E} \\ &= \frac{2.25\times10^{-3}}{1.8\times10^{11}\times0.5} \\ &= 2.5\times10^{-14}\ \mathrm{s} \end{aligned} \)
NCERT (2026-27):
Summary – “If such collisions occur on an average at a time interval τ, vd = aτ = eEτ/m”
Ohm’s Law – “Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its ends is El.”