A potential difference of 0.5 V is applied across a metal wire 5 m long. Take \( e/m=1.76\times10^{11}\ \mathrm{C\,kg^{-1}} \) for an electron. Between two collisions, the size of the acceleration of a free electron in the wire is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.76\times10^{10}\ \mathrm{m\,s^{-2}}\)
The field in the wire pulls each free electron with a force \( eE \). Between two collisions, this force gives the electron an acceleration of size \( eE/m \). For a uniform field in a wire of length \( l \), the field is \( E=V/l \).
\( \begin{aligned} E &= \frac{V}{l} \\ &= \frac{0.5}{5} \\ &= 0.1\ \mathrm{V\,m^{-1}} \end{aligned} \)
\( \begin{aligned} a &= \frac{e}{m}E \\ &= 1.76\times10^{11}\times0.1 \\ &= 1.76\times10^{10}\ \mathrm{m\,s^{-2}} \end{aligned} \)
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “Consider now the situation when an electric field is present. Electrons will be accelerated due to this field by”
\( \displaystyle a=\frac{-eE}{m} \)\( \qquad (3.15) \)
“where –e is the charge and m is the mass of an electron.”
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. …”