A potential difference of 5 V is applied across a conductor of length 0.1 m. If the drift velocity of the electrons is \( 2.5\times10^{-4}\,\mathrm{m\,s^{-1}} \), calculate the electron mobility.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5\times10^{-6}\,\mathrm{m^2\,V^{-1}\,s^{-1}} \)
The mobility of a charge carrier is the magnitude of its drift velocity per unit electric field, \( \mu=|v_d|/E \). For a uniform conductor the field inside is the potential difference divided by the length.
The electric field in the conductor is:
\( \begin{aligned} E &= \frac{V}{l} \\ &= \frac{5\,\mathrm{V}}{0.1\,\mathrm{m}} \\ &= 50\,\mathrm{V\,m^{-1}} \end{aligned} \)
The electron mobility is:
\( \begin{aligned} \mu &= \frac{|v_d|}{E} \\ &= \frac{2.5\times10^{-4}\,\mathrm{m\,s^{-1}}}{50\,\mathrm{V\,m^{-1}}} \\ &= 5\times10^{-6}\,\mathrm{m^2\,V^{-1}\,s^{-1}} \end{aligned} \)
NCERT (2026-27):
Mobility – “An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field:
\( \displaystyle \mu=\frac{|\mathbf{v}_d|}{E} \qquad (3.24) \)”