A metal has \( 8\times10^{28} \) free electrons per cubic metre. The mobility of these electrons is \( 50\ \mathrm{cm^{2}\,V^{-1}\,s^{-1}} \). Take \( e=1.6\times10^{-19}\ \mathrm{C} \). The conductivity of the metal is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(6.4\times10^{7}\ \mathrm{S\,m^{-1}}\)
Conductivity is \( \sigma=ne^{2}\tau/m \) and mobility is \( \mu=e\tau/m \). Putting the second into the first gives \( \sigma=ne\mu \).
\( \begin{aligned} \sigma &= \frac{ne^{2}\tau}{m} \\ &= ne\left(\frac{e\tau}{m}\right) \\ &= ne\mu \end{aligned} \)
The mobility must first be changed to SI units. One SI unit of mobility is \( 10^{4} \) times the practical unit \( \mathrm{cm^{2}\,V^{-1}\,s^{-1}} \).
\( \begin{aligned} \mu &= 50\times10^{-4}\ \mathrm{m^{2}\,V^{-1}\,s^{-1}} \\ &= 5\times10^{-3}\ \mathrm{m^{2}\,V^{-1}\,s^{-1}} \end{aligned} \)
\( \begin{aligned} \sigma &= ne\mu \\ &= 8\times10^{28}\times1.6\times10^{-19}\times5\times10^{-3} \\ &= 6.4\times10^{7}\ \mathrm{S\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… if we identify the conductivity σ as”
\( \displaystyle \sigma=\frac{ne^{2}}{m}\tau \)\( \qquad (3.23) \)
Mobility – “The SI unit of mobility is m2/Vs and is 104 of the mobility in practical units (cm2/Vs). Mobility is positive. … Hence,”
\( \displaystyle \mu=\frac{v_d}{E}=\frac{e\tau}{m} \)\( \qquad (3.25) \)