Match List-I with List-II
| List-I | List-II |
| Formula | Physical quantity |
| (A) \( neAv_d \) | (I) Mobility |
| (B) \( \displaystyle\frac{v_d}{E} \) | (II) Current |
| (C) \( \displaystyle\frac{ne^{2}\tau}{m} \) | (III) Conductivity |
| (D) \( \displaystyle\frac{m}{ne^{2}\tau} \) | (IV) Resistivity |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
These are standard results of current electricity in terms of the electron density \( n \), the charge \( e \), the drift velocity \( v_d \) and the relaxation time \( \tau \).
(A) \( neAv_d \) — this matches (II) Current. The current through a conductor of area \( A \) is \( I = neAv_d \).
(B) \( \displaystyle\frac{v_d}{E} \) — this matches (I) Mobility. Mobility is the drift velocity per unit electric field, \( \mu = v_d/E \).
(C) \( \displaystyle\frac{ne^{2}\tau}{m} \) — this matches (III) Conductivity. The conductivity is \( \sigma = ne^{2}\tau/m \).
(D) \( \displaystyle\frac{m}{ne^{2}\tau} \) — this matches (IV) Resistivity. The resistivity is the reciprocal of the conductivity, \( \rho = 1/\sigma = m/(ne^{2}\tau) \).
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… The amount of charge crossing the area A in time Δt is by definition [Eq. (3.2)] I Δt, where I is the magnitude of the current. Hence,
\( \displaystyle I\,\Delta t = +n\,e\,A\,|v_d|\,\Delta t \qquad (3.18) \)”
Drift of Electrons and the Origin of Resistivity – “… Comparison with Eq. (3.13) shows that Eq. (3.22) is exactly the Ohm’s law, if we identify the conductivity σ as
\( \displaystyle \sigma = \frac{n e^2 \tau}{m} \qquad (3.23) \)”
Mobility – “… An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field:
\( \displaystyle \mu = \frac{|v_d|}{E} \qquad (3.24) \)”
Temperature Dependence of Resistivity – “… From this equation, resistivity of a material is given by
\( \displaystyle \rho = \frac{1}{\sigma} = \frac{m}{n e^{2} \tau} \qquad (3.27) \)”