Match List-I with List-II (The physical quantities pertain to a metallic conductor in an external electric field. Symbols in the formulae have their usual meaning)
| LIST-I Physical Quantity | LIST-II Formula |
| A. E (Electric Field) | I. \( \displaystyle\frac{eE\tau}{m} \) |
| B. vd (Drift Velocity) | II. \( \displaystyle\frac{V}{l} \) |
| C. I (Current) | III. \( neAv_d \) |
| D. ρ (Resistivity) | IV. \( \displaystyle\frac{m}{ne^{2}\tau} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A - II, B - I, C - III, D - IV
In the free-electron model of a metal, the field makes the electrons drift, and collisions limit their average velocity; \( \tau \) is the relaxation time (average time between collisions).
(A) E (Electric Field) — this matches (II) \( V/l \). A potential difference V across a conductor of length l sets up a uniform field along it.
(B) vd (Drift Velocity) — this matches (I) \( eE\tau/m \). Each electron has acceleration eE/m, and on average it gains velocity for a time \( \tau \) before a collision randomises its motion.
\( \displaystyle v_d = a\tau = \frac{eE}{m}\,\tau \)
(C) I (Current) — this matches (III) \( neAv_d \). In time \( \Delta t \) the electrons in a length \( v_d\Delta t \) cross the section, so the charge crossing is \( ne(Av_d\Delta t) \).
(D) ρ (Resistivity) — this matches (IV) \( m/ne^{2}\tau \). Putting (B) into (C) gives the current density in terms of E; the factor multiplying E is the conductivity, and the resistivity is its reciprocal.
\( \displaystyle j = \frac{I}{A} = ne\left(\frac{eE\tau}{m}\right) = \frac{ne^{2}\tau}{m}\,E \)
\( \displaystyle \rho = \frac{1}{\sigma} = \frac{m}{ne^{2}\tau} \)
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… This is the phenomenon of drift and the velocity \( \mathbf{v}_d \) in Eq. (3.17) is called the drift velocity.
\( \displaystyle \mathbf{v}_d = -\frac{e\mathbf{E}}{m}\,\tau \qquad (3.17) \)”
Summary – “… If j is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is \( nev_dA \).”
Summary – “… 8. Using \( E = V/l \), \( I = nev_dA \), and Ohm’s law, one obtains
\( \displaystyle \frac{eE}{m} = \rho\,\frac{ne^{2}}{m}\,v_d \)”
Temperature Dependence of Resistivity – “… From this equation, resistivity of a material is given by
\( \displaystyle \rho = \frac{1}{\sigma} = \frac{m}{n e^{2} \tau} \qquad (3.27) \)”