Match List-I with List-II
| List-I | List-II |
| Physical Quantity | Formula |
| (A) \( j \) (current density) | (I) \( \frac{e\tau}{m} \) |
| (B) \( v_d \) (drift velocity) | (II) \( \frac{ne^{2}}{m}\tau \) |
| (C) \( \sigma \) (conductivity) | (III) \( \frac{e}{m}\tau E \) |
| (D) \( \mu \) (mobility) | (IV) \( \frac{ne^{2}}{m}\tau E \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
All four quantities come from the drift of free electrons: in a field \( E \) an electron accelerates at \( eE/m \) for an average relaxation time \( \tau \) between collisions.
(A) \( j \) (current density) — this matches (IV) \( \frac{ne^{2}}{m}\tau E \). The current density is the charge carried per second per unit area.
\( \displaystyle j = n e v_d = n e \cdot \frac{e\tau}{m}E = \frac{ne^{2}}{m}\tau E \)
(B) \( v_d \) (drift velocity) — this matches (III) \( \frac{e}{m}\tau E \). It is the acceleration \( eE/m \) times the relaxation time \( \tau \).
(C) \( \sigma \) (conductivity) — this matches (II) \( \frac{ne^{2}}{m}\tau \). From Ohm's law in the form \( j = \sigma E \), the conductivity is the current density per unit field.
(D) \( \mu \) (mobility) — this matches (I) \( \frac{e\tau}{m} \). Mobility is the drift velocity per unit field, \( \mu = v_d/E \).
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… By definition I is related to the magnitude |j| of the current density by
\( \displaystyle I = |\mathbf{j}|A \qquad (3.20) \)
Hence, from Eqs.(3.19) and (3.20),
\( \displaystyle |\mathbf{j}| = \frac{ne^{2}}{m}\,\tau\,|\mathbf{E}| \qquad (3.21) \)”
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
\( \displaystyle \sigma = \frac{ne^{2}\tau}{m} \qquad (3.23) \)”
Mobility – “… An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field:
\( \displaystyle \mu = \frac{|\mathbf{v}_d|}{E} \qquad (3.24) \)
… From Eq. (3.17), we have
\( \displaystyle v_d = \frac{e\tau E}{m} \)
Hence,
\( \displaystyle \mu = \frac{v_d}{E} = \frac{e\tau}{m} \qquad (3.25) \)”