Match List-I with List-II
(Symbols have their usual meaning)
| List-I | List-II |
| (A) Drift velocity | (I) \( \displaystyle\frac{m}{ne^{2}\rho} \) |
| (B) Electrical Resistivity | (II) \( nev_d \) |
| (C) Relaxation time | (III) \( \displaystyle\frac{eE\tau}{m} \) |
| (D) Current density | (IV) \( \displaystyle\frac{E}{J} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
In the free-electron (Drude) model, an electric field E gives each electron an acceleration eE/m, and on average an electron travels for the relaxation time \( \tau \) between collisions.
(A) Drift velocity — this matches (III) \( eE\tau/m \). The average velocity gained in time \( \tau \) is the drift velocity.
\( \displaystyle v_d = a\tau = \frac{eE\tau}{m} \)
(B) Electrical Resistivity — this matches (IV) \( E/J \). Ohm’s law in microscopic form is \( J = \sigma E \), and resistivity is \( \rho = 1/\sigma \).
\( \displaystyle \rho = \frac{E}{J} \)
(C) Relaxation time — this matches (I) \( m/ne^{2}\rho \). Combining the results for J and \( v_d \) gives the resistivity in terms of \( \tau \), which is then rearranged.
\( \displaystyle \rho = \frac{m}{ne^{2}\tau} \Rightarrow \tau = \frac{m}{ne^{2}\rho} \)
(D) Current density — this matches (II) \( nev_d \). The current through area A is \( I = neAv_d \), so the current per unit area is
\( \displaystyle J = \frac{I}{A} = nev_d \)