Question

Match the LIST-I with LIST-II 

LIST-I	LIST-II
A. Mobility of electrons (\(\mu\))	I. \( Ne^2\tau/m \)
B. Drift velocity of electrons (\(v_d\))	II. \( \mu E \)
C. Electrical conductivity of conduction electrons (\(\sigma\))	III. \( \mu m/e \)
D. Relaxation time of electrons (\(\tau\))	IV. \( 1/\rho ne \)

Choose the correct answer from the options given below:

• A. A - I, B - II, C - IV, D - III
• B. A - I, B - III, C - II, D - IV
• C. A - IV, B - II, C - I, D - III
• D. A - III, B - IV, C - I, D - II

Hint:

Start with the most fundamental definitions you remember, like \(v_d = \mu E\) and \(\sigma = ne\mu\). Use these to derive the other relationships. Be prepared for potential typos (like N for n) in exam questions and use the process of elimination to find the best-fitting answer.

Answer 1

The Correct Option is C

Step 1: Concept Overview:
The question requires matching electron transport parameters in a conductor with their corresponding formulas. We will derive the correct relationships and identify the best matches, accounting for potential typos in the provided options.
Step 2: Detailed Formula Explanations:


B. Electron Drift Velocity (\(v_d\)): Drift velocity is the average velocity of charge carriers due to an electric field (E). It is proportional to the electric field, with mobility (\(\mu\)) as the proportionality constant: \( v_d = \mu E \). Therefore, B matches II.
A. Electron Mobility (\(\mu\)): Mobility indicates how quickly an electron moves through a material. The Drude model defines it as \( \mu = \frac{e\tau}{m} \), where \(\tau\) is the relaxation time. No LIST-II option directly matches this initially.
C. Electrical Conductivity (\(\sigma\)): Conductivity measures a material's ability to conduct current and is given by \( \sigma = ne\mu \), where n is the electron density. Substituting for \(\mu\), we have \( \sigma = \frac{ne^2\tau}{m} \). This matches option I, assuming 'N' is a typo for 'n'. Thus, C matches I.
D. Electron Relaxation Time (\(\tau\)): This is the average time between electron collisions. Rearranging the mobility formula, \( \tau = \frac{\mu m}{e} \), which perfectly matches option III. Thus, D matches III.

Revisiting A. Mobility (\(\mu\)), option IV gives \(1/(\rho ne)\), where \(\rho\) is resistivity. Since conductivity \(\sigma = 1/\rho\), this becomes \( \sigma / (ne) \). Given \( \sigma = ne\mu \), then \( \sigma/(ne) = \mu \). Therefore, A matches IV.
Step 3: Final Matching:
The correct pairings are:

A \(\rightarrow\) IV
B \(\rightarrow\) II
C \(\rightarrow\) I
D \(\rightarrow\) III
This sequence corresponds to option (C).