On applying an electric field of intensity 10 V/cm across a semiconductor at a certain temperature the average drift velocity of free electrons is measured to be 70 m/s. Then the electron mobility is

This question was previously asked in
ESE Electronics 2011 Paper 1: Official Paper
View all UPSC IES Papers >
  1. 7 × 104 cm2/Vs
  2. 700 cm2/Vs
  3. 7 cm2/Vs
  4. 700 cm/Vs

Answer (Detailed Solution Below)

Option 2 : 700 cm2/Vs
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
20 Qs. 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

Concept:

Mobility: It denotes how fast is the charge carrier is moving from one place to another.

It is demoted by μ.

Mobility is also defined as:

 cm2/Vs     ------(1)

Where,

Vd = Drift velocity

E =  field intensity

Calculation:

E = 10 V/cm

Vd = 70 m/s = 7000 cm/s

From equation (1):

μ = 700 cm2/Vs

Note: 

Electron mobility is always greater than hole mobility, 

Hence electron can travel faster and contributes more current than a hole.

Latest UPSC IES Updates

Last updated on Jul 2, 2025

-> ESE Mains 2025 exam date has been released. As per the schedule, UPSC IES Mains exam 2025 will be conducted on August 10. 

-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Carrier Transport Questions

Hot Links: teen patti rummy teen patti chart teen patti stars teen patti master king