The equation for f(E) is the Fermi-Dirac probability function and is

This question was previously asked in
UGC NET Paper 2: Electronic Science June 2019 Official Paper
View all UGC NET Papers >
  1. \(f\left( E \right) = \frac{1}{{1 + \exp \left( {\frac{{E - {E_F}}}{{{K_B}T}}} \right)}}\)
  2. \(f\left( E \right) = \frac{1}{{1 + \exp \left( {\frac{{{E_F-E}}}{{{K_B}~~T}}} \right)}}\)
  3. \(f\left( E \right) = \frac{1}{{1 - \exp \left( {\frac{{E - {E_F}}}{{{K_B}~~T}}} \right)}}\)
  4. \(f\left( E \right) = \frac{1}{{1 - \exp \left( {\frac{{{E_F-E}}}{{{K_B}~~T}}} \right)}}\)

Answer (Detailed Solution Below)

Option 1 : \(f\left( E \right) = \frac{1}{{1 + \exp \left( {\frac{{E - {E_F}}}{{{K_B}T}}} \right)}}\)
Free
UGC NET Paper 1: Held on 21st August 2024 Shift 1
15.1 K Users
50 Questions 100 Marks 60 Mins

Detailed Solution

Download Solution PDF

Fermi Level:

The probability that an electron occupies an energy level E is given by the Fermi-Dirac probability function and It can be expressed by

\({{\rm{f}}_{\rm{}}}\left( {\rm{e}} \right) = \frac{1}{{1 + \exp \left[ {\frac{{E - {E_F}}}{{KT}}} \right]}}\)

The probability function of a hole existing below Fermi level = 1 - f (e)

\(=1- \frac{1}{{1 + \exp \left[ {\frac{{E - {E_F}}}{{KT}}} \right]}}\)

\(1-{{\rm{f}}_{\rm{}}}\left( {\rm{e}} \right) = \frac{1}{{1 + \exp \left[ {\frac{{-(E - {E_F)}}}{{KT}}} \right]}}\)

26 June 1

Fermi level for n-type:

The density of electrons in a semiconductor is related to the density of available states and the probability that each of these states is occupied, i.e.

n(E) = gc(E) f(E)

gc(E) = Density of states in the conduction band

f(E) = Probability of electron at Energy Level, E

\(f\left( E \right) = \frac{1}{{1 + \exp \left( {\frac{{E - {E_F}}}{{{K_B}T}}} \right)}}\)

The above can be approximated as:

\(f\left( E \right) \cong \exp \left( { - \frac{{\left( {{E} - E_F} \right)}}{{{K}T}}} \right)\)   ---(1)

The majority carrier density in a semiconductor is obtained by integrating the product of the density of states gc(E) and the probability density function, over all possible states.

For electrons in the conduction band, the integral is taken from the conduction band, labeled EC, to the top, i.e.

\(n = \mathop \smallint \limits_{ - \infty }^{{E_v}} {g_c}\left( E \right)\left[ {f\left( E \right)} \right]dE\)

Using Equation (1), the above equation becomes:

\(n = \mathop \smallint \limits_{ - \infty }^{{E_v}} {g_c}\left( E \right) \exp \left( { - \frac{{\left( {{E} - E_F} \right)}}{{{K}T}}} \right)\)

Assuming all the density of states to be concentrated at the top of the conduction band with effective density given by Nc, the above Equation becomes:

\(n = {N_c}\exp \left[ {\frac{{ - \left( {{E_c} - {E_F}} \right)}}{{KT}}} \right]\)

Since for an n-type semiconductor, ND is the majority carriers, we can write:

\(N_D = {N_c}\exp \left[ {\frac{{ - \left( {{E_c} - {E_F}} \right)}}{{KT}}} \right]\)

\(\frac{N_D}{N_C} = \exp \left[ {\frac{{ - \left( {{E_c} - {E_F}} \right)}}{{KT}}} \right]\)

Taking natural log on both the sides, we get:

\(ln(\frac{N_D}{N_C}) = -{\frac{{\left( {{E_c} - {E_F}} \right)}}{{KT}}}\)

\(KT~ln(\frac{N_C}{N_D}) = E_C-E_F\)

\(E_F=E_C-KT~ln(\frac{N_C}{N_D})\)

Similarly for p-type:

\(E_F=E_V+KT~ln(\frac{N_V}{N_A})\)

Latest UGC NET Updates

Last updated on Jun 22, 2025

-> The UGC Net Admit Card has been released on its official website today.

-> The UGC NET June 2025 exam will be conducted from 25th to 29th June 2025.

-> The UGC-NET exam takes place for 85 subjects, to determine the eligibility for 'Junior Research Fellowship’ and ‘Assistant Professor’ posts, as well as for PhD. admissions.

-> The exam is conducted bi-annually - in June and December cycles.

-> The exam comprises two papers - Paper I and Paper II. Paper I consists of 50 questions and Paper II consists of 100 questions. 

-> The candidates who are preparing for the exam can check the UGC NET Previous Year Papers and UGC NET Test Series to boost their preparations.

Get Free Access Now
Hot Links: teen patti sequence teen patti rummy 51 bonus real cash teen patti teen patti master purana teen patti master old version