Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q. 14.9 In an intrinsic semiconductor the energy gap $E_{g}$ is $1.2\; eV.$ Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600K$ and that at $300K$ Assume that the temperature dependence of intrinsic carrier concentration $n_{i}$ is given by

$n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$

Where, $n_{0}$ is constant.

Answers (1)

best_answer

Energy gap of given intrinsic semiconductor = E_g = 1.2eV

temperature dependence of intrinsic carrier concentration $n_{i}$ is given by

$n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$

Where is constant, $K_B$ is Boltzmann constant = $8.862 * 10^{-5}eV/K$ ,

T is temperature

Initial temperature = T1 = 300K

the intrinsic carrier concentration at this temperature :

$n_{i1} = n_0exp[\frac{-E_g}{2K_B*300}]$

Final temperature = T2 = 600K

the intrinsic carrier concentration at this temperature :

$n_{i2} = n_0exp[\frac{-E_g}{2K_B*600}]$

the ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures

$\frac{n_{i2}}{n_{i2}} = \frac{n_0exp[\frac{-E_g}{2K_B*600}]}{n_0exp[\frac{-E_g}{2K_B*300}]}$

$= exp\frac{E_g}{2K_B}[\frac{1}{300}-\frac{1}{600}]=exp[\frac{1.2}{2*8.62*10^{-5}}* \frac{2-1}{600}]$

$= exp[11.6] = 1.09 * 10^{5}$

Therefore the ratio between the conductivities is $1.09 * 10^{5}$ .

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question