
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 600 K and that at 300 K?
Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by $n_i = n_0 \exp \left( - \dfrac{E_g}{2 k_B T} \right)$, where $n_0$ is a constant. $k_B = 8.62 \times 10^{-5} eV K^{-1}$
Answer
571.8k+ views
Hint: In intrinsic semiconductor there are two charge carriers, electrons and holes. These charge carriers are generated only because of the temperature. For intrinsic semiconductors the concentration of electrons and holes is equal.
Formula used:
The conductivity $\sigma$ can be written in terms of mobility $\mu$ as:
$\sigma = n e \mu$
e is the charge of the electron or hole and n is the concentration.
The concentration for charge carriers at a temperature T is given as:
$n_i = n_0 \exp \left( - \dfrac{E_g}{2 k_B T} \right)$
Complete answer:
We are given that the mobility of holes does not depend on temperature. Therefore the only temperature dependent factor is the carrier concentration and remaining two factors i.e., e and mobility will stay constant even if we increase the temperature.
Therefore, the ratio of two conductivities can be written as:
$\dfrac{\sigma_{600}}{\sigma_{300}} = \dfrac{n_{i(600)} e \mu}{n_{i(300)} e \mu} = \dfrac{n_0 e^{ \left( - \dfrac{E_g}{1200 k_B} \right)}}{n_0 e^{\left( - \dfrac{E_g}{600 k_B} \right) }}$
Simplification of the exponential can be done in the following manner:
First,
$\dfrac{E_g}{k_B} = \dfrac{1.2 eV}{8.62 \times 10^{-5} eV/K} = 0.139 \times 10^5 K$
Now, as
$\dfrac{e^{ma}}{e^{mb}} = e^{m(a - b)}$ ,
we will require:
$\dfrac{e^{ -(1/1200)}}{e^{-(1/600)} } = e^{\dfrac{1}{600} - \dfrac{1}{1200}} = e^{\dfrac{1}{1200}} $
Therefore, the ratio of conductivities can now be written as:
$\dfrac{\sigma_{600}}{\sigma_{300}} = \dfrac{n_{i(600)}}{n_{i(300)}} = e^{\dfrac{0.139 \times 10^5 K}{1200 K}} = e^{11.58} = 1.059 \times 10^5 $
Therefore it should be noted that as the temperature was changed from 300 K to 600k the conductivity increased by order of about 5. This is a magnificent increase. This is the reason why doping is used to create an extrinsic semiconductor where the conductivity does not increase in an uncontrolled manner.
Note:
We were given that the whole mobility stays constant with temperature and we were given more information about the electron mobility but it happens that with temperature the mobility of the electron decreases due to increase in the concentration. In the expression for conductivity, the dominant factor stays the carrier concentration has temperature is varied. The change in concentration is quite drastic as compared to change in mobility with temperature.
Formula used:
The conductivity $\sigma$ can be written in terms of mobility $\mu$ as:
$\sigma = n e \mu$
e is the charge of the electron or hole and n is the concentration.
The concentration for charge carriers at a temperature T is given as:
$n_i = n_0 \exp \left( - \dfrac{E_g}{2 k_B T} \right)$
Complete answer:
We are given that the mobility of holes does not depend on temperature. Therefore the only temperature dependent factor is the carrier concentration and remaining two factors i.e., e and mobility will stay constant even if we increase the temperature.
Therefore, the ratio of two conductivities can be written as:
$\dfrac{\sigma_{600}}{\sigma_{300}} = \dfrac{n_{i(600)} e \mu}{n_{i(300)} e \mu} = \dfrac{n_0 e^{ \left( - \dfrac{E_g}{1200 k_B} \right)}}{n_0 e^{\left( - \dfrac{E_g}{600 k_B} \right) }}$
Simplification of the exponential can be done in the following manner:
First,
$\dfrac{E_g}{k_B} = \dfrac{1.2 eV}{8.62 \times 10^{-5} eV/K} = 0.139 \times 10^5 K$
Now, as
$\dfrac{e^{ma}}{e^{mb}} = e^{m(a - b)}$ ,
we will require:
$\dfrac{e^{ -(1/1200)}}{e^{-(1/600)} } = e^{\dfrac{1}{600} - \dfrac{1}{1200}} = e^{\dfrac{1}{1200}} $
Therefore, the ratio of conductivities can now be written as:
$\dfrac{\sigma_{600}}{\sigma_{300}} = \dfrac{n_{i(600)}}{n_{i(300)}} = e^{\dfrac{0.139 \times 10^5 K}{1200 K}} = e^{11.58} = 1.059 \times 10^5 $
Therefore it should be noted that as the temperature was changed from 300 K to 600k the conductivity increased by order of about 5. This is a magnificent increase. This is the reason why doping is used to create an extrinsic semiconductor where the conductivity does not increase in an uncontrolled manner.
Note:
We were given that the whole mobility stays constant with temperature and we were given more information about the electron mobility but it happens that with temperature the mobility of the electron decreases due to increase in the concentration. In the expression for conductivity, the dominant factor stays the carrier concentration has temperature is varied. The change in concentration is quite drastic as compared to change in mobility with temperature.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

