Question

A semiconductor having electron and linear mobilities ${\mu _n}$ and ${\mu _p}$ respectively.
If its intrinsic carrier density is ${n_i}$ , then what will be the value of hole concentration $P$ for which the conductivity will be maximum at a given temperature?
(A) ${n_i}\sqrt {\dfrac{{{\mu _n}}}{{{\mu _p}}}}$
(B) ${n_h}\sqrt {\dfrac{{{\mu _n}}}{{{\mu _p}}}}$
(C) ${n_i}\sqrt {\dfrac{{{\mu _p}}}{{{\mu _n}}}}$
(D) ${n_h}\sqrt {\dfrac{{{\mu _p}}}{{{\mu _n}}}}$

Answer 1

A semiconductor material has an electrical conductivity which lies between that of a conductor and an insulator. Its resistivity falls as its temperature rises; metals are the opposite Semiconductors are employed in the manufacture of various kinds of electronic devices, including diodes, transistors, and integrated circuits.Such gadgets have discovered wide application in light of their minimization, dependability, power effectiveness, and ease.

Formula used:
$\sigma = {n_e}e{\mu _n} + {n_p}e{\mu _p}$

Complete step by step solution:
We know that the Total conductivity of semiconductor is given by
$\sigma = {n_e}e{\mu _n} + {n_p}e{\mu _p}$
This shows that the conductivity depends upon electrons and hole concentration and their mobilities.
In this equation $e$ is common after taking it common the equation becomes
$\sigma = e\left( {{n_e}{\mu _n} + {n_p}{\mu _p}} \right)$
We know that for an intrinsic semiconductor
${n_e}{n_p} = {n^2}$
Or it can also be written as
${n_e} = \dfrac{{{n^2}}}{{{n_p}}}$
Now we have to substitute the value of ${n_e}$in the total conductivity equation
Hence $\sigma = e\left[ {\dfrac{{n_i^2}}{{{n_p}}}{\mu _n} + {n_p}{\mu _p}} \right]......\left( i \right)$
If $\sigma$is minimum then the value of $\dfrac{{d\sigma }}{{d{n_p}}}$is $0$
Differentiating (i) with respect to${n_p}$, we get
$\dfrac{{d\sigma }}{{d{n_p}}} = e\left[ { - \dfrac{{n_i^2}}{{n_p^2}}{\mu _n} + {\mu _p}} \right] = 0$
Or the equation can be written as
${\mu _p} = \dfrac{{n_i^2}}{{n_p^2}}\mu {}_n$
Therefore the above equation can be written as
$n_p^2 = n_i^2\dfrac{{\mu {}_n}}{{{\mu _p}}}$
Then the equation can be written as
${n_p} = \sqrt {n_i^2} \times \sqrt {\dfrac{{\mu {}_n}}{{{\mu _p}}}}$
Here square and square root get cancel then the equation become
${n_p} = {n_i}\sqrt {\dfrac{{\mu {}_n}}{{{\mu _p}}}}$

Hence the correct answer is option (A)

Note In strong state material science, the electron portability portrays how rapidly an electron can travel through a metal or semiconductor when pulled by an electric field. Hole concentration refers to the free electrons and holes. They carry the charges (electron negative and hole positive), and are responsible for electrical current in the semiconductor. Concentration of electrons. (=$n$) and hole (=$p$) is measured in the unit of $cm$ . The electron mobility is greater than the whole mobility.