Saturday, June 29, 2019

Question from Material Science (Summer 2019) AMIE Exam

At room temperature, the electrical conductivity of PbS is 25 (Ω.m)-1, whereas the electron and hole mobilities are 0.06 and 0.02 m2/V.s, respectively. Calculate the intrinsic carrier concentration for PbS at room temperature. (AMIE, Material Science, Summer 2019)

Solution
\[\begin{array}{l}\sigma  = 25{(\Omega m)^{ - 1}}\\{\mu _e} = 0.06{m^2}/Vs\\{\mu _h} = 0.02{m^2}/Vs\end{array}\]
In the intrinsic region
\[n = p = {n_i}\]
So, \[\sigma  = {n_i}\left| e \right|({\mu _e} + {\mu _h})\]
\[\therefore {n_i} = \frac{\sigma }{{\left| e \right|({\mu _e} + {\mu _h})}} = \frac{{25}}{{1.6x{{10}^{ - 19}}(0.06 + 0.02)}} = 1.95x{10^{21}}{m^{ - 3}}\]


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