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Q. Given that the mobility of electrons in Ge is $0.4\, m ^{2}\, V ^{-1}\, s ^{-1}$ and electronic charge is $1.6 \times 10^{-19} C$. The number of donor atom $(\text{per}\, m ^{3})$ semiconductor of conductivity $500\, mho / m$ is

AIIMSAIIMS 2013Semiconductor Electronics: Materials Devices and Simple Circuits

Solution:

Using, $\sigma = n_{e}q\mu_{e}$
Here, $\sigma = 500\, mho/m$
e = 1.6 $\times 10^{-19} C$
$\mu_{e}=0.4\, m^{2}\, V^{ -1} s^{ -1}$
$\therefore n_{e}=\frac{500}{1.6\times10^{-19}\times0.4}$
$= 7.8 \times 10^{21} \simeq 8 \times 10^{21} m^{-3}$