Semiconductor Electronics Materials Devices and Simple Circuits - Result Question 18
18. Pure $Si$ at $500 K$ has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 \times 10^{16} m^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} m^{-3}$. The doped semiconductor is of
[2011M]
(a) n-type with electron concentration
$ n_e=5 \times 10^{22} m^{-3} $
(b) p-type with electron concentration
$ n_e=2.5 \times 10^{10} m^{-3} $
(c) n-type with electron concentration
$ n_e=2.5 \times 10^{23} m^{-3} $
(d) p-type having electron concentration
$ n_e=5 \times 10^{9} m^{-3} $
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Answer:
Correct Answer: 18. (d)
Solution:
- (d) $n_i{ }^{2}=n_e n_h$
$ \begin{aligned} & (1.5 \times 10^{16})^{2}=n_e(4.5 \times 10^{22}) \\ & \Rightarrow \quad n_e=0.5 \times 10^{10} \\ & \text{ or } \quad n_e=5 \times 10^{9} \\ & \text{ Given } \quad n_h=4.5 \times 10^{22} \\ & \Rightarrow n_h»n_e \end{aligned} $
$\therefore$ Semiconductor is p-type and
$n_e=5 \times 10^{9} m^{-3}$.
