Question

A semiconductor has the electron concentration 0.45 X 10^12 m^-3and hole concentration 5 x 10^20m^-3
Find its conductivity. Given: electron mobility=0.135 m^2 v-1s^-1 and hole mobility =0.048 m^2V^-1s-1;e=1.6x10 coulomb
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Answer 1

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Answer 2

Answer:

The conductivity of the semiconductor is 3.85Ω⁻¹m⁻¹.

Explanation:

From the question, we have the following data,

The electron concentration($n_e$)=0.45×10¹²

The hole concentration($n_h$)=5×10²⁰

The mobility of electron($\mu_e$)=0.135m²V⁻¹s⁻¹

The mobility of hole ($\mu_h$)=0.048m²V⁻¹s⁻¹

The charge on an electron(e)=1.6×10⁻¹⁹C

Here we can see that the concentration of the hole is more than that of the electron. So, the conductivity(σ) will be due to holes. Which is calculated as,

$\sigma=en_h\mu_h$    (1)

When all the values are entered into equation (1), we obtain;

$\sigma=1.6\times 10^-^1^9\times 5\times 10^2^0\times 0.048$

$\sigma=3.85\ \Omega^-^1m^-^1$

So, the conductivity of the semiconductor is 3.85Ω⁻¹m⁻¹.

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