An intrinsic semiconductor has a resistivity of 0.50 `Omega` m at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are `0.39 m^2 V^(-1) s^(-1)` and `0.11 m^2 V^(-1) s^(-1)` respectively.

Options

A. `1.2xx10^18 m^(-3)`
B. `2.5xx10^19 m^(-3)`
C. `1.9xx10^20 m^(-3)`
D. `3.1xx10^21 m^(-3)`

Explanation

To find the intrinsic carrier concentration (ni) of the semiconductor, we can use the formula for the resistivity (ρ) of an intrinsic semiconductor:

\[ \rho = \frac{1}{q(n \mu_e + p \mu_h)} \]

In an intrinsic semiconductor, the concentration of electrons (n) is equal to the concentration of holes (p), which is equal to the intrinsic carrier concentration (ni). Thus, we can rewrite the equation as:

\[ \rho = \frac{1}{q(n_i \mu_e + n_i \mu_h)} = \frac{1}{q n_i (\mu_e + \mu_h)} \]

Where:
- ρ is the resistivity (0.50 Ω·m)
- q is the charge of an electron (approximately \(1.6 \times 10^{-19} C\))
- \(n_i\) is the intrinsic carrier concentration
- \(\mu_e\) is the mobility of electrons (0.39 m²/V·s)
- \(\mu_h\) is the mobility of holes (0.11 m²/V·s)

Now, we can rearrange the formula to solve for \(n_i\):

\[ n_i = \frac{1}{\rho q (\mu_e + \mu_h)} \]

Now, let's substitute the known values into the equation:

1. Calculate \(\mu_e + \mu_h\):
\[ \mu_e + \mu_h = 0.39 + 0.11 = 0.50 \, m^2/V·s \]

2. Substitute the values into the equation for \(n_i\):
\[ n_i = \frac{1}{0.50 \times (1.6 \times 10^{-19}) \times 0.50} \]

3. Calculate the denominator:
\[ 0.50 \times (1.6 \times 10^{-19}) \times 0.50 = 0.50 \times 0.8 \times 10^{-19} = 0.4 \times 10^{-19} \]

4. Now calculate \(n_i\):
\[ n_i = \frac{1}{0.4 \times 10^{-19}} = 2.5 \times 10^{18} \, m^{-3} \]

Thus, the intrinsic carrier concentration \(n_i\) is approximately \(2.5 \times 10^{18} \, m^{-3}\).