Properties of Semiconductors: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Properties of Semiconductors. Understanding the underlying formula is key to arriving at the correct answer below:

The resistivity (ρ) of a pure (intrinsic) semiconductor is given by the formula:

$$\rho = \frac{1}{q (n_i) (\mu_n + \mu_p)}$$

ρ=q(ni​)(μn​+μp​)1​

Where:

• 
  $\rho = 0.5 \, \Omega\text{m}$ 

  $\rho =0.5\Omega \text{m}$ (resistivity),

• 
  $q = 1.6 \times 10^{-19} \, \text{C}$ 

  q=1.6×10−19C (charge of an electron),

• 
  $n_i$ 

  ni​ = intrinsic carrier concentration (to be calculated),

• 
  $\mu_n = 0.39 \, \text{m}^2/\text{V-s}$ 

  μn​=0.39m2/V-s (electron mobility),

• 
  $\mu_p = 0.19 \, \text{m}^2/\text{V-s}$ 

  μp​=0.19m2/V-s (hole mobility).

Rearranging the formula to solve for

$n_i$

ni​:

$$n_i = \frac{1}{q \rho (\mu_n + \mu_p)}$$

​

Substitute the values:

$$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.39 + 0.19)}$$

​

$$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.58)}$$

$$n_i=\frac{1}{4.64\times 10^{-20}}=2.16\times 10^{19}{\text{m}}^{-3}$$

Thus, the intrinsic carrier concentration

$n_i$

ni​ is

$2.16 \times 10^{19} \, \text{m}^{-3}$