The conductivity of a semiconductor sample having electron concentration of $$5 \times 10^{18}$$ electrons m$$^{-3}$$, hole concentration of $$5 \times 10^{19}$$ holes m$$^{-3}$$, electron mobility of 2.0 m$$^2$$ V$$^{-1}$$ s$$^{-1}$$ and hole mobility of 0.01 m$$^2$$ V$$^{-1}$$ s$$^{-1}$$ is: (Take charge of an electron as $$1.6 \times 10^{-19}$$ C)

For a semiconductor the electrical conductivity $$\sigma$$ is given by the well-known relation

$$\sigma = q\,(n\mu_n + p\mu_p),$$

where

$$q = 1.6 \times 10^{-19}\ \text{C}$$ is the electronic charge,

$$n = 5 \times 10^{18}\ \text{m}^{-3}$$ is the electron (negative charge carrier) concentration,

$$p = 5 \times 10^{19}\ \text{m}^{-3}$$ is the hole (positive charge carrier) concentration,

$$\mu_n = 2.0\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the electron mobility,

$$\mu_p = 0.01\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the hole mobility.

First we calculate the contribution of electrons to conductivity:

$$n\mu_n = (5 \times 10^{18}) \times (2.0) = 10 \times 10^{18} = 1.0 \times 10^{19}.$$

Next we calculate the contribution of holes to conductivity:

$$p\mu_p = (5 \times 10^{19}) \times (0.01) = 5 \times 10^{17}.$$

Now we add the two contributions inside the bracket:

$$n\mu_n + p\mu_p = 1.0 \times 10^{19} + 5 \times 10^{17}.$$

To combine, we rewrite the smaller term with the same power of ten:

$$5 \times 10^{17} = 0.05 \times 10^{19}, \qquad\text{so}$$

$$n\mu_n + p\mu_p = 1.0 \times 10^{19} + 0.05 \times 10^{19} = 1.05 \times 10^{19}.$$

Substituting this result and the value of $$q$$ back into the formula for $$\sigma$$, we obtain

$$\sigma = (1.6 \times 10^{-19}) \times (1.05 \times 10^{19}).$$

We now multiply the numerical factors and add the exponents of ten:

$$1.6 \times 1.05 = 1.68,$$

$$(10^{-19}) \times (10^{19}) = 10^{0} = 1.$$

So,

$$\sigma = 1.68 \times 1 = 1.68\ (\Omega\ \text{m})^{-1}.$$

Because of rounding in the given data, this value is best matched by the option 1.65 $$(\Omega\ \text{m})^{-1}.$$

Hence, the correct answer is Option B.