Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an N-type semiconductor, the density of electrons is $$10^{19}$$ m$$^{-3}$$ and their mobility is 1.6 m$$^2$$ V$$^{-1}$$ s$$^{-1}$$, then the resistivity of the semiconductor (since it is an N-type semiconductor contribution of holes is ignored) is close to:

We recall the basic relation between conductivity and carrier parameters in a semiconductor. For one kind of charge carrier (only electrons in an N-type material) the conductivity is given by the formula $$\sigma = n\,e\,\mu$$ where

$$n = 10^{19}\ \text{m}^{-3}$$ is the electron concentration, $$e = 1.6 \times 10^{-19}\ \text{C}$$ is the electronic charge, and $$\mu = 1.6\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the mobility of the electrons.

Substituting these values, we obtain

$$\sigma = (10^{19}\ \text{m}^{-3})(1.6 \times 10^{-19}\ \text{C})(1.6\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}).$$

First we multiply the numerical parts of the first two factors:

$$10^{19} \times 1.6 \times 10^{-19} = 1.6.$$

Now we include the mobility factor:

$$\sigma = 1.6 \times 1.6 = 2.56\ \text{S m}^{-1}.$$

Resistivity $$\rho$$ is the reciprocal of conductivity, so

$$\rho = \frac{1}{\sigma} = \frac{1}{2.56}\ \Omega\,\text{m}.$$

Carrying out the division gives

$$\rho \approx 0.390625\ \Omega\,\text{m} \approx 0.4\ \Omega\,\text{m}.$$

Hence, the correct answer is Option D.