A cylindrical conductor of length 2 m and area of cross-section $$0.2mm^{2}$$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $$\alpha \times 10^{-3}m^{2}/V.s.$$ The value of $$\alpha$$ is :
(electron concentration = $$5 \times 10^{28}/m^{3}$$ and electron charge = $$1.6 \times 10^{-19}C$$)

We need to find the mobility of electrons in a cylindrical conductor.

The length of the conductor is $$L = 2$$ m, its cross-sectional area is $$A = 0.2$$ mm$$^2 = 0.2 \times 10^{-6}$$ m$$^2$$, the current through it is $$I = 1.6$$ A, the applied voltage is $$V = 2$$ V, the electron density is $$n = 5 \times 10^{28}$$/m$$^3$$, and the charge of an electron is $$e = 1.6 \times 10^{-19}$$ C.

Since resistance is given by $$R = \frac{V}{I} = \frac{2}{1.6} = 1.25 \text{ } \Omega$$, we obtain the resistance.

Substituting this into the formula for resistivity yields $$\rho = \frac{RA}{L} = \frac{1.25 \times 0.2 \times 10^{-6}}{2} = \frac{0.25 \times 10^{-6}}{2} = 1.25 \times 10^{-7} \text{ } \Omega \cdot m$$.

Mobility is defined by $$\mu = \frac{1}{ne\rho}$$, and substituting the known values gives

$$\mu = \frac{1}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.25 \times 10^{-7}}$$

$$= \frac{1}{5 \times 1.6 \times 1.25 \times 10^{28-19-7}} = \frac{1}{10 \times 10^{2}} = \frac{1}{1000} = 10^{-3} \text{ m}^2/\text{Vs}$$

So $$\alpha = 1$$.

The answer is 1.