A metal wire of cross-sectional area 0.5 mm² and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 Ω. The density, atomic mass and electrical conductivity of the metal are 6.35 × 10³ kg m⁻³, 63.5 gm/mole and 2 × 10⁸ mho m⁻¹, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s⁻¹) of the electrons in the wire is:
(Take Avogadro’s number as 6 × 10²³ and charge of the electron as 1.6 × 10⁻¹⁹ C.)

The correct option is 0.208.

Step-by-Step Solution:

1. Resistance of the metal wire (R):
The resistance of a wire is given by the formula:

$R=\frac{L}{\sigma A}$

$R=\frac{100}{(2\times {10}^8)\times (0.5\times {10}^{-6})}=1\text{ Ω}$

2. Current through the circuit (I):
Using Ohm's Law for a complete circuit with internal resistance r:

$I=\frac{E}{R+r}$

$I=\frac{2}{1+1}=1\text{ A}$

3. Number density of conduction electrons (n):
Assuming 1 conduction electron per atom, the number of electrons per unit volume is:

$n=\frac{d\times N_A}{M}$

$n=\frac{6.35\times {10}^6\times 6\times {10}^{23}}{63.5}=6\times {10}^{28}\text{ m}{\hspace{0.17em}}^{-3}$

4. Drift Velocity (v_d):
The current is related to drift velocity by:

$I=nAe{v}_d$

$v_d=\frac{I}{nAe}$

$v_d=\frac{1}{(6\times {10}^{28})\times (0.5\times {10}^{-6})\times (1.6\times {10}^{-19})}$

$v_d=\frac{1}{4.8\times {10}^3}=2.083\times {10}^{-4}\text{ m s}{\hspace{0.17em}}^{-1}$

$v_d=2.083\times {10}^{-4}\times {10}^3\text{ mm s}{\hspace{0.17em}}^{-1}=0.208\text{ mm s}{\hspace{0.17em}}^{-1}$