Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0\times {10}^{-7}{m}^2$ carrying a current of $1.5A$. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $9.0\times {10}^3{kgm}^{-3}$ and its atomic mass is $63.5amu$.

$\text{Step 1: No. of conduction electron per unit volume}$

Molar mass of copper $M=63.5$ gram $=63.5\times 10^{-3}kg$ 

Density of copper $\rho =9\times 10^{3}Kg/m^3$

$\text{No. of copper atoms per unit volume:}$ 

        $N=$ No. of moles in unit volume $\times$ No. of atoms in 1 mole$(N_A)$                   $....(1)$

$\text{No. of moles in unit volume}$ $=\frac{\text{Mass of unit volume}}{\text{Mass of one mole}}$

                                                    $=\frac{\text{Density}}{\text{Molar Mass}}$ $=\frac{\rho }{M}$

Therefore, Using Equation $(1)$, We get:

                       $N=\frac{\rho }{M}\times N_A$                 $\text{Where }{N}_A=6.023\times 10^{23}$

               $\Rightarrow N=\frac{9\times 10^3\times 6.023\times 10^{23}}{63.5\times 10^{-3}}$ $=8.54\times 10^{28}{m}^{-3}$ 

According to question one copper atom contributes one conduction electron.

So, No. of conduction electron per unit volume $=$ No. of copper atoms per unit volume

                        $\therefore n=N=8.54\times 10^{28}{m}^{-3}$ 

$\text{Step 2: Drift velocity}$

Given: Current $I=1.5A$, Cross sectional area $A=1\times 10^{-7}{m}^2$ 

We know that,                     

                      $I=neA{v}_d$ 

So, Drift Velocity $v_d=\frac{I}{neA}$ 

   

    $\Rightarrow v_d=\frac{1.5A}{8.5\times 10^{28}\times (1.6\times 10^{-19}C)\times (10^{-7}{m}^{-2})}$ $=1.1\times 10^{-3}m/s$

Hence, option B is correct.