Question

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 $63.5amu$.

• A. $3.1\times {10}^3{ms}^{-1}$
• B. $1.1\times {10}^{-3}{ms}^{-1}$
• C. $4.1\times {10}^3{ms}^{-1}$
• D. $2.1\times {10}^3{ms}^{-1}$

Answer 1

Answer: (B)

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

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$\left(N_A\right)$ $....\left(1\right)$

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

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

Therefore, Using Equation $\left(1\right)$, 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 $v_d=\frac{I}{neA}$

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

Hence, option B is correct.