Question

Potential difference of $100V$ is applied to the ends of a copper wire one metre long. Find the ratio of average drift velocity and thermal velocity of electrons at ${27}^{\circ }C$. (Consider there is one conduction electron per atom. The density of copper is $9.0\times {10}^3$; Atomic mass of copper is $63.5g$.
$N_A=6.0\times {10}^{23}$ per gram-mole, conductivity of copper is $5.81\times {10}^7{\Omega }^{-1}$.$K=1.38\times {10}^{-23}J{K}^{-1}$).

• A. $3.67\times {10}^{-6}$
• B. $4.3\times {10}^{-6}$
• C. $6\times {10}^{-5}$
• D. $5.6\times {10}^{-6}$

Answer 1

The correct option is A $3.67\times {10}^{-6}$

Given,

Potential Difference is $100V$

Length of Wire is $1m$

Atomic Mass of copper is $63.5g$

Density of copper is $9\times {10}^3$

$N_A=6.0\times {10}^{23}Pergrammole$

$K=1.38\times {10}^{-23}J{K}^{-1}$

As given one conduction electron per atom therefore Conduction electron density is calculated by,

$\frac{N_A}{massofcopper}\times density$

Substituting all value in equation

$=\frac{6.0\times {10}^{23}}{63.5}\times 9\times {10}^3$

$=\frac{54\times {10}^{26}}{63.5}=8.5\times {10}^{22}c{m}^{-3}=8.5\times {10}^{28}{m}^{-3}$

Now Electric field is

$E=\frac{V}{l}=\frac{100}{1}=100V/m$

We know that Current density is

$J=\sigma E=ne{v}_d$

Here $v_d$ is current density

$v_d=\sigma \frac{E}{ne}$

Substituting all value in above equation

$v_d=\frac{5.81\times {10}^7\times 100}{8.5\times {10}^{28}\times 1.61\times {10}^{-19}}=\frac{5.81}{13.69}=0.42m/s$

And

$Thermalvelocity\left(V_{rms}\right)=\sqrt{\frac{3K_{B}T}{m_e}}$

$=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times (273+27)}{9.1\times {10}^{-31}}}=116826.16\approx 1.17\times {10}^{5}m/s$

$Ratio=\frac{DriftVelocity}{ThermalVelocity}=\frac{0.43}{1.17\times {10}^5}=3.67\times {10}^{-6}$

Hence, Option(A) is correct option