The density of copper is$\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^3\frac{\mathrm{kg}}{{\mathbf{m}}^3}$. Its femi energy is 7.0 eV, and it has one conduction electron per atom. At liquid nitrogen temperature (77 K), its resistivity is about $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{-9}\mathbf{\Omega }\cdot \mathbf{m}$. Estimate how far a conduction electron would travel before colliding and how many copper ions it would pass.

Consider the formula for the number of copper ions as:

$\mathbf{Number}\mathbf{of}\mathbf{copper}\mathbf{ions}\mathbf{=}\frac{\mathrm{distance}}{a}$

Consider the formula for the density of atoms:

$\eta =\frac{density}{atomicmass}$

Solve for the mass of the copper:

$\begin{array}{c}m=\left(63.546\text{\hspace{0.17em}u}\right)\left(\frac{1.66\times {10}^{-27}\text{\hspace{0.17em}kg}}{\text{1\hspace{0.17em}u}}\right)\\ =1.055\times {10}^{-25}\text{\hspace{0.17em}kg}\end{array}$

Solve for the number of atoms per unit volume as:

$\begin{array}{c}\eta =\frac{density}{atomicmass}\\ =\frac{8.9\times {10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}}{1.055\times {10}^{-25}\text{\hspace{0.17em}kg}}\\ =8.437\times {10}^{28}\frac{\text{atoms}}{{\text{m}}^3}\end{array}$

Solve for the relaxation time as:

$\begin{array}{c}\tau =\frac{m}{e^2\eta \rho }\\ \tau =\frac{9.11\times {10}^{-31}\text{\hspace{0.17em}kg}}{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})(8.437\times {10}^{28}{\text{\hspace{0.17em}m}}^{-3})(4\times {10}^{-9}\Omega \cdot \text{m})}\\ \tau =1.054\times {10}^{-13}\text{\hspace{0.17em}s}\end{array}$

Determine the speed of the electron as:

$\begin{array}{c}v=\sqrt{\frac{2\left(KE\right)}{\text{m}}}\\ =\sqrt{\frac{2\left(7.0\text{\hspace{0.17em}eV}\right)(1.6\times {10}^{-19}\text{\hspace{0.17em}J})}{9.11\times {10}^{-31}\text{\hspace{0.17em}kg}}}\\ =1.568\times {10}^6\frac{\text{m}}{\text{s}}\end{array}$

Solve for the distance travelled by the electrons as:

localid="1660041976047" $\begin{array}{c}d=v\tau \\ =\left(1.568\times {10}^6\hspace{0.17em}\frac{m}{s}\right)(1.054\times {10}^{-13}s)\\ =0.17\hspace{0.17em}\mu m\end{array}$

Consider the inverse of $\eta$is the volume per atom and is solved as:

$\begin{array}{c}a=\frac{1}{\eta }\\ =2.28\times {10}^{-10}\text{\hspace{0.17em}m}\end{array}$

Solve for the number of ions as:

$\begin{array}{c}n=\frac{\text{distance}}{a}\\ =\frac{1.7\times {10}^{-7}\text{\hspace{0.17em}m}}{2.28\times {10}^{-10}\text{\hspace{0.17em}m}}\\ \approx 700\end{array}$

Therefore, the distance travelled$0.17\mu \text{m}$is and the number of ions is 700.