Answer to Question #162967 in Physics for Ajay

Question

The density of copper is 8.92 × 10³

kg m^-3

and its atomic weight is 63.5. Assuming 

that each copper atom provides one conduction (free) electron, calculate the density 

of conduction electrons. Also estimate the relaxation time for free electrons 

applying classical free electron theory. What is the corresponding mean free path? 

Take the electrical conductivity of copper to be 6.8×10⁷ Ω^-1 m^-1.

.

Answer 1

$$n=\frac{N_A\rho}{M}=\frac{8.92(6.025\cdot10^{26})}{63.5}=8.46\cdot10^{25}\ m^{-3}$$

$$\tau=\frac{\sigma m}{ne^2}\\\tau=\frac{(6.8\cdot10^{7})(9.11\cdot10^{-31})}{(8.46\cdot10^{25})(1.6\cdot10^{-19})^2}=2.86\cdot10^{-11}s$$

$$\lambda=v_F\tau\\\lambda=1.57\cdot10^{6}\cdot2.86\cdot10^{-11}=4.5\cdot10^{-5}m$$