Question

The number of free electrons per unit volume ($\text{m}^3$) in a wire made of copper (resistivity $1.72 \times 10^{-8}\,\Omega\,\text{m}$) is $8.5 \times 10^{28}$. Calculate the average number of collisions made by an electron in one second when an electric field is maintained in the wire.

Hint:

Relaxation time is tau = m / (n e^2 rho). The number of collisions per second is 1/tau.

Answer 1

Step 1: Understanding the Concept:
Free electrons in a metal collide with ions and lose their drift speed. The average time between collisions is the relaxation time \(\tau\). The number of collisions per second is \(1/\tau\).

Step 2: Key Formula or Approach:
Resistivity of a conductor: \(\rho = \dfrac{m}{n e^2 \tau}\). So \(\tau = \dfrac{m}{n e^2 \rho}\).
Use the standard values \(m = 9.1 \times 10^{-31}\) kg and \(e = 1.6 \times 10^{-19}\) C.

Step 3: Calculate the relaxation time:
\[ \tau = \frac{9.1 \times 10^{-31}}{(8.5 \times 10^{28})(1.6 \times 10^{-19})^2 (1.72 \times 10^{-8})} \]
Denominator: \(8.5 \times 10^{28} \times 2.56 \times 10^{-38} \times 1.72 \times 10^{-8} = 3.74 \times 10^{-17}\).
\[ \tau = \frac{9.1 \times 10^{-31}}{3.74 \times 10^{-17}} = 2.43 \times 10^{-14}\ \text{s} \]

Step 4: Collisions per second:
\[ \frac{1}{\tau} = \frac{1}{2.43 \times 10^{-14}} = 4.11 \times 10^{13}\ \text{s}^{-1} \]

Final Answer:
An electron makes about \(4.1 \times 10^{13}\) collisions every second. \[ \boxed{\approx 4.1 \times 10^{13}\ \text{s}^{-1}} \]