Question

5. The electron drift speed is $2.0\times {10}^{-4}\mathrm{m}/\mathrm{s}$ in a metal with a mean time between collisions of $5.0\times {10}^{-14}\mathrm{s}$. What is the electric field strength?

Short answer

The electric field strength is$22.8\mathrm{mV}/\mathrm{m}$.

Step-by-step solution

Given Information

Electron drift speed$=2.0\times {10}^{-4}\mathrm{m}/\mathrm{s}$

Time between collisions$=5.0\times {10}^{-14}\mathrm{s}$

Explanation

The drift speed is given as a function of the electric field as

$v_d=\frac{e\tau }{m}E,$

The electric field as

$E=\frac{m{v}_d}{e\tau }.$

Mass of the electron is $m=9.1·{10}^{-31}\mathrm{kg}$and the charge, the elementary charge, is $m=9.1·{10}^{-31}\mathrm{kg}$we can find the value of the electric field strength to be

role="math" localid="1648877656088" $E=\left(\frac{9.1\times {10}^{-31}kg\times 2\times {10}^{-4}m/s}{1.6\times {10}^{-19}c\times 5\times {10}^{-14}s}\right)$

$E=22.75\mathrm{mV}/\mathrm{m}$

Final Answer 

Hence, the electric field strength is$22.8\mathrm{mV}/\mathrm{m}.$