Question

A molecule of a gas, filled in a discharge tube, gets ionised when an electron is detached from it. An electric field of 5.0 kV m⁻¹ exists in the vicinity of the event. (a) Find the distance travelled by the free electron in 1 µs, assuming there's no collision. (b) If the mean free path of the electron is 1.0 mm, estimate the time of transit of the free electron between successive collisions.

Answer 1

Given:
Electric field inside discharge tube, E = 5.0 kV/m
$$\begin{gathered} t=1\mathrm{\mu s}=1\times {10}^{-6}\mathrm{S} \\ F=qE=(1.6\times {10}^{-19})\times 5\times {10}^3\mathrm{N} \\ a=\frac{qE}{m}=\frac{8\times {10}^{-16}}{9.1\times {10}^{-31}} \end{gathered}$$

(a)
Let S be distance travelled by the free electron in 1 µs.
$$\begin{gathered} S=\frac{1}{2}a{t}^2 \\ S=\frac{1}{2}\times \frac{8}{9.1}\times {10}^{-16+31}\times ({10}^{-6}{)}^2 \\ S=0.43956\times {10}^3\mathrm{m}\approx 440\mathrm{m} \end{gathered}$$

(b)
Let the time of transit of the free electron between successive collisions be t.
$$\begin{gathered} S=1\mathrm{mm}=1\times {10}^{-3}\mathrm{m} \\ \mathrm{Using},\mathit{}S=\frac{1}{2}\times \left(\frac{qE}{m}\right)\times t^2,\text{we get}: \\ 1\times {10}^{-3}=\frac{1}{2}\times \frac{1.6\times 5}{9.1}\times {10}^{16}\times (t{)}^2 \\ \Rightarrow t^2=\frac{2\times 9.1\times {10}^{-3}}{1.6\times 5\times {10}^{16}} \\ \Rightarrow t^2=\frac{9.1}{0.8\times 5}\times {10}^{-19} \\ \Rightarrow t=0.4802\times {10}^{-9}s=0.5\mathrm{ns} \end{gathered}$$