#### 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.

#### Solution

Given:-

Electric field inside discharge tube, E = 5.0 kV/m

\[t = 1 \mu s = 1 \times {10}^{- 6} S\]

\[F = qE = (1 . 6 \times {10}^{- 19} ) \times 5 \times {10}^3 N\]

\[a = \frac{qE}{m} = \frac{8 \times {10}^{- 16}}{9 . 1 \times {10}^{- 31}}\]

(a) Let S be distance travelled by the free electron in 1 µs.

\[S = \frac{1}{2}a t^2 \]

\[S = \frac{1}{2} \times \frac{8}{9 . 1} \times {10}^{- 16 + 31} \times ( {10}^{- 6} )^2 \]

\[S = 0 . 43956 \times {10}^3 m \approx 440 m\]

(b) Let the time of transit of the free electron between successive collisions be t.

\[S = 1\text{ mm} = 1 \times {10}^{- 3}\text{ m}\]

Using, \[S = \frac{1}{2} \times \left( \frac{qE}{m} \right) \times t^2 ,\] we get

\[1 \times {10}^{- 3} = \frac{1}{2} \times \frac{1 . 6 \times 5}{9 . 1} \times {10}^{16} \times (t )^2 \]

\[ \Rightarrow t^2 = \frac{2 \times 9 . 1 \times {10}^{- 3}}{1 . 6 \times 5 \times {10}^{16}}\]

\[ \Rightarrow t^2 = \frac{9 . 1}{0 . 8 \times 5} \times {10}^{- 19} \]

\[ \Rightarrow t = 0 . 4802 \times {10}^{- 9} s = 0 . 5\text{ ns}\]