Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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. - Physics

#### 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

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

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

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

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

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Solution 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. Concept: Dual Nature of Radiation.
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