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## Q. 3.4

In a parallel diode, the cathode and anode are spaced 5 mm apart and the anode is kept at 200 V d.c. with respect to cathode. Calculate the velocity and the distance travelled by an electron after a time of 0.5 nsec, when (a) the initial velocity of an electron is zero and (b) the initial velocity is 2 × $10^{6}$ m/sec in the direction towards the anode.

## Verified Solution

Given V = 200 V, d = 5 mm, Time of travel by an electron, t = 0.5 nsec.

Therefore,          $E=\frac{V}{d}=\frac{200}{5 \times 10^{-3} } =40\times \frac{10^{3}V }{m}$

$a_{x}=\frac{qE}{m}=\frac{1.602\times 10^{-19}\times 40\times 10^{3} }{9.1\times 10^{-31} } =70.33\times 10^{14}m/\sec ^{2}$

We know that

Velocity of an electron,       $v_{x} = v_{ox} + a_{x}t$

Distance travelled by an electron, $x = v_{ox} + 0.5at^{2}$

(a) When the initial velocity of an electron is zero

$v_{x} = v_{ox} + a_{x}t = 0 + 70.33 × 10^{14} × 0.5 × 10^{–9} = 3.5165 × 10^{6} m/sec$

$x = v_{ox}t + 0.5 at^{2} = 0 + 0.5 × 70.33 × 10^{14} × (0.5 × 10^{–9})^{2} = 0.879 × 10^{–3} m$

(b) When the initial velocity of an electron is $v_{ox} = 2\times \frac{10^{6}m }{sec}$

$v_{x} = v_{ox} + a_{x}t = 2 × 10^{6} + 70.33 × 10^{14} × 0.5 × 10^{–9} = 5.51 × 10^{6} m/sec$

$x = v_{ox}t + 0.5 at^{2} = 2 × 10^{6} × 0.5 × 10^{–9} + 0.5 × 70.33 × 10^{14} × (0.5 × 10^{–9})^{2}$

$= 1.879 × 10^{–3} m$