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## Q. 3.3

In a parallel plate diode, anode is made 350 V positive with respect to the cathode and is 6 mm from it. An electron is emitted from the cathode with an initial velocity of $3 × 10^{6}$ m/s in the direction of the anode. Calculate the velocity and the time of travel of the electron (i) when it is midway between cathode and anode, and (ii) on reaching the anode.

## Verified Solution

Given           $V = 350 V, d = 6 mm, v_{ox} = 3 × 10^{6} m/s$

Therefore,           $E = V/d = 350/(6 × 10^{–3}) = 58.33 × 10^{3} V/m$

$a_{x} = qE/m = 1.026 × 10^{16} m/s^{2}$

We know that,

$x = v_{ox}t + 0.5 at^{2}$

$v_{x} = v_{ox} + a_{x}t$

(i) Consider $x = 3 × 10^{–3} m$

$3 × 10^{–3} = 3 × 10^{6}t + 5.13 × 10^{15} t^{2}$

$t^{2} + 5.85 × 10^{–10} t – 5.85 × 10^{–19} = 0$

Solving this equation,

$t = 5.26 × 10^{–10} s$

Therefore,             $v_{x} = v_{ox} + a_{x} t = 3 × 10^{6} + 1.026 × 10^{16} (5.264 × 10^{–10})$

$= 8.4 × 10^{6} m/s$

(ii) Consider $x = 6 × 10^{–6} m$

$t^{2} + 5.85 × 10^{–10} t – 1.17 × 10^{–18} = 0$

Solving this equation,         $t = 8.28 × 10^{–10} s$

Therefore,                              $v_{x} = 3 × 10^{6} + 8.28 × 10^{–10} (1.026 × 10^{16}) = 11.5 × 10^{6} m/s$