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DanielleElmas [232]

2 years ago

13

A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the ve

locity of an electron midway between the electrodes and at the instant of striking the plate, if the electrons are emitted from the cathode with zero velocity?

1 answer:

Alex_Xolod [135]2 years ago

6 0

Answer:

Velocity of the electron mid way = v = 6.47×10⁶ m/s

Explanation:

Electric potential varies with distance as $V(x) = C x^\frac{4}{3}$

At a distance 5 mm , the constant C is evaluated.

C = $V/x^\frac{4}{3}$

= $300/(0.005)^\frac{4}{3}$

= 3.50×10⁵

Now at the mid point, x = 2.5 mm = 0.0025 m

Potential = V' = $C x^\frac{4}{3}$

= $3.50\times 10^5\times (0.0025)^\frac{4}{3}$

= 119 V

Kinetic energy per unit charge is equal to the electric potential.

or 0.5 m v^2 = V' q

Here m is the mass of the electron and q is the charge of the electron.

m= 9.1×10⁻³¹ kg

q = 1.6×10⁻¹⁹ coulombs

⇒ $v^2 = \frac{V q}{0.5 m}$

⇒ $v^2 = \frac{(119)(1.6\times 10⁻¹⁹)}{0.5 (9.1\times 10⁻³¹)}$

⇒ v = 6.47×10⁶ m/s

⇒ Velocity of the electron mid way = v = 6.47×10⁶ m/s

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