Question

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 velocity 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?

Answer 1

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