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

Step-by-step explanation:

There is electric field between the plates whose value is given by the following expression

electric field E = V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

= 60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s = 2.5 x 10⁻³ m

s = 1\2 a t²

t =
`\sqrt{(2s)/(a\\ ) }`

=
`\sqrt{(2*2.5*10^(-3))/( 10.55*10^(15))`

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t =
`[tex]\sqrt{(2*5*10^(-3))/( 10.55*10^(15))`[/tex]

t = 0.67 x 10⁻¹⁸ s