Question

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a television tube. It is constantly accelerated while travelling 0.01 m, and leaves the gun at 5.4 x 106 m/s. What was the acceleration of the electron

Answer 1

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Step-by-step explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

`a = (2.91375 \ * \ 10^(13))/(2 \ * \ 0.01) \\\\a = 1.457 \ * \ 10^(15) \ m/s^2`

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².