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

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

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

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

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

d) physical means can be used to separate the components of air.