Question

An electron is accelerated through a potential difference of 18 kV in a colour TV cathode ray tube. Calculate the kinetic energy and speed of electron.

Answer 1

`K.E = 1/2 xx  m xx  v^2 = eV`

E = eV

`= 1.6 xx 10^(-19)  xx 18 xx  10^3`

`= 28.8 xx 10^(-16) J`

`K.E = 1/2 m v^2`

`v = sqrt((2E)/m)`

`= sqrt(2  xx  28.8  xx  10^(−16))/( 9.1   xx   10^(−31))`

`= 7.95  xx  10^7 m/s`

Kinetic energy of electron = 28.8 x 10^-16 J 

Speed of electron = 7.95 x 10⁷ m/s