Question

The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen

Answer 1

The speed is
`v =10.27 *10^(7) \ m/s`

Step-by-step explanation:

From the question we are told that

The voltage is
`V = 30 kV = 30*10^(3) V`

The initial velocity of the electron is
`u = 0 \ m/s`

Generally according to the law of energy conservation

Electric potential Energy = Kinetic energy of the electron

So

`PE = KE`

Where

`KE = (1)/(2) * m* v^2`

Here m is the mass of the electron with a value of
`m = 9.11 *10^(-31) \ kg`

and

`PE = e * V`

Here e is the charge on the electron with a value
`e = 1.60 *10^(-19) \ C`

=>
`e * V = (1)/(2) * m * v^2`

=>
`v = \sqrt{ (2 * e * V)/(m) }`

substituting values

`v = \sqrt{ (2 * (1.60*10^(-19)) * 30*10^(3))/(9.11 *10^(-31)) }`

`v =10.27 *10^(7) \ m/s`