Question

An electron in a TV is accelerated toward the screen. The potential between the back of the tube and the screen is 22,000 V (which is why it is a VERY good idea not to play around TV picture tubes). How much kinetic energy does the electron lose when it strikes the TV screen?

Answer 1

The kinetic energy lost by the electron ΔK.E is 3.5244 × 10⁻¹⁵ J

Step-by-step explanation:

Here we have;

Change in electric potential, ΔV, of the screen = 25000 V

The energy gained by the electron can be derived from the relation of change in electrical potential energy, ΔEPE, as follows;

ΔEPE = -Charge, q × ΔV

Therefore, since charge, q = -1.602 × 10⁻¹⁹ C we have;

ΔEPE = -(-1.602 × 10⁻¹⁹ C) × 25,000 V = 3.5244 × 10⁻¹⁵ C·V =

Also, since the electron is accelerated to the screen by the electrical potential energy, we have;

From the principle of conservation of energy in a given system;

The change in potential energy, ΔEPE of the electrons = Change in kinetic energy, ΔK.E, of the electrons3.5244 × 10⁻¹⁵ J

`\Delta KE = (1)/(2) \cdot m \cdot v_f^2 - (1)/(2) \cdot m \cdot v_i^2`

Where:

m = Mass of the electron = 9.11 × 10⁻³¹ kg

`v_i` = Initial velocity of the electrons = 0 m/s (electrons at rest)

`v_f` = Final velocity of the electrons

`\therefore \Delta KE = (1)/(2) \cdot m * (v_f^2 - \cdot v_i^2)`

`\therefore \Delta KE = (1)/(2) \cdot m * (v_f^2 - 0) = (1)/(2) \cdot m \cdot v_f^2`

Hence;

The kinetic energy lost by the electron ΔK.E = ΔEPE = 3.5244 × 10⁻¹⁵ J.