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

Answer:

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

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

$\Delta KE = \frac{1}{2} \cdot m \cdot v_f^2 - \frac{1}{2} \cdot m \cdot v_i^2$

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

$\therefore \Delta KE = \frac{1}{2} \cdot m \times (v_f^2 - \cdot v_i^2)$

$\therefore \Delta KE = \frac{1}{2} \cdot m \times (v_f^2 - 0) = \frac{1}{2} \cdot m \cdot v_f^2$

Hence;

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