Question

An electron is to be accelerated from a velocity of 4.50×106 m/s to a velocity of 9.00×106 m/s. Through what potential difference must the electron pass to accomplish this?

Answer 1

-2.85 * 10^(-17) J

Step-by-step explanation:

Parameters given:

Final velocity, v = 9 * 10^6 m/s

Initial velocity, u = 4.5 * 10^6 m/s

Using the conservation of energy formula, total energy is conserved:

K.Ein + PEin = KEf + PEf

K.Ef - K.Ein = P.Ein - P.Ef

=> -∆P.E = K.Ef - K.Ein

∆P.E = K.Ein - K.Ef

∆P.E = ½mu² - ½mv²

∆P.E = ½m[(4.5 * 10^6)² - (9 * 10^6)²]

∆P.E = ½ * 9.31 * 10^(-31) * (-61.25 * 10¹²)

∆P.E = -2.85 * 10^(-17) J