Question

What is the speed of a proton after being accelerated from rest through a 5.6x107 V potential difference?

Answer 1

Answer:

1.04 x 10⁸ m/s

Explanation:

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

ΔV = Potential difference through which proton is accelerated = 5.6 x 10⁷ Volts

m = mass of the proton = 1.67 x 10⁻²⁷ kg

v = speed of the proton = ?

Using conservation of energy

Kinetic energy gained by the proton = Electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (5.6 x 10⁷)

v = 1.04 x 10⁸ m/s