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
