Question

A proton with a speed of 2 X10^5m/s goes through potential difference of 110 volts gaining speed. What is the speed its reached?

Show your work in detail. ( the ans is 2.9 x10^5m/s. i know the right answer but i just dont know how they worked it out) Please work the problem in detail. thanks

Answer 1

Solution :

Given :

Initial speed (u) = 2 x 10⁵ m/ s
potential difference (V) = 110 Volts
charge of electron (q) = 1.6 x 10^-19 C
mass of proton (m) = 1.67 x 10^-27 kg

According to the law of conservation of energy :
(1/ 2) mu² + q V = (1/ 2) mv²
(0.5)(1.67 x 10^-27 kg)(2 x 10⁵ m/ s)² + (1.6 x 10^-19 C)(110 V) = (0.5)(1.67 x 10^-27 kg) v²
v² = 8.41 x 10¹⁰
v = 2.9 x 10⁵ m/s