Q. When a proton is accelerated with $1\ volt$ potential difference, then its kinetic energy is
A) $\frac{1}{1840}eV$
B) $1840\ eV$
C) $1\ eV$
D) $1840\ c^{2}\ eV$
Explanation
(c) $K = Q.V = 1e \times 1\ Volt = 1eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.