Question

A proton is accelerated through one volt the increase in its kinetic energy is approximately:

• A. $\frac{1}{1870}eV$
• B. $1870eV$
• C. $1eV$
• D. $300eV$

Answer 1

Answer: (C)

Gain in $K.E=$ charge $\times$ potential difference.
$=1e\times 1V=1eV$