Speed of a Proton After it Accelerates Through a Potential Difference

Problem Statement

A proton is accelerated over a potential difference of 100 V. What is the final speed of the proton?

Solution

We can use the law of Conservation of Energy to solve this problem. If a proton with charge $e$ travels over a potential difference ($\Delta V$) the gain in energy of the proton is $\Delta U=e\Delta V$. This gain in energy will become kinetic energy:

$$e\Delta V=\frac{1}{2}m{v}^2$$

Solving for speed (v) yields:

$$\frac{1}{2}m{v}^2=e\Delta V$$ $$v^2=\frac{2e\Delta V}{m}$$ $$v=\sqrt{\frac{2e\Delta V}{m}}$$

Plugging in values yields:

$$v=\sqrt{\frac{2(1.6\times 10^{-19}C)(100V)}{1.67\times 10^{-27}kg}}=138425.7m/s$$