# The speed of an alpha particle is determined to be 4.75 \times 10^6 m/s. If all of its kinetic...

## Question:

The speed of an alpha particle is determined to be {eq}4.75 \times 10^6 {/eq} m/s. If all of its kinetic energy is acquired by passing through an electric potential, what is the magnitude of that potential?

## Conservation Law of Energy:

Let us assume a system that contains some energy and the system does not allow the trapped energy to be transformed with its surroundings, then the trapped energy will be always unchanged or remain constant.

Let the magnitude of the potential be V.

From the energy conservation law, the kinetic energy of the alpha particle will be converted into electric potential energy.

Thus,

{eq}qV = \frac{1}{2}mv^{2} {/eq}

{eq}V = \frac{mv^{2}}{2q} {/eq}

Here,

q = charge of the alpha particle

m = mass of the alpha particle

v = speed of the alpha particle.

Thus,

{eq}V = \frac{6.68 \times 10^{-27} \times (4.75 \times 10^{6})^{2}}{2 \times 3.2 \times 10^{-19}} {/eq}

On solving,

{eq}\boxed{V = 2.35 \times 10^{5} \ \rm V.} {/eq}