🤓 Based on our data, we think this question is relevant for Professor Ene's class at UH.

Work done

$\overline{)\begin{array}{rcl}\mathbf{W}& \mathbf{=}& \mathbf{q}\mathbf{\xb7}\mathbf{\u2206}\mathbf{V}\end{array}}$

Kinetic energy,

$\overline{)\begin{array}{rcl}\mathbf{KE}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}\end{array}}$

ΔV = -3.45x10^{-3} V

q_{a} = 3.20x10^{ -19} C

m_{a} 6.68x10^{-27 }kg

**Part a)**

Change in kinetic energy is equal to work done.

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of -3.45×10^{-3} V.

The charge and the mass of an alpha particle are q_{α} = 3.20x10^{ -19} C and m_{α} 6.68x10^{-27 }kg, respectively.

What is the value of the change in potential energy, Δ*U*=*U*_{f}−*U*_{i}, of the alpha particle? Express your answer in joules using three significant figures.