ΔV = -3.45x10-3 V
qa = 3.20x10 -19 C
ma 6.68x10-27 kg
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 qα = 3.20x10 -19 C and mα 6.68x10-27 kg, respectively.
What is the value of the change in potential energy, ΔU=Uf−Ui, of the alpha particle? Express your answer in joules using three significant figures.
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