Final answer:
The alpha particle was accelerated through a potential difference of approximately 4.68 kV, which is calculated using the relationship E = qV where E is energy, q is charge, and V is potential difference.
Step-by-step explanation:
To calculate the potential difference an alpha particle was accelerated through, given its gain in kinetic energy, one must apply the relationship between energy (E) and potential difference (V), which is E = qV, where q is the charge of the particle and V is the potential difference. For an alpha particle which has a charge of +2e (where e is the elementary charge, approximately 1.60 x 10^-19 coulombs), and given that it has gained 1.50 x 10^-15 joules of kinetic energy, the potential difference can be calculated as follows:
V = E/q
Substituting the given values, we get:
V = (1.50 x 10^-15 J) / (2 x 1.60 x 10^-19 C) = 4.68 x 10^3 volts or 4.68 kV. So, the alpha particle was accelerated through a potential difference of approximately 4.68 kV.