Final answer:
The magnitude of the potential difference needed to give an alpha particle 239.0 keV of kinetic energy is approximately 746 kV. This is calculated by dividing the kinetic energy by the charge of two electrons, since the alpha particle carries a charge equivalent to that of 2 protons.
Step-by-step explanation:
The magnitude of the potential difference needed to give an alpha particle 239.0 keV of kinetic energy can be calculated using the relationship between electric potential energy and kinetic energy. Since the alpha particle is composed of 2 protons and 2 neutrons, it carries a charge equivalent to that of 2 protons. Since 1 electron volt (eV) is the amount of kinetic energy acquired by an electron when it accelerates through an electric potential difference of 1 volt, and the alpha particle has twice the charge of an electron, the needed potential difference would simply be the kinetic energy divided by the charge of 2 electrons (two elementary charges).
Given:
- Kinetic Energy (KE) = 239.0 keV = 239,000 eV
- Charge of an electron (e) = 1.602 x 10-19 C
- Charge of an alpha particle = 2e
Thus, the potential difference (V) required to accelerate the alpha particle is:
V = KE / (2e)
V = 239,000 eV / (2 * 1.602 x 10-19 C)
V = 239,000 eV / (3.204 x 10-19 C)
V = 7.46 x 105 Volts
Therefore, the alpha particle would require a potential difference of about 746 kV to be given 239.0 keV of kinetic energy.