What is the magnitude of the potential difference needed to give an alpha particle (composed of 2 protons and 2 neutrons) 464 keV of kinetic energy?    kV

The kinetic energy (KE) of a particle is given by the equation KE = qV, where q is the charge of the particle and V is the potential difference.

An alpha particle is composed of 2 protons and 2 neutrons. However, only the protons carry charge, and each proton has a charge of +1.6 x 10^-19 Coulombs. Therefore, the total charge of an alpha particle is q = 2 * 1.6 x 10^-19 C = 3.2 x 10^-19 C.

The kinetic energy given in the problem is 464 keV, which is equivalent to 464 x 10^3 eV. Since 1 eV = 1.6 x 10^-19 Joules, the kinetic energy in Joules is KE = 464 x 10^3 * 1.6 x 10^-19 J = 7.424 x 10^-14 J.

Substituting the values of KE and q into the equation KE = qV gives:

7.424 x 10^-14 J = (3.2 x 10^-19 C) * V

Solving for V gives:

V = 7.424 x 10^-14 J / 3.2 x 10^-19 C = 2.32 x 10^5 Volts = 232 kV.

Therefore, the magnitude of the potential difference needed to give an alpha particle 464 keV of kinetic energy is 232 kV.