Final answer:
The potential near the surface of the Van de Graaff generator is 4.5 x 10^7 V. The distance from the center where the potential is 1.00 MV is 45 km. The kinetic energy of the oxygen atom is 1.92 MeV.
Step-by-step explanation:
To calculate the potential near the surface of a Van de Graaff generator, we use the formula V = kQ / r, where V is the potential, k is the electrostatic constant (9 x 10^9 N·m^2/C^2), Q is the charge on the sphere, and r is the radius of the sphere. The radius is half the diameter, so r = 1 m. Plugging in the values, we get V = (9 x 10^9 N·m^2/C^2) x (5 x 10^-3 C) / 1 m = 45 x 10^6 V = 4.5 x 10^7 V.
To find the distance from the center where the potential is 1.00 MV, we rearrange the formula to solve for r: r = kQ / V. Plugging in the values, we get r = (9 x 10^9 N·m^2/C^2) x (5 x 10^-3 C) / (1 x 10^6 V) = 45 x 10^3 m = 45 km.
For part (c) of the question, we need to calculate the kinetic energy of the oxygen atom. We can use the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the oxygen atom, and v is its velocity. Assuming the oxygen atom has a mass of 16 atomic mass units (AMU), we can convert it to kilograms: 16 AMU x (1.67 x 10^-27 kg / 1 AMU) = 2.67 x 10^-26 kg.
We can then calculate the velocity using the equation v = sqrt(2qV / m), where q is the charge on the atom (-3e), V is the potential, and m is the mass of the atom. Plugging in the values, we get v = sqrt((2 x -3 x 1.6 x 10^-19 C x 1 x 10^6 V) / (2.67 x 10^-26 kg)) = 1.52 x 10^6 m/s. Finally, we can calculate the kinetic energy: KE = (1/2) x (2.67 x 10^-26 kg) x (1.52 x 10^6 m/s)^2 = 3.07 x 10^-17 Joules = 1.92 MeV.