A single electron orbits a lithium nucleus that contains three protons (+3e). The radius of the orbit is 1.76*10^-11m. Determine the kinetic energy of the electron.

Answer 1

For a rotating object the equation for acceleration is a = v^2 / r. The force is F = m*a = m*v^2 / r. The force in this problem is the coulomb force.

F = k*3q*q / r^2 = m*v^2 / r

Since the kinetic energy is just m*v^2 / 2, and that's all we need, let's multiply both sides by r / 2:

k*3q*q / (2*r) = m*v^2 / 2

Plug in numbers on the left and we'll get the kinetic energy:

8.99*10^9 N*m^2 / c^2 * 3*(1.60*10^-19 c)*(1.60*10^-19 c) / (2*1.76*10^-11 m) = 1.96*10^-17 Joules

1.96*10^-17 Joules = about 122 electron volts (eV), which is probably a better answer.