Answer to Question #156101 in Physics for mj

Question

A gold nucleus emanates an electric field, holding all its electrons in place. How strong is the electric field experienced by the electron from its outermost orbital, located at 136pm? How fast is it moving on its circular orbit, assuming it has a hypothetical plane orbit? How much kinetic energy is gained by the electron?

Answer 1

Calculate the electric field strength:

$$E=\frac{kQ}{r^2},$$

where Q - total charge created by the nucleus:

$$E=\frac{(9\cdot10^9)(79\cdot1.609\cdot10^{-19})}{136\cdot10^{-12}}=6.15\cdot10^{12}\text{ V/m}.$$

Calculate how fast it is moving:

$$F=Ee=m_e\frac{v^2}{r},\\\space\\ v=\sqrt{\frac{Eer}{m_e}},\\\space\\ v=\sqrt{\frac{6.15\cdot10^{12}\cdot1.609\cdot10^{-19}\cdot136\cdot10^{-12}}{9.109\cdot10^{-31}}}=12.13\cdot10^6\text{ m/s}.$$

The kinetic energy is

$$E_K=\frac12m_ev^2=5.52\cdot10^{-24}\text{ J}.$$