## College Physics (4th Edition)

The speed of the electron in this model is $~5.0\times 10^6~m/s$
We can find the magnitude of the electric force: $F = \frac{k~\vert q_p \vert~\vert q_e \vert}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)~(1.6\times 10^{-19}~C)(1.6\times 10^{-19}~C)}{(5.3\times 10^{-11}~m)^2}$ $F = 4.347\times 10^{-7}~N$ We can find the speed of the electron in this model: $F = \frac{mv^2}{r}$ $v^2 = \frac{F~r}{m}$ $v = \sqrt{\frac{F~r}{m}}$ $v = \sqrt{\frac{(4.347\times 10^{-7}~N)(5.3\times 10^{-11}~m)}{9.109\times 10^{-31}~kg}}$ $v = 5.0\times 10^6~m/s$ The speed of the electron in this model is $~5.0\times 10^6~m/s$