# College Physics 2.15 – Velocity of Electrons in Planetary Model

## Solution:

### Part A

The formula to be used is

$\displaystyle \text{average speed}=\frac{\text{distance}}{\text{time}}$

Rearraging the formula–solving for the distance

$\text{distance}=\text{speed}\times \text{time}$

Substituting the given values

$\text{distance}=\left(2.20\times 10^6\:\text{m/s}\right)\left(1\:\text{s}\right)=2.20\times 10^6\:\text{meters}$

We know the total distance traveled in 1 sec. We know that in 1 revolution, the total distance is

$\displaystyle 1\:\text{revolution}=2\pi \text{r}=\pi \text{d}=\pi \left(1.06\times 10^{-10}\text{m}\right)$

Therefore, the total number of revolutions traveled in 1 second is

$\displaystyle \text{no. of revolutions}=\frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)}=6.61\times 10^{15}$

### Part B

In one complete revolution, the electrons will go back to its original position. Thus, there is no net displacement. Therefore,

$\text{v}=0\:m/s$