College Physics 2.15 – Velocity of Electrons in Planetary Model


The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model, you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10-10 m in diameter.

(a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus.

(b) What is the electron’s average velocity?


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