Problem:

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

\begin{align*}
\text{average speed} & =\frac{\text{distance}}{\text{time}} \\
r & = \frac{d}{t}
\end{align*}

Rearranging the formula–solving for the distance

\begin{align*}
d=r\times t
\end{align*}

Substituting the given values for 1 second period

\begin{align*}
d & = \left(2.20\times 10^6\:\text{m/s}\right)\left(1\:\text{s}\right) \\
& =2.20\times 10^6\:\text{meters}
\end{align*}

This is the total distance traveled in 1 sec.

With the given radius, the total distance traveled in 1 revolution is

\begin{align*}
1\:\text{revolution} & =2\pi \text{r} \\
& =\pi \text{d} \\
&=\pi \left(1.06\times 10^{-10}\text{m}\right)
\end{align*}

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

\begin{align*}
\text{no. of revolutions} & = \frac{\text{total distance}}{\text{distance in 1 revolution}} \\
& = \frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)} \\
& =6.61\times 10^{15} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

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

\begin{align*}
\overline{v} & =\frac{\Delta x}{\Delta t} \\
\overline{v} & =0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}