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*}
Advertisements
Advertisements