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*}

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity

(a) for each of the three intervals and

(b) for the entire motion.

Solution:

Part A

The average velocity for each interval is computed using the formula

\overline{v}=\frac{\Delta x}{\Delta t}

For the first interval

 \overline{v_1}=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the second interval

\overline{v_2}=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the third interval

\overline{v_3}=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For the entire motion, we need displacement from the beginning to the end.

\begin{align*}
\overline{v}& =\frac{\Delta x}{\Delta t} \\
& =\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}} \\
& =\frac{33\:\text{m}}{9.45\:\text{s}} \\
& =3.49\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light 3.00×10⁸ m/s.

Solution:

The total distance traveled is computed using the formula

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

Therefore, the total distance traveled is

\begin{align*}
\text{distance} & =\left(3.00\times 10^8\:\text{m/s}\right)\left(2.56\:\sec \right) \\
& =768\:000\:000\:\text{meters}
\end{align*}

Hence, the distance between the Earth and Moon is equal to the total distance traveled divided by 2.

\begin{align*}
\text{Earth-Moon Distance} & =\frac{768\:000\:000\:\text{m}}{2} \\
& =384\:000\:000\:\text{meters} \\
& = 384 \ 000 \  \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The distance between the Earth and Moon is approximately 384 thousand kilometers.

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by 3.84×10⁶ m(1%)?

Solution:

From the formula \overline{v}=\frac{\Delta x}{\Delta t}, we can solve for \Delta t as follows

\begin{align*}
\Delta \text{t} & = \frac{\Delta x}{\overline{v}} \\
& = \frac{3.84\times 10^6\:\text{m}}{4\:\text{cm/year}}\times \frac{100\:\text{cm}}{1\:\text{m}} \\
& =96\:000\:000\:\text{years} \\
& =96.0\times 10^6\:\text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

It will take about 96 million years.