College Physics Problem 2.15

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\times 10^{-10}\:m in diameter.
(a) If the average speed of the electron in this orbit is known to be 2.20\times 10^6\:m/s, calculate the number of revolutions per second it makes about the nucleus.
(b) What is the electron’s average velocity?

SOLUTION:

A) The formula to be used is

average\:speed=\frac{distance}{time}

Rearraging the formula–solving for the distance

distance=speed\times time

Substituting the given values

d=\left(2.20\times 10^6\:\frac{m}{s}\right)\left(1\:s\right)\:

d=2.20\times 10^6

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

1\:revolution=2\pi r

1\:revolution=2\pi \left(\frac{d}{2}\right)=\pi d

1\:revolution=\pi \left(1.06\times 10^{-10}\:m\right)

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

no.\:of\:revolutions=\frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)}

no.\:of\:revolutions=6.61\times 10^{15}

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

v=0\:m/s

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