Answer:
a) The electron gives 6.62x10¹⁵ revolutions in 1 second (hertz) around the nucleus.
b) The electron’s average velocity is 2.20x10⁶ m/s.
Explanation:
(a) If the average speed of the electron in this orbit is known to be 2.20×10^6m/s, calculate the number of revolutions per second it makes about the nucleus.
The number of revolution per time is define as frequency:
[tex]f = \frac{rev}{s}[/tex]
f can be related with the period by means of the next expression:
[tex]f = \frac{1}{T}[/tex] (1)
So it is necessary first to find out the period. That can be done by using the definition of average velocity in a circular motion:
[tex]v = \frac{2\pi r}{T}[/tex] (2)
Where r is the radius of the trajectory and T is the orbital period.
Equation (2) can be rewrite in terms of T:
[tex]T = \frac{2\pi r}{v}[/tex] (3)
As it can be see is need it the radius of the circular orbit.
The diameter can be expressed as:
[tex]d = 2r[/tex]
[tex]r = \frac{d}{2}[/tex]
Remembering that the radius is the distance to any part of the circumference from a central point.
[tex]r = \frac{1.06x10^{-10} m}{2}[/tex]
[tex]r = 5.3x10^{-11} m[/tex]
Replacing the value of r in equation (3) it is get:
[tex]T = \frac{2\pi (5.3x10^{-11} m)}{(2.20x10^{6} m/s)}[/tex]
[tex]T = 1.51x10^{-16} s[/tex]
Now the orbital period (T) can be replace in equation (1)
[tex]f = \frac{1}{1.51x10^{-16} s}[/tex]
[tex]f = 6.62x10^{15} rev/s[/tex] or [tex]f = 6.62x10^{15} hertz[/tex]
Hence the electron gives 6.62x10¹⁵ revolutions in 1 second around the nucleus.
b) What is the electron’s average velocity?
The average velocity is given in the statement, however it can be calculated using equation (2) and replacing the value found for the orbital period.
[tex]v = \frac{2 \pi r}{T}[/tex]
[tex]v = \frac{2 \pi (5.3x10^{-11} m)}{1.51x10^{-16} s}[/tex]
[tex]v = 2.20x10^{6} m/s[/tex]
So the electron’s average velocity is 2.20x10⁶ m/s.
