a) The angular velocity of the electron is [tex]4.12\cdot 10^{16} rad/s[/tex]
b) The number of revolutions per second is [tex]6.54\cdot 10^{15}[/tex] rev/s
c) The centripetal acceleration of the electron is [tex]8.98\cdot 10^{22} m/s^2[/tex]
Explanation:
a)
This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.
The angular velocity of an object in uniform circular motion is given by
[tex]\omega = \frac{v}{r}[/tex]
where
v is the linear speed
r is the radius of the trajectory
For the electron orbiting around the proton, we have
[tex]v=2.18 \cdot 10^6 m/s[/tex]
[tex]r=5.29\cdot 10^{-11} m[/tex]
Therefore, the angular velocity is
[tex]\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s[/tex]
b)
The period of revolution of the electron is given by
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = 4.12\cdot 10^{16}rad/s[/tex] is the angular velocity
Substituting,
[tex]T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s[/tex]
The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:
[tex]f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}[/tex] rev/s
c)
The centripetal acceleration of an object in circular motion is
[tex]a=\frac{v^2}{r}[/tex]
where
v is the linear speed
r is the radius of the circle
For the electron, we have
[tex]v=2.18 \cdot 10^6 m/s[/tex]
[tex]r=5.29\cdot 10^{-11} m[/tex]
Therefore, the centripetal acceleration is
[tex]a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2[/tex]
Learn more about circular motion:
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