Question

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proton. In the lowest-energy orbit the distance from the proton to the electron is 5.29 x 10-11 m, and the linear speed of the electron is 2.18 x 106 m/s. a) What is the angular velocity of the electron? b) How many orbits about the proton does the electron make each second? c) What is the electron❝s centripetal acceleration?

Answer 1

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

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

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

b)

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

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:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s

c)

The centripetal acceleration of an object in circular motion is

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circle

For the electron, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the centripetal acceleration is

$a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2$

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