Question

research group uses short bursts of laser light to measure the motion of electrons moving within and among atoms. In the planetary model of the atom, the hydrogen atom can be viewed as having a single electron in a circular orbit with a diameter of approximately 10-10 m. If the average speed of the electron in the orbit is known to be 2,200,000 m/s, calculate the number of revolutions per second it makes around the nucleus.

Answer 1

ω = 7 x 10¹⁵ rev/s

Step-by-step explanation:

The linear and angular velocities are related through the following formula:

`v = r\omega`

where,

v = linear speed of electron = 2200000 m/s

r = radius of orbit =
`(diameter)/(2) = (10^(-10)\ m)/(2) = 0.5\ x\ 10^(-10)\ m`

therefore, substituting these values in the formula, we get:

`2200000\ m/s = (0.5\ x\ 10^(-10)\ m)(\omega)\\\omega = (2200000\ m/s)/(0.5\ x\ 10^(-10)\ m) \\`

ω = 4.4 x 10¹⁶ rad/s

`\omega = (4.44\ x\ 10^(16)\ rad/s)(1\ revolution)/(2 \pi\ rad)`

ω = 7 x 10¹⁵ rev/s