In a simple model of the hydrogen atom, the electron moves in a circular orbit, of radius 0.053nm, around a stationary proton. how many revolutions per second does the electron make?

f  =  6.57 × 10 ¹⁵H z

Explanation:

Given:

r = 0.053 n m  =  5.3 × 10 ⁻¹¹m   is the electron orbit radius

In order to determine the frequency of the electron motion, we first equate the Coulomb force due to the electrostatic attraction between proton and electron to the centripetal force of the circular motion. We have:

mv²/r² = kq²/r²

Here we want to get v so we isolate it:

v = √ k q ²/m r

We then recall that the speed in uniform circular motion is:

v = 2 π r f

So we replace everything and substitute:

2 π r f  = √ k q ²/ m r

We isolate the frequency here:

f = ( 1 /2 π r ) ×  ( √ k q ²/m r )

We insert the term inside the square root to get a singular term of the form:

f = √ k q ²/4 π ²m r ³

We substitute and use the charge and mass of an electron here:

f = ⎷ ( 9 × 10 ⁹ N m ²/ C ²) ( 1.602 × 10 ⁻¹⁹C )²/4 π ² ( 9.109 × 10 ⁻³¹k g ) ( 5.3 × 10 ⁻¹¹m ) ³

We will get:

f =  6.57 × 10 ¹⁵H z

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

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