25: Problem 25.51

INTRO:
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton.

Part A: 
How many revolutions per second does the electron make?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
6.6×10¹⁵ rev/s

UPDATE, 09/12/2016

The first thing I always do is convert my givens to SI units 

r = 0.053 nm = 5.3×10^-2 nm = 5.3×10^-11 m

Now. We know an ELECTRON is moving around a PROTON, so we can look up the charges for these two items

q₁ = q_PROTON = 1.602×10^-19 C
q₂ = q_ELECTRON = -1.602×10^-19 C

Next we need to do a basic dynamics problem. 

There is an electromagnetic force pulling the electron around the neutron. F = ma
the acceleration is tangential to the orbit, so a = a_n = ω²⋅r
ω = v/r ∴ ω² = v² / r² ∴ a_n = v² / r² ⋅ r = v² / r
⇒ F = m⋅a_n = m⋅v² / r

Since we're trying to find the velocity and know everything else except the force (and this entire section has to do with Force & Coulomb's Law...)

F = k_e⋅q₁⋅q₂/r²
∴ k_e⋅q₁⋅q₂/r² = m⋅v² / r

⇒ k_e⋅q₁⋅q₂ / (r⋅m) = v²
∴ v = (k_e⋅q₁⋅q₂ / (r⋅m) )^½

​

The mass of an electron is m_e = 9.109×10^-31 kg
v = (k_e⋅q₁⋅q₂ / (r⋅m) )^½ 

= SQRT{ (8.99×10⁹ Nm²/C²) ⋅ (1.602×10^-19 C) ⋅ (-1.602×10^-19 C) ⋅ 1/(9.109×10^-31 kg) ⋅ 1/(5.3×10^-11 m) ⋅ (kg⋅m/s²)/N }

= SQRT{ -4.779 × 10¹² m² / s² }

--> v = 2.186 × 10⁶ m/s

convert to rev/s -> 1 rev / (2 π r) = 1 rev / 3.33×10^-10 m

v = 6.565 rev/s

*update (07/26/2017 @ Kirtan Jani) * v = 6.565×10¹⁵ rev/s =  6.6×10¹⁵ rev/s

This was wrong because I missed an exponent while explaining. However, if you look at the purple highlighted answer above.... you can see that that was only wrong because you were being too GD judgemental.  Remember this is a free service bruh..