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

• A. 1.2×10^14 rev/s
• B. 4.2×10^16 rev/s
• C. 6.6×10^15 rev/s
• D. 3.3×10^12 rev/s

Answer: (C)

The electron in a circular orbit is held in place by the Coulomb attraction, which acts as the centripetal force. So mv^2/r = ke^2/r^2, giving the orbital speed v = sqrt( ke^2 /( m r ) ). The revolutions per second is f = v/(2πr). Combining these, f = (1/(2π)) sqrt( ke^2 /( m r^3 ) ).

Plug in r = 0.053 nm = 5.3×10^-11 m, e = 1.60×10^-19 C, m = 9.11×10^-31 kg, and k ≈ 8.99×10^9 N·m^2/C^2. Compute ke^2 ≈ 2.31×10^-28, r^3 ≈ 1.49×10^-31, and m r^3 ≈ 1.36×10^-61. The ratio is about 1.70×10^33, its square root ≈ 4.1×10^16, and dividing by 2π gives f ≈ 6.6×10^15 s^-1.

So the electron makes about 6.6×10^15 revolutions per second.