Bohr's Radii

According to the Bohr model, the hydrogen atom consists of a proton of charge +e and an electron in circular orbit around it with charge -e. The electron can exist only in orbits for which the angular momentum is an integral multiple of (h/2$\pi$) = where h is Planck's constant. The angular momentum of a mass m in a circular orbit with radius r is

L = n

= Iw = mr²w

so that Bohr's condition requires

m r_n²w = m v r_n = n

; n = 1, 2, 3, ...

where r_n is the orbit radius corresponding to the integer n. If we now solve for v and then square v, we get

v² = [(n

)/(mr_n)]².

From Newtonian mechanics entering the derivation of the Kepler Laws (this analogy works, because the gravitational and electrostatic forces have the same dependence on r) we know that the centripetal force is provided by the Coulomb attraction between the electron and the nucleus (charge Ze) and is given by

k (Ze) e / r² = m v²/ r

so we get

v²= (kZe²) / (mr).

From this result, we can obtain the radius of the n^th orbit by substituting the classical expression for v² with Z =1 into the Bohr quantum model

r_n = (n²

²) / (mke²)

where k = 9.0·10⁹ N·m²/C².

We can rewrite this formula as r_n = n²a_o where

a_o = (

²)/(mke²) = 5.29·10^-11 m

which is called the Bohr radius.

In the drawing to the right, the first four orbits are drawn to scale. The orbit corresponding to n = 1 has the radius r₁ = a_o. The electron may not reside in any orbit between these orbits. The radii are quantized meaning that when the electron is in one of these orbits, it does not radiate away energy by gradually slowing down.