Results of Bohr's Theory

Solving out Bohr's postulates for the hydrogen atom gives several concrete, testable results. The allowed stationary states are labelled by the positive integers $n = 1,2,3,\ldots$, called principal quantum numbers. Their radii follow $r_n = n^2 a_0$, where $a_0 = 52.9$ pm; the smallest orbit ($n=1$), called the Bohr radius, therefore has radius 52.9 pm. The energy of the stationary state of principal quantum number n is $E_n = -R_H\left(\frac{1}{n^2}\right)$, where $R_H$, the Rydberg constant expressed in energy units, equals $2.18\times10^{-18}$ J; working this out for the lowest state (the ground state, $n=1$) gives $E_1 = -2.18\times10^{-18}\ \text{J}$, and for $n=2$ gives $E_2 = -2.18\times10^{-18}\times\frac{1}{4} = -0.545\times10^{-18}$ J — a smaller magnitude, i.e. a less negative, higher energy, exactly as expected for an orbit farther from the nucleus. The negative sign itself is a convention: a free electron at rest infinitely far from the nucleus (n → ∞) is assigned zero energy, and since the electron in an atom is bound MORE tightly than that free, zero-energy state, its energy is lower, i.e. negative — the smaller the value of n, the closer the electron sits to the nucleus and the more negative (lower) its energy becomes. Bohr's theory extends directly to hydrogen-LIKE species — ions such as He⁺, Li²⁺, Be³⁺ that, like hydrogen, have only one electron outside the nucleus — simply by including the nuclear charge Z: $E_n = -2.18\times10^{-18}\left(\frac{Z^2}{n^2}\right)$ J and $r_n = 52.9\left(\frac{n^2}{Z}\right)$ pm. As Z increases (for the same n …