Atoms and Quantum Mechanics

Question

Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? When we talk about atomic size it can be anything from some finite value to infinity as the electrons can be excited to any value in accordance with the conservation of angular momentum?

Answer 1

Classical Physics: Thinking in the context of classical physics (Newtonian mechanics+Maxwell Theory of Electromagnetism), the atom is not stable due to the radiation, however, assume the electron does not radiate. Then, solving the Newton equations in the potential $V(r)=-\frac{ke^2}{r}$ with bounded movements, leads to elliptic circuits for electron. In the circular case, there is the radius $r_c=\frac{L^2}{mke^2}$ where $L$ is the angular momentum.

Quantum Mechanics: In the context of quantum mechanics, solving the Schrödinger equation for the wave function $\psi$ leads to a specific length called Bohr radius $a_0$. The Hilbert space of the solutions tensor product to spin space of electrons, has (Hilbert) basis indexed by four numbers: $n$, $l$, $m$, and $m_s$. It is known that the wave function $\psi_{n,l,m,m_s}$ is $e^{-\frac{r}{n a_0}}$ times some polynomials in $r$ and $Y_{l,m}(\theta,\phi)$ that are known as spherical harmonics. For s-orbital ($l=0,m=0$) Bohr radius has a specific characteristic: the most probable radius of $|\psi|^2$ occurs at $r=a_0$. So again, there is a length which is very probable for the electron to be.

Chemical bonds: In the context of chemical bonds, when two atoms of the same type makes a bond ($X_2$), the typical length for the atomic radius then would be the expected value of bonds in quantum sense.

So, in quantum mechanics, the electron has no preferred radius, however, there are some probable radius like Bohr radius which can be used to adopt an order of magnitude to the radius of atom.