Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom

Question

Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ($\sim 10^{-10}\ 	ext{m}$).
(a) Construct a quantity with the dimensions of length from the fundamental constants $e$, $m_e$, and $c$. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves $c$. But energies of atoms are mostly in non-relativistic domain where $c$ is not expected to play any role. This is what may have suggested Bohr to discard $c$ and look for 'something else' to get the right atomic size. Now, the Planck's constant $h$ had already made its appearance elsewhere. Bohr's great insight lay in recognising that $h$, $m_e$, and $e$ will yield the right atomic size. Construct a quantity with the dimension of length from $h$, $m_e$, and $e$ and confirm that its numerical value has indeed the correct order of magnitude.

Accepted Answer

> [!TLDR]
> The only length buildable from $e$, $m_e$, $c$ ($\approx2.8	imes10^{-15}$ m, the classical electron radius) is $10^5$ times too small to be an atom; replacing $c$ with $h$ gives $\approx5.3	imes10^{-11}$ m -- the Bohr radius -- exactly the right order of magnitude, which is precisely the insight that led Bohr to build his model around $h$ rather than $c$.

**Step 1 -- What quantity, built from $e$, $m_e$, $c$, has dimensions of length?**
The Coulomb potential energy is $U = \dfrac{e^2}{4\pi\varepsilon_0r}$, so the combination $\dfrac{e^2}{4\pi\varepsilon_0}$ has dimensions of (energy $	imes$ length). Dividing by an energy -- the only energy nature offers us from $m_e$ and $c$ alone is the rest energy $m_ec^2$ -- leaves a pure length:
$$\ell_1 = \frac{e^2}{4\pi\varepsilon_0m_ec^2}$$
This combination is essentially forced: it's the only way to cancel every unit except length using just these three constants.

**Step 2 -- Evaluate $\ell_1$ numerically.**
$$\frac{e^2}{4\pi\varepsilon_0} = (8.99	imes10^9)(1.6	imes10^{-19})^2 \approx 2.30	imes10^{-28}\ 	ext{J m}$$
$$m_ec^2 = (9.11	imes10^{-31})(3	imes10^8)^2 \approx 8.2	imes10^{-14}\ 	ext{J}$$
$$\ell_1 = \frac{2.30	imes10^{-28}}{8.2	imes10^{-14}} \approx 2.8	imes10^{-15}\ 	ext{m}$$
This is the **classical electron radius** -- but it's about $10^5$ times *smaller* than the known atomic size ($\sim10^{-10}\ 	ext{m}$). It also explicitly involves $c$, a signature of relativistic physics -- yet atomic binding energies ($\sim$ eV) are minuscule compared to $m_ec^2$ ($\sim0.5\ 	ext{MeV}$), so atoms are a thoroughly non-relativistic problem. Both of these are strong hints that $c$ is the *wrong* constant to be building atomic size from.

**Step 3 -- Try $h$, $m_e$, $e$ instead.** …

Concept understanding — Scale Analogy

Scale Analogy

Why we need an analogy at all

The intuition: the atom is mostly empty space

Making the number imaginable

Density: the flip side of the analogy

Why this formula?

Scale Analogy

The Sizes Involved

Bringing It to Human Scale

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