QNT-04

Bohr radius

a0 = 4π ε0 ħ² /(m e²) ≈ 0.0529 nm.

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HydrogenBohr 1913

Governing equation

a_n=\dfrac{4\pi\varepsilon_0\hbar^2}{m e^2}\dfrac{n^2}{Z}=a_0\dfrac{n^2}{Z}

$n \in \left[1, 6\right]$

where

n: n (—)
Z: Atomic number (—)
a_n: Orbital radius (nm)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

a_n=\dfrac{4\pi\varepsilon_0\hbar^2}{m e^2}\dfrac{n^2}{Z}=a_0\dfrac{n^2}{Z}

$n \in \left[1, 6\right]$

$n = 1$ $a_n = 0.053$

Along the curve

②

Domain of the variable — blue

n_{\min} = 1,\quad n_{\max} = 6

$n \in \left[1, 6\right]$

$n = 1$ $n = 1$

Along the curve

③

Response curve — green

a_n=\dfrac{4\pi\varepsilon_0\hbar^2}{m e^2}\dfrac{n^2}{Z}=a_0\dfrac{n^2}{Z}

$n \in \left[1, 6\right]$

$n = 1$ $a_n = 0.053$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}a_n}{\mathrm{d}n}

$n \in \left[1, 6\right]$

$n = 1.031$ $a_n^{\prime} = 0.109$

Along the curve

Lecture brief

Historical brief

Planck (1900), Einstein’s photoelectric law, Bohr, de Broglie, Heisenberg and Schrödinger’s 1926 equation rebuilt matter as amplitude. These sheets are the first solvable models: wells, spin, tunneling, uncertainty. This sheet (QNT-04 — Bohr radius) is the form associated with Bohr 1913. Working symbols: $n$ , $Z$ $\rightarrow$ $a_n$ . Quantising angular momentum as n ħ in a Coulomb orbit gives a discrete radius a0 n²/Z, with a0 the hydrogen ground-state radius.

Purpose

Purpose: compute

a_n

from

n

Z

in Quantum mechanics via

a_n=\dfrac{4\pi\varepsilon_0\hbar^2}{m e^2}\dfrac{n^2}{Z}=a_0\dfrac{n^2}{Z}

a0 = 4π ε0 ħ² /(m e²) ≈ 0.0529 nm. Use it when a real quantum mechanics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

n = 1.000\,\mathrm{—}

Z = 1.000\,\mathrm{—}

, the governing relation

a_n=\dfrac{4\pi\varepsilon_0\hbar^2}{m e^2}\dfrac{n^2}{Z}=a_0\dfrac{n^2}{Z}

yields

a_n = 0.05292\,\mathrm{nm}

. Hydrogen-like ion, reduced mass ≈ me. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

n n1 —Atomic number Z1 —

Outputs

Orbital radius a_n0.05292 nm

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Narration of this film

Hydrogen-like ion, reduced mass ≈ me.

Quantising angular momentum as n ħ in a Coulomb orbit gives a discrete radius a0 n²/Z, with a0 the hydrogen ground-state radius.

Reading speed

Watch on YouTube