I saw online that the radius for each level is defined by the formula:

$$r_n=n^2 r_1$$

Where $r_1$ is approximately $53\ \mathrm{pm}$.

However, on the Barron's AP chemistry book 7th edition, it says "The radii of the other orbits are whole-number multiples of the Bohr radius" (Page 89). It even gives an example of when $n=2$ that the radius is $106\ \mathrm{pm}$.

But that's different from what the formula above says, which when $n=2$ radius should be $212\ \mathrm{pm}$. Is this an error in the book? I'm quite confused by the discrepancy.

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    – user7951
    Dec 20, 2016 at 0:44
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    $\begingroup$ Hmm...looks to me like they're equating the fine-structure constant with the principal quantum number $n$, but that's just from eyeballing the expression you've written. That said, if it is indeed a formula for estimating the Bohr radii for different values of $n$, then it does look like a mistake in the text. $\endgroup$ Dec 20, 2016 at 0:53

1 Answer 1


Bohr's theory of the atom precedes quantum theory but does describe the Balmer and other lines in atomic hydrogen. I've outlined the calculation so that you can see where the Bohr radius comes from.

He assumed that the electron orbits the nucleus in a circle with a radius r and is determined by the balance of centripetal acceleration and Coulomb attraction towards the nucleus. For an electron of mass $m_e$ and speed $v$ this is $$\frac{m_ev^2}{r}=\frac{e^2/(4\pi\epsilon _0)}{r^2}$$ where e is the charge on the electron and $4\pi\epsilon_0$ puts us into SI units, ( $\epsilon_0$ is the permittivity of free space).

The total energy is the sum of kinetic and potential parts $$E=\frac{m_ev^2}{2}-\frac{(e^2/4\pi\epsilon _0)}{r}$$ and using the first equation gives $$E=\frac{e^2/(4\pi\epsilon _0)}{2r}$$

Bohr had to make the assumption that electrons circling in an orbit do not radiate energy, which is contrary to classical electrodynamics, but he then needed to determine which orbits were allowed. He assumed that the angular momentum is in quantised units of Plancks constant h and a (principal) quantum number n. This gives $$ m_evr = n\frac{h}{2\pi}$$

which when combined with the first equation gives

$$ r=a_0n^2$$

which is the equation you ask about and $a_0$ is conventionally the symbol for the Bohr radius and is 0.0529 nm. It is calculated as
$$a_0 = \frac{\hbar^2}{e^2/(4\pi\epsilon_0)m_e}$$

where $\hbar = h/2\pi$. Re-expressing the energy using $a_0$ gives $$E=-\frac{e^2/(4\pi\epsilon_0)}{2a_0}\frac{1}{n^2}$$ which is the famous Bohr equation and explains the energy levels in hydrogen atoms and so the transitions between them.

  • $\begingroup$ So does that mean it was an error in the book to say radii of other orbitals are whole number multiples of 53pm? $\endgroup$
    – vmoe
    Dec 20, 2016 at 19:18
  • $\begingroup$ I don't have the book you refer to but 'yes' it seems so from what you have written in your question. $\endgroup$
    – porphyrin
    Dec 20, 2016 at 21:52

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