# Atomic radius definition disproved - Bohr's model to Schroedinger's wave equation

According to the Schrödinger wave equation, Bohr's model of atomic orbitals is disproved and a new formalism that describes atomic orbitals is introduced.

It follows that the definition of atomic radius - "the distance between the center of the nucleus to the outermost orbit" - doesn't make sense, given that the new theory does not describe atomic orbitals as discrete entities in space.

If an orbital has no exact boundary, then what exactly is the definition of atomic radius and how can I calculate it?

• The most probable distance of the distribution is used. – MaxW Sep 15 '16 at 16:30
• How do you measure radius of elements in their elemental stste??? – Bharat Sep 15 '16 at 17:20
• I'm going to keep this simple and not give a book length explanation. A Cl atom would have many kinds of radii. First what would be the radius of a Cl atom out in free space. Second what would the nuclear distance be between Cl atoms in the $\ce{Cl2}$ molecule? Half that distance would be another radius. Third what would be the distance be between a $\ce{Cl^-}$ ion and various cations such as $\ce{Na^+}$, $\ce{K^+}$, $\ce{Rb^+}$, $\ce{Ca^{2+}}$ and so forth. // For diatomic gases you could measure the distance by the rotational moments of the molecules. For solids you'd use x-ray diffraction. – MaxW Sep 15 '16 at 18:59

As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system. In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus ($r$) and two angles ($\phi$ and $\theta$). [If you pick a particular $r$, the angles $\phi$ and $\theta$ work in a way similar to how we locate points on the surface of the earth (latitude and longitude).]

This being said, the image below may clarify things and answer your question (source): The graph deals only with s atomic orbitals of hydrogen, the 1s highlighted in red.

See the x-axis? It is measured in multiples of $a_0$ a.k.a. Bohr radius,

$$a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} = \frac{\hbar}{m_e c \alpha}$$

This is the radius obtained from Bohr's theory for the 1s orbital of hydrogen. How can it the useful as a unit in this graph?

Well, the square of a wave function is a probability distribution, $\rho(r, \phi, \theta) = \Psi^2 (r, \phi, \theta)$. If you had a dice with probability distribution $\bar{\rho}(n) = 1/6$ (a fair dice) for each $n = 1,..., 6$, you would get, throwing that dice, on average,

$$\sum_{n=1}^6 \bar{\rho}(n) \times n = \sum_{n=1}^6 \frac{n}{6} \\= \frac{1}{6} + \frac{2}{6} + \cdots + \frac{6}{6} = 3.5$$

So $3.5$ is the expected value of throwing a dice. See how the last equality is just the usual average we're familiar with? But the first one is most useful in our context. Since our variables $r$, $\phi$ and $\theta$ are all continuous, in order to get an average radius, we integrate instead of sum:

$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [\rho(r, \phi, \theta) \times r] r^2 \sin(\phi) d r d \phi d \theta$$

$r^2 \sin(\phi) d r d \phi d \theta$ is just the volume element in spherical coordinates (in our dice example we had an implicit $1$ there as "volume"), what matters is just what is in squared brackets.

Now 1s ($n = 1$, $\ell = 0$) orbitals do not depend on angles, they are perfectly spherical. That means that $\Psi_{n \ell m_\ell}(r, \phi, \theta) = R_{n \ell} (r)$ with $n = 1$, $\ell = 0$ and $R_{n \ell} (r)$ is some function describing the radial electron wave function. Our integration simplifies to:

$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 \sin(\phi) d r d \phi d \theta \\ = \int_0^{2 \pi} \int_0^{\pi} \sin(\phi) d \phi d \theta \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\ = 4 \pi \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\ = \int_0^\infty [ (4 \pi r^2 R^2_{n \ell} (r)) \times r ] d r$$

This surely looks like the expected value of throwing a dice! The variable being averaged is $r$ and the new probability distribution is $\rho_\text{radial} (r) = 4 \pi r^2 R^2_{n \ell} (r)$, called radial probability distribution. That's what's being plotted in the figure above.

If you go further and carry out the last integral ($R_{1 0} (r)$ can be found here), you'll see the answer is just $a_0$, the Bohr radius. That's expected, after all, since Bohr's theory was quite successful for the hydrogen atom. Nothing has been disproved.

Thus, the atomic radius of an atom can be obtained from quantum mechanics by calculating the expected value of the distance between the outermost electron and the nucleus, as MaxW has pointed out.