4.3.4.1 So­lu­tion hydd-a

Ques­tion:

At what dis­tance $r$ from the nu­cleus does the square of the ground state wave func­tion be­come less than one per­cent of its value at the nu­cleus? Ex­press it both as a mul­ti­ple of the Bohr ra­dius $a_0$ and in Å.

An­swer:

The square wave func­tion is

$$\vert\psi_{100}(r)\vert^2 = \frac{1}{\pi a_0^3} e^{-2r/a_0}$$

and the value at the nu­cleus $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is then

$$\vert\psi_{100}(0)\vert^2 = \frac{1}{\pi a_0^3}$$

For the value at $r$ above to be one per­cent of this, you must have

$$e^{-2r/a_0} = 0.01$$

or tak­ing log­a­rithm, $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2.3 $a_0$. Ex­pressed in Å, $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.53 Å, so $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.22 Å.