# Schrödinger wave function for 3s orbital of hydrogen atom [closed]

I am unable to understand what these terms in the wave function represent and how to solve it.

For 3s orbital of hydrogen atom, the normalised wave function is

$$Ψ_\mathrm{3s} = \frac{1}{81\sqrt{3π}}\left(\frac{1}{a_0}\right)^{3/2}\left[27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2}\right]\mathrm e^{\displaystyle\frac{-r}{3a_0}}$$

If distance between the radial nodes is $$d,$$ calculate the value of $$\displaystyle\frac{d}{1.73a_0}.$$

• The individual terms do not represent anything. As for the solution, you might want to find the first radial node, then the second one, then the distance between them. – Ivan Neretin Nov 9 '19 at 6:04

To solve the problem, think about what is the condition for a radial node- what happens to the density $$|\Psi_{\textrm{3s}}|^2$$ at a node? First some definitions: $$r$$ is the distance from the nucleus and $$a_0$$ is a constant distance called the Bohr radius (=0.5291 Å).

The density, given by $$|\Psi_{\textrm{3s}}|^2 = k\left[27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2}\right]^2 \mathrm{e}^{\displaystyle\frac{-2r}{3a_0}}$$ goes to zero (k is some constant which as we'll soon find out is not important) .

How can this go to zero?

There are two ways this can happen: for large r the exponential term will go to zero. At intermediate r the polynomial in brackets can go to zero. The latter is the condition to solve for.

Now what?

We can solve for the zeroes of the expression in brackets, it is a quadratic polynomial in r:$$27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2} = 0$$$$r^2 - 9ra_0 + \left(\frac{27}{2}\right)a_0^2 = 0$$that we know how to solve for two roots using a closed form expression.

The solution for the position of the nodes is

\begin{align} r_{node} &= \frac{9}{2}a_0 \pm \sqrt{\frac{81}{4}a_0^2 - \frac{54}{4}a_0^2 } \\ &= (4.5 \pm \sqrt{27/4})a_0 \\ &= (1.01, 3.76)Å \end{align}

The distance between nodes is then

$$d=\sqrt{27}a_0=5.20a_0$$

so that

$$\frac{d}{1.73a_0}=\frac{5.20}{1.73}=3.00$$

• Obviously a homework problem. . . . – Andrew Nov 9 '19 at 12:55