Schrödinger's Equation and the Hydrogen Atom — Question 5

Question 5

The 2s radial node separates two radial-probability maxima

Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. The 2s radial amplitude is proportional to $(2-r/a)e^{-r/(2a)}$. Thus with $u=r/a$, the radial probability is proportional to $u^2(2-u)^2e^{-u}$.

1. Locate the positive-radius radial node and state the number of angular nodes.

2. Find both positive radial-probability maxima.

Question 5 - Solution

The 2s radial node separates two radial-probability maxima

(a) Method and calculation

$$2-r/a=0\Rightarrow\boxed{r_{\mathrm{node}}=2a},\qquad\boxed{\ell=0\Rightarrow0\text{ angular nodes}}.$$$$\frac d{du}\ln P_r=\frac2u-\frac2{2-u}-1=0\quad(u\ne0,2).$$

(b) Interpretation and further calculation

$$2(2-u)-2u-u(2-u)=0\Rightarrow u^2-6u+4=0.$$$$\boxed{r_{\max,\pm}=(3\pm\sqrt5)a=.763932a,\ 5.23607a}.$$

Physical check

The origin is a radial-density zero from the shell-area factor, not a radial node of the 2s amplitude. The node at 2a divides two positive probability lobes.