Schrödinger's Equation and the Hydrogen Atom — Question 6

Question 6

The angular probability of a p orbital concentrates in polar cones

Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. A hydrogen state with $\ell=1,m_\ell=0$ has normalized angular density $|Y_{10}|^2=3\cos^2\theta/(4\pi)$, where $\theta$ is measured from +z. Integrate with $d\Omega=\sin\theta\,d\theta\,d\phi$.

1. Derive the probability inside the cone $0\le\theta\le\theta_0$.

2. Evaluate it at $\theta_0=\pi/3$ and find the probability in the union of the two corresponding polar cones.

Question 6 - Solution

The angular probability of a p orbital concentrates in polar cones

(a) Method and calculation

$$P_{+}=\frac3{4\pi}\int_0^{2\pi}d\phi\int_0^{\theta_0}\cos^2\theta\sin\theta\,d\theta.$$$$\boxed{P_+=(1-\cos^3\theta_0)/2}.$$

(b) Interpretation and further calculation

$$\theta_0=\pi/3:\quad\boxed{P_+=(1-1/8)/2=7/16}.$$$$\boxed{P_{\mathrm{two\ cones}}=2P_+=7/8}.$$

Physical check

The radial factor is already normalized and cancels from this angular probability. The equatorial plane is an angular node, while equal upper and lower cones have equal probability.