Schrödinger's Equation and the Hydrogen Atom — Question 3

Question 3

Radial integration gives the probability inside a sphere and outside another

Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. For the 1s state use $P_r(r)=4r^2e^{-2r/a}/a^3$. The probability inside radius R is the integral from zero to R.

1. Derive a closed expression for $P(r<R)$.

2. Find $P(r<a)$ and $P(r>2a)$.

Question 3 - Solution

Radial integration gives the probability inside a sphere and outside another

(a) Method and calculation

$$u=2r/a\Rightarrow P(r<R)=\frac12\int_0^{2R/a}u^2e^{-u}du.$$$$\boxed{P(r<R)=1-e^{-2R/a}\left(1+\frac{2R}a+\frac{2R^2}{a^2}\right)}.$$

(b) Interpretation and further calculation

$$\boxed{P(r<a)=1-5e^{-2}=.323324}.$$$$\boxed{P(r>2a)=13e^{-4}=.238103}.$$

Physical check

The radius a is the mode of the radial distribution, not a boundary enclosing most of the electron probability. The distribution has a substantial exterior tail.