For \(\displaystyle n = 2, l = 0\) state of hydrogen,
When \(\displaystyle l = 0\), it means \(\displaystyle m = 0\), so the wave function becomes:
\(\displaystyle \psi_{nlm} = \psi_{200} = \frac{1}{\sqrt{32\pi r^3_0}}\left(2 - \frac{r}{r_0}\right)e^{-r/2r_0}\)
Then, the probability is:
\(\displaystyle P = \int P_r(r) \ dr = \int 4\pi r^2|\psi_{200}|^2 \ dr = \int_{4r_0}^{5r_0} r^2\frac{1}{8r^3_0}\left(2 - \frac{r}{r_0}\right)^2e^{-r/r_0} \ dr\)
\(\displaystyle u = \frac{r}{r_0}\)
\(\displaystyle du = \frac{1}{r_0} \ dr\)
\(\displaystyle P = \frac{1}{8}\int_{4}^{5} u^2\left(2 - u\right)^2e^{-u} \ du\)
I am lazy to solve this integral so let us see what Lady Alpha says!
\(\displaystyle P = 0.173\)
This means that the probability is \(\displaystyle 17.3\%\) to find the electron.


