state of hydrogen

logistic_guy

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For \(\displaystyle n = 2, l = 0\) state of hydrogen, what is the probability of finding the electron within a spherical shell of inner radius \(\displaystyle 4.00r_0\) and outer radius \(\displaystyle 5.00r_0\)?
 
For \(\displaystyle n = 2, l = 0\) state of hydrogen, what is the probability of finding the electron within a spherical shell of inner radius \(\displaystyle 4.00r_0\) and outer radius \(\displaystyle 5.00r_0\)?
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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.

💪😈😈
 
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