1
$\begingroup$

The solution to Schrodinger's equation are wave functions $\Psi (r,\theta ,\phi )$ of the form, $\Psi (r,\theta ,\phi )= R(r)\Theta(\theta)\Phi(\phi)$ Where, the probability of finding an electron in a region, V:$0, is found using the volume integral;

$N^2\int_{0}^{\infty }r^2R^2(r)dr\int_{0}^{2\pi }\Phi^2(\phi)d\phi\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta =1$

Now, consider the $2p_{y}$ orbital; $\Psi _{2p_{y}}=\frac{1}{4\sqrt{2\pi a_{0}^{5}}}re^{-\frac{r}{2a_{0}}}sin \theta sin \phi$

And now the question itself; Evaluate the three integrals,

One. $\int_{0}^{\infty }r^2R^2(r)dr$

Two. $\int_{0}^{2\pi }\Phi^2(\phi)d\phi$

Three. $\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta$

and show $\int_{V}^{}\Psi ^2dV=1$

I understand this may be alot but any help would be greatly appreciated. Integration is something I really need to work on. Thanks guys.

  • 2
    Note that in many languages that use diacritical marks (including German) the letters with and without them bear only a historical relationship and don't represent similar phonemes. Thus, leaving them off is about as bad as replacing the letter by a completely different one; that is, when you write "Schrodinger", you might as well write "Schridinger". If you can't produce letters with diacritics on your keyboard, you can always copy them from the Web, e.g. in this case from the Wikipedia article on Schrödinger (which you can find by searching for "Schrodinger" :-).2012-09-22

1 Answers 1

3

From your expression for $\Psi$ one can deduce

\begin{align} R(r) &= re^{-r/(2a_0)} \\ \Phi(\phi) &= \sin\phi \\ \Theta(\theta) &= \sin\theta \\ N &= \frac{1}{4\sqrt{2\pi a_0^5}} \end{align}

so that

\begin{align} &\int_0^{\infty}r^2R^2(r)dr = \int_0^{\infty}r^4e^{-r/a_0}dr = \\ &\quad= -a_0^5\left.\left[ \left(\frac{r}{a_0}\right)^4 +4\left(\frac{r}{a_0}\right)^3 +12\left(\frac{r}{a_0}\right)^2 +24\left(\frac{r}{a_0}\right) +24 \right]e^{-r/a_0}\right|_0^{\infty}=24a_0^5\\ &\int_0^{2\pi}\Phi^2(\phi)d\phi=\int_0^{2\pi}\sin^2\phi d\phi=\\ &\quad=\left.\frac{1}{2}(\phi-\sin\phi\cos\phi)\right|_0^{2\pi}=\pi\\ &\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\int_0^{\pi}\sin^3\theta d\theta=\\ &\quad=\left.\left(-\cos\theta+\frac{1}{3}\cos^3\theta\right)\right|_0^{\pi}=\frac{4}{3} \end{align}

Putting all together:

$ N^2\int_0^{\infty}r^2R^2(r)dr\int_0^{2\pi}\Phi^2(\phi)d\phi\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\frac{1}{16(2\pi a_0^5)}\cdot24a_0^5\cdot\pi\cdot\frac{4}{3}=1 $

  • 0
    The first two integrals don't really require antiderivatives -- $\int_0^\infty r^n\mathrm e^{-r}\mathrm dr=n!$ (by $n$-fold integration by parts, or the definition of the gamma function), and $\sin^2\phi$ averages to $\frac12$ over any quarter period.2012-09-22