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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

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

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.

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    The tag `complex-integration` does not seem to fit.2012-09-22
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    @Fabian It's a complex problem about integration.2012-09-22
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    The solutions to the Schrödinger equation are _linear combinations_ of such things.2012-09-22
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    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

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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:

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

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    Damn, I was just warming up mu wxmaxima.2012-09-22
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    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