Solving Normalization Factor for 1s Atomic Orbital of H

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SUMMARY

The normalization factor for the 1s atomic orbital of hydrogen (H) is definitively established as 1/(\Pi a_o^3)^\frac{1}{2}. The wavefunction is expressed as ψ(r) = N exp(-r / a_o). To derive the normalization factor, one must square the wavefunction and integrate using spherical coordinates, specifically applying the volume element dt = r^2 sin Θ dr dΘ dΦ. The integral ∫_{0}^{\infty} x^n e^{-ax} dx = n!/a^{n+1} is crucial for solving this problem.

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Hi all! I hope somebody is able to help me on my way with this question.

I have been asked to show that the Normalization factor for the 1s atomic orbital of H is [tex]1/(\Pi a_o^3)^\frac{1}{2}[/tex]. The wavefunction is [tex]\psi(r) = N exp(-r / a_o)[/tex]

I'm given [tex]dt = r^2 sin \Theta[/tex] and [tex]dr d\Theta d\Phi[/tex] and [tex]\int_{0}^{\infty}x^n e^a^x dx=n!/a^n+1[/tex]

I must admit I'm clueless which direction to go. It was mentioned to me, that squaring the wavefunction is the first step, but I cannot arrive at the given constant. Am I starting off on the wrong foot?

Thanks for any input, in advance.

geronimo
 
Last edited:
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geronimo123 said:
Hi all! I hope somebody is able to help me on my way with this question.

I have been asked to show that the Normalization factor for the 1s atomic orbital of H is [tex]1/(\Pi a_o^3)^\frac{1}{2}[/tex]. The wavefunction is [tex]\psi(r) = N exp(-r / a_o)[/tex]

I'm given [tex]dt = r^2 sin \Theta[/tex] and [tex]dr d\Theta d\Phi[/tex] and [tex]\int_{0}^{\infty}x^n e^ax dx=n!/a^n+1[/tex]

I must admit I'm clueless which direction to go. It was mentioned to me, that squaring the wavefunction is the first step, but I cannot arrive at the given constant. Am I starting off on the wrong foot?

Thanks for any input, in advance.

geronimo
If you make mistakes in a post, you can go back and make changes by clicking on the EDIT button (instead of QUOTE).
 
Sorry about this - I don't know how it managed to post twice. It's my first time working with tex and I was having evident difficulties :) Thanks for the pointer!
 

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