# Quick Normalisation Question

White Ink

## Homework Statement

A hydrogen atom in the ground state can be described by the following wavefunction:

$$\Psi(r) = \frac{C}{\sqrt{4\pi}}e^{- \frac{r}{a_{0}}}$$

Normalise this wavefunction.

## The Attempt at a Solution

I did this and got:

$$C = \sqrt{\frac{8\pi}{a_{0}}}$$

I have no way of checking this, so I was wondering if anybody could tell me whether I am right or wrong.

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## Answers and Replies

Homework Helper
Not quite, I don't think. Can you show us how you got that?

White Ink
$$\int^{\infty}_{0}\left|\Psi(r)\right|^{2}dr = 1$$

$$\frac{C^{2}}{4\pi}\int^{\infty}_{0}e^{\frac{-2r}{a_{0}}}dr = 1$$

$$u = \frac{2r}{a_{0}}$$

$$du = \frac{2}{a_{0}}dr$$

$$dr = \frac{a_{0}}{2}du$$

$$\frac{C^{2}}{4\pi}\frac{a_{0}}{2}\int^{\infty}_{0}e^{-u}du = 1$$

Using:

$$\int^{\infty}_{0}x^{n}e^{-x}dx = n!$$

$$\int^{\infty}_{0}e^{-u}du = 0! = 1$$

Thus:

$$\frac{C^{2}}{4\pi}\frac{a_{0}}{2} = 1$$

$$C^{2} = \frac{8\pi}{a_{0}}$$

Proggle
Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.

White Ink
Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.

I know exactly what you mean (integrating in the $$\theta$$ & $$\phi$$ directions becomes equivalent to multiplying to constant in front of the integration with respect to r by $$2\pi^{2}$$, since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.

Homework Helper
I know exactly what you mean (integrating in the $$\theta$$ & $$\phi$$ directions becomes equivalent to multiplying to constant in front of the integration with respect to r by $$2\pi^{2}$$, since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.

In addition there will be an extra factor of r^2 in the integrand, right?

White Ink
In addition there will be an extra factor of r^2 in the integrand, right?

Now that one beats me. I can't see where an additional $$r^{2}$$ would come from.

Ooops, I feel stupid now. I forgot about the metric coefficients and took my volume element to be $$dr$$$$d\theta$$ . At least I won't be making that mistake again. Thanks.