Quantum Physics: Probability density

frankR

For the ground state of the hydrogen atom, evaluate the probabilty density psi^2(r) and the radial probability density of P(r) for the positions.

a) r = 0
b) r = rb

I confused how this probability function is used. What's the technique here?

Thanks

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Tom Mattson

Staff Emeritus
Gold Member
In this problem, you are only being asked to evaluate functions of a single variable variable at two different points. It's no different than if you were asked to evaluate f(x) and g(x) at x1 and x2 in Algebra II, for instance.

The wavefunction &Psi;(r) should be in your book. You already stated the definition of the probability density (|&Psi;|2), and the definition of the radial probability density is bound to be in your book, too. Compute those functions, and then insert r=0 and r=rb (the Bohr radius). No integration is required.

frankR

Okay, what do I use for A, n, L and x?

If I evaluate x=0 I get 0. But the answer is non-zero.

Tom Mattson

Staff Emeritus
Gold Member
Originally posted by frankR
Okay, what do I use for A, n, L and x?
Whoa: What are A, n, and L? Also, don't you mean r instead of x?

If I evaluate x=0 I get 0. But the answer is non-zero.
If you evaluate what at r=0? The probability density is nonzero there, but the radial probability density is zero.

frankR

Halliday gives:

|&Psi;|2(x) = A2Sin2(n[pi]/L * x), n = 1, 2, 3,...

Tom Mattson

Staff Emeritus
Gold Member
Originally posted by frankR
Halliday gives:

(|&Psi;|2) = A2Sin2(n:pi:/L * x), n = 1, 2, 3,...

That's the wavefunction for a particle in a box. It isn't applicable to the hydrogen atom. You need to look up that wavefunction, which will certainly have a radial coordinate.

frankR

So I want this:

&Psi;(r) = 1/(sqrt[[pi]a3/2) e-r/a

Square that and evaulate, or can I just evaluate and squre?

Tom Mattson

Staff Emeritus
Gold Member
Yes, that's the one. It makes no difference what order you do the evaluating and squaring in. Don't forget that you also have to do it for the radial probability density.

frankR

For some reason the answer I'm getting is no where close to the given answer: 2150(nm)-3

I get: 6.121e^9

Staff Emeritus
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frankR

I just plugged it into my calculator.

Since a=5.29x10^-11m the answer should be very large.

This must not be the correct formula or something.

Tom Mattson

Staff Emeritus
Gold Member
Originally posted by frankR
Since a=5.29x10^-11m the answer should be very large.
You should ask yourself:
It should be very large in what units?

You are working in meters, and the answer was given in inverse cubic nanometers.

frankR

HAHHAHA!

That's what happens when you do physics for 10 hours straight.

Edit: BTW, I've never seen that type of unit used before so my brain must have dismissed it. Last edited:

frankR

Oh BTW: Thanks for your help Tom.

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