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 = r_b




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

Thanks

 

[quantumdude]

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 Ψ(r) should be in your book. You already stated the definition of the probability density (|Ψ|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=r_b (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.

 

[quantumdude]

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:

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

No radial coordinate.

 

[quantumdude]

Originally posted by frankR
Halliday gives:

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

No radial coordinate.

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:

Ψ(r) = 1/(sqrt[[pi]a^3/2) e^-r/a

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

 

[quantumdude]

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

 

[quantumdude]

And you got your answer how...?

 

[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.

 

[quantumdude]

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[/color]?

You are working in meters, and the answer was given in inverse cubic nanometers[/color].

 

[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.

 

[frankR]

Oh BTW: Thanks for your help Tom.