- #1

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a) r = 0

b) r = r

_{b}

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

Thanks

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- Thread starter frankR
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- #1

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a) r = 0

b) r = r

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

Thanks

- #2

Tom Mattson

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

- #3

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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.

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

- #4

Tom Mattson

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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

- #5

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|Ψ|

No radial coordinate.

- #6

Tom Mattson

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Originally posted by frankR

Halliday gives:

(|Ψ|^{2}) = A^{2}Sin^{2}(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.

- #7

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Ψ(r) = 1/(sqrt[[pi]a

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

- #8

Tom Mattson

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- #9

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I get: 6.121e^9

- #10

Tom Mattson

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And you got your answer how....?

- #11

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

This must not be the correct formula or something.

- #12

Tom Mattson

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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.

- #13

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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.

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:

- #14

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Oh BTW: Thanks for your help Tom.

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