- #1

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P(r) = |rR(r)|^2 = [r^2 / (8a_o^3)] [(2-r/a_o)^2] exp(-r/a_o)

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- Thread starter jaejoon89
- Start date

- #1

- 195

- 0

P(r) = |rR(r)|^2 = [r^2 / (8a_o^3)] [(2-r/a_o)^2] exp(-r/a_o)

- #2

lanedance

Homework Helper

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the probability of finding the electron in between r and r = dr is P(r).dr

So the integral of P(r).dr must be 1.

Although the limit of the integral goes from 0 to infinity, P(r) is only really non-zero for a few a_0. Thinking of the intergal in terms of area, as the span of r is so small, P(r) must be large for the integral to add up to one

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