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natugnaro

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**[SOLVED] Hydrogen atom (finding electron probability)**

## Homework Statement

For electron in eigensate of Hydrogen we have these expectation values

<r>= 6a , <r^-1>= 1/4a

(a is Bohr radius)

a) find that eigenstate.

b) find the probability of finding electron in region 0 < phi < Pi/6 , 0 < theta < Pi/8

and 0 < r < rm

where rm is the first minimum of radial wavefunction in which electron is.

## Homework Equations

## The Attempt at a Solution

a) Using <nl|r^-1|nl>=1/((n^2)*a) it follows that n=2

then using <nl|r|nl>=1/2*[3*n^2-l(l+1)]a I got l=0 ,

so the eigenstate must be Psi(200).

b) To find the probability, I have to integrate

[tex]P=\int|\psi_{200}|^{2}r^{2}Sin\theta drd\phi d\theta = \frac{1-Cos(\frac{\pi}{8})}{(2a)^{3}6}\int^{r_{m}}_{0}(1-\frac{r}{2a})^{2}e^{\frac{-r}{a}}r^{2}dr[/tex]

I can do the integral using tables, but can this integral be simplified. I only have to integrate from 0 to rm so if rm is much less than Bohr radius a, I can make an approximation (series expansion of e^x).

I'm not shure is rm<<a ? , also how are rm and a (Bohr radius) related ?

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