Hydrogen atom (finding electron probability)

[natugnaro]

[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

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

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 ?

 

[dextercioby]

How do you determine "r_{m}" ? From what equation ?

 

[natugnaro]

$$\frac{\partial}{\partial r}(R_{20}(r))=0 \Rightarrow \frac{e^{-r/2a}(-4a+r)}{4\sqrt{2}a^{7/2}}=0 \Rightarrow r_{m}=4a$$

So, rm is four times smaller than a.
Well, I think this is ok, since I'm just trynig to avoid some integral, not build something that suppose to work .

 

[PhysiSmo]

You can also expand the $$(1-\frac{r}{2a})^{2}$$ term, multiply with $$r^2$$ and integrate by parts. Quite painful, but exact!

 

[natugnaro]

You can also expand the $$(1-\frac{r}{2a})^{2}$$ term, multiply with $$r^2$$ and integrate by parts. Quite painful, but exact!

and multiply by e^(-r/a), then I have integral of r^4*e^(-r/a)

 

[natugnaro]

I'll just use $$e^{\frac{-r}{a}} \approx 1-\frac{r}{a}$$. $$r_{m}$$ is small enough for me.
thanks.

 

[PhysiSmo]

I meant that you could integrate by parts the integral

$$\int_0^{r_m} r^4e^{-r/a}dr=r^4\frac{e^{-r/a}}{-1/a}\left|_0^{r_m}-\int_0^{r_m} 4r^3e^{-r/a}\left(-\frac{1}{a}\right)dr=\cdots$$

and so on...

 

[natugnaro]

ah, yes , tahnks again.