Calculating Probabilities for Ground State Hydrogen Atom

[fluidistic]

Homework Statement

The normalized wave-function corresponding to the ground state hydrogen atom has the form $$\Psi _0 = \frac{1}{\sqrt \pi} \cdot \frac{e^{-r/a_0}}{a_0 ^{3/2}}$$ where $$a_0$$ is Bohr's radius.
1)Graph the density of probability to find the electron in a spherical edge between r and r+dr. (Sorry "spherical edge" might not be the words but I hope you can understand. If not, let me know).
2)Calculate the probability that an electron in this state can be found at a greater distance than $$a_0$$.
3)Calculate the most probable distance between the electron and the nucleus.
4)Calculate $$\langle r \rangle$$.

Homework Equations

$$\int _0 ^{\infty} \Psi _0 ^2 dr =1$$. I omit the absolute value since the function seems positive on all its interval of definition.

The Attempt at a Solution

For 2), my intuition tells me "$$1-\int _0 ^{a_0} \Psi _0 ^2 dr$$" or "$$\int _{a_0} ^ \infty \Psi _0 ^2 dr$$".
I'm having a hard time with 1).
I'm trying to find out the probability to find the electron between any radius of length dr. By any radius I mean a radius of length dr that is far away from the nucleus by an arbitrary number.
I've done an attempt on it, lead to nothing. I called $$C=\frac{1}{\sqrt \pi a_0 ^{3/2}}$$ and so what I'm looking for should be of the form $$C^2 \int _r^{r+dr}e^{-\frac{2r}{a_0}} dr$$. I must be confusing variables calling all r's... Anyway I could never get rid of "dr" and I know my result must depend on r and not on dr.

Any insight is appreciated.

 

[vela]

Homework Equations

$$\int _0 ^{\infty} \Psi _0 ^2 dr =1$$. I omit the absolute value since the function seems positive on all its interval of definition.

This isn't quite correct. In spherical coordinates, the normalization condition is

$$\int_0^\infty \int_0^\pi \int_0^{2\pi} \Psi^2(r,\theta,\phi) r^2\sin\theta\,d\phi\,d\theta\,dr = 1$$

Your answer for part 1 will depend on dr.

 

[fluidistic]

This isn't quite correct. In spherical coordinates, the normalization condition is

$$\int_0^\infty \int_0^\pi \int_0^{2\pi} \Psi^2(r,\theta,\phi) r^2\sin\theta\,d\phi\,d\theta\,dr = 1$$

Your answer for part 1 will depend on dr.

Ah yes you're right. Thanks for pointing this out.
Actually I've shown in my draft that $$\int_0^\infty \int_0^\pi \int_0^{2\pi} \Psi^2(r,\theta,\phi) r^2\sin\theta\,d\phi\,d\theta\,dr = 1$$. The algebra involved 2 integrations by part.
So I've done the same algebra but with the integral limits r' and dr' instead of 0 to $$\infty$$.
What I get is a huge expression, not sure it simplifies nicely.
For 1), I get $$-\frac{1}{\sqrt a_0} \{ (r'^2 +2r' dr')e^{-\frac{2(r'+dr')}{a_0}} - r'^2 e ^{-\frac{2 r'}{a_0}} - a_0 \{ (r'+dr') e^{- \frac{2 (r'+dr')}{a_0}} + r' e^{-\frac{2r'}{a_0}} + \frac{a_0}{2} \left [ e^{-\frac{2 (r'+dr')}{a_0}} - e^{- \frac{2r'}{a_0}} \right ] \} \}$$.
I must graph that . Also I depreciated the terms $$dr'^2$$ as I've been thaught in this forum some months ago.

 

[vela]

Well, you could keep going, and as long as you don't make an algebra mistake, it will eventually work out. But it would be a lot quicker to recognize that

$$\int_r^{r+dr} f(r)dr = f(r)dr$$

 

[fluidistic]

Well, you could keep going, and as long as you don't make an algebra mistake, it will eventually work out. But it would be a lot quicker to recognize that

$$\int_r^{r+dr} f(r)dr = f(r)dr$$

Oh, I still don't see it. If I take it as true, I reach $$-\frac{2}{a_0 ^{3/2}} (r^2 e^{ \frac{-2r}{a_0} } dr )$$, not sure it's what I should get?

I tried part 2). I get a non sense result (negative probability).
What I did was $$\int_{a_0}^\infty \int_0^\pi \int_0^{2\pi} \Psi^2(r,\theta,\phi) r^2\sin\theta\,d\phi\,d\theta\,dr = e^{-2} \left ( -1 - \frac{3}{a_0} \right )$$.
My intuition tells me I should reach 1/2. Now I must sleep so I'll redo the algebra tomorrow. I'm somehow curious what do you get on this.

 

[vela]

I should have been a little more careful. The probability of finding the electron between r and r+dr will be

$$P(r<R<r+dr) = \int_r^{r+dr} \int_0^\pi \int_0^{2\pi} \Psi^2(r,\theta,\phi) r^2\sin\theta\,d\phi\,d\theta\,dr = f(r) dr$$

where f(r) is the density that you're asked to find. Does that make what you're trying to find in part 1 clearer? Something went wrong with your calculation above since you have a negative sign and the units don't work out. The density should have units of 1/length.

In part 2, your setup is fine. You apparently messed up the integration. You should get 5/e² overall. The angular integrals by themselves should give you a factor of 4π.

 

[vela]

Oh, I still don't see it.

Intuitively, you can consider f(r) to be constant between r and r+dr since dr is infinitesimally small, so the area under the curve will just be the height, f(r), multiplied by the width, dr.

You could also get this result using the fundamental theorem of calculus.

 

[fluidistic]

Ok thank you for the explanation.
Now I indeed reach $$5 e^{-2}$$. I had made an error with the limits of integration (I interchanged theta for phi).
This means the electron has a probability of around 0.68 to be farer than Bohr's radius. Then why does Bohr's model works well to explain transition emission?
Hmm maybe because even though the electron has a greater chance to be farer than Bohr's radius than closer to the radius, the mean value of expectation to find the electron can still be $$a_0$$.
Now looking at part 3). It seems like I'm looking for a maximum, therefore I should find the maximum of $$\Psi _0$$? Because it's also the max of $$\Psi _0 ^2$$ and thus of the probability to find the electron.
Hmm it would give r=0... Ok I'll look into part 1) maybe.