# Homework Help: Expectation value r^2 for a radial wave function

1. Nov 23, 2009

### blaksheep423

1. The problem statement, all variables and given/known data
The ground state (lowest energy) radial wave function for an electron bound to a proton to form a hydrogen atom is given by the 1s (n=1, l=0) wave function:

R10 = (2 / a3/2) exp(-r / a)

where r is the distance of the electron from the proton and a is a constant.

a) Sketch the wave function, and
b) What is the expectation value of <r2> for the electron?

2. Relevant equations
i think the equation for <r2> is

$$\int$$ R*r2R dr
from -infinity to infinity

3. The attempt at a solution

for a) i am completely blanking on how to sketch the wave function. for b) i was able to use an integral table to evaluate <r2> and i got 8 / $$\sqrt{}a$$

however, i have a feeling that my original equation for <r2> is wrong, and should involve some sort of Y and $$\phi$$ and $$\Theta$$
also, should i be integrating from 0 - a, or -infinity to infinity?

2. Nov 23, 2009

### cepheid

Staff Emeritus
It decays exponentially. It's an exponential function. You just have to sketch it, not even plot it. You should have some idea of what it looks like.

It's true that in general you need to evaluate〈ψ|r2|ψ〉using the entire wavefunction, which in this case means doing the full 3D integral. However, in this case, because l = 0, the angular part of the wavefunction, Y00 is constant (i.e. it is independent of θ and ϕ), making things really easy. You can look up what it is.

3. Nov 23, 2009

### cepheid

Staff Emeritus
Well, you should integrate over all possible r values. But you can't have a negative radius, can you?

4. Nov 23, 2009

### jdwood983

You're off by a little bit. In the 1s state, there is actually no angular dependence, but you do need the $4\pi r^2$ coefficient of $dr$. Thus, your integral should be

$$\langle r^2\rangle=\int_0^\infty R^*r^2R\,4\pi r^2dr$$

5. Nov 23, 2009

### blaksheep423

OK, so i understand that the second r2 term came from r d$$\Theta$$ r sin$$\Theta$$ d$$\phi$$, but where does the 4$$\pi$$ come from? $$\phi$$ goes from 0-2$$\pi$$, but what about the extra factor of 2? Is it because the function is symmetric, so you're now integrating from 0- infinity and multiplying the result by 2?

And what happens to sin$$\Theta$$ d$$\Theta$$?

man, i need a laTex tutorial.....

6. Nov 23, 2009

### blaksheep423

ok, i think ive got it now. if anyone sees an error in this post, let me know.

the integral above comes from integrating sin($$\Theta$$) from 0 to $$\pi$$ = 1, and d $$\phi$$ from 0 - 2$$\pi$$ = 2$$\pi$$. then, integrating from 0 - infinity gives the extra factor of 2, yielding

<r2> = (4 $$\pi$$)(4/a3) $$\int$$ r4 exp(-2r/a) dr -------------- from 0 to infinity

for this particular radial wave function. Right?

Well, with my handy table of integrals, i evaluated this to an obscenely complicated function, which because of the exponential ended up evaluating fairly easily from 0 to infinity. i got

<r2> = 12 $$\pi$$ a2

ignore the fact that the pi's are all superscripts. i dont know why thats happening.

7. Nov 23, 2009

### Pengwuino

The $$sin(\theta) d\theta$$ contributes a 2 which results in the $$4 \pi$$

8. Nov 24, 2009

### jdwood983

I get that value as well. As for your work not aligning correctly, this is because you are not writing everything in the tex environment. Rather than writing "tex \int /tex r sup4/sup etc "etc (with braces, I'm dropping them so as to show you what i mean) write "tex \int_{bottom limit}^{upper limit} r^4e^{-2r/a}dx /tex" and it'll look pretty:

$$\langle r^2\rangle=\frac{3\pi}{a^3}\int_0^\infty r^4\exp\left[-\frac{2r}{a}\right]dx=12\pi a^2$$

Also:

$$\int_0^\pi\sin[\theta]d\theta\int_0^{2\pi}d\phi=\int_{-1}^1d(\cos\theta)\int_0^{2\pi}d\phi=(1-(-1))\cdot(2\pi-0)=4\pi$$