# Expectation value r^2 for a radial wave function

• blaksheep423
In summary, the conversation discusses the ground state radial wave function for an electron bound to a proton in a hydrogen atom, given by the equation R10 = (2 / a3/2) exp(-r / a), where r is the distance between the electron and proton and a is a constant. The conversation also includes a sketch of the wave function and finding the expectation value of <r2>, which is evaluated to be 12πa2. The conversation also addresses concerns about the angular dependence of the wave function and the use of LaTeX in writing the equations.
blaksheep423

## Homework Statement

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?

## Homework Equations

i think the equation for <r2> is

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

## 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?

blaksheep423 said:
for a) i am completely blanking on how to sketch the wave function.

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.
blaksheep423 said:
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?

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.

blaksheep423 said:
also, should i be integrating from 0 - a, or -infinity to infinity?

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

blaksheep423 said:

## 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?

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

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

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...

ok, i think I've 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 don't know why that's happening.

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

blaksheep423 said:
<r2> = 12 $$\pi$$ a2

ignore the fact that the pi's are all superscripts. i don't know why that's happening.

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:

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

## What is the expectation value r^2 for a radial wave function?

The expectation value r^2 for a radial wave function is a measure of the average distance of an electron from the nucleus in an atom. It is calculated by squaring the radial distance and multiplying it by the probability density function of the wave function.

## Why is the expectation value r^2 important in quantum mechanics?

The expectation value r^2 is important in quantum mechanics because it provides information about the spatial distribution of electrons in an atom. This is crucial for understanding the behavior and properties of atoms, molecules, and other systems at the atomic and subatomic level.

## How is the expectation value r^2 related to the size of an atom?

The expectation value r^2 is directly proportional to the size of an atom. As the value of r^2 increases, the average distance of an electron from the nucleus also increases, indicating a larger atom. This is a fundamental concept in atomic theory and helps us understand the size and structure of atoms.

## Can the expectation value r^2 be used to predict the behavior of electrons in an atom?

Yes, the expectation value r^2 can be used to predict the behavior of electrons in an atom. It provides valuable information about the probability of finding an electron at a certain distance from the nucleus, which is essential for understanding the electronic structure and chemical properties of atoms.

## How is the expectation value r^2 calculated?

The expectation value r^2 is calculated by integrating the square of the radial distance multiplied by the probability density function of the wave function. This involves using advanced mathematical techniques such as calculus and complex numbers, making it a fundamental concept in quantum mechanics.

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