# What determines the probability pattern of an orbital?

##### 1 Answer

*It just turned out that way. We can't define equations to influence nature, only describe it.*

You can imagine anything you want for what "determines" the probability distribution, but ultimately, nature *turned out* that way, and we described these orbitals **after-the-fact**.

*DISCLAIMER: This is going to be a long answer.*

**WHAT IS THE WAVE FUNCTION FOR AN ORBITAL?**

Erwin Schrodinger published the **wave function**

For each orbital, its *radial density distribution* describes the regions with particular probabilities for finding an electron in that particular orbital.

In general, the wave function for spherical harmonics coordinates can be written as:

#\mathbf(psi_(nlm_l)(r,theta,phi) = R_(nl)(r)Y_(l)^(m_l)(theta,phi))# where:

#r# ,#theta# , and#phi# are spherical coordinates:

#psi_(nlm_l)# is awave functionthat can be constructed to describe what the orbital's electron distribution looks like.

It depends on the quantum numbers#\mathbf(n)# ,#\mathbf(l)# , and#\mathbf(m_l)# .

#R_(nl)# is theradial componentof the wave function, describing the variation in the distance from the center of the orbital (the radius!).

It depends on the quantum numbers#\mathbf(n)# and#\mathbf(l)# .

#Y_(l)^(m_l)# is theangular componentof the wave function, describing the aspects of the orbital electron distribution that can possibly give it anon-sphericalshape.

It depends on the quantum numbers#\mathbf(l)# and#\mathbf(m_l)# .

**OKAY, WHAT DOES IT HAVE TO DO WITH PROBABILITY PATTERNS??**

Focus on the

When we plot the radial density distribution (or "probability pattern") of an orbital, we plot

The wave functions themselves, as we defined them, generate these "probability patterns".

**HOW DO I KNOW YOU'RE NOT MAKING THIS UP?**

Well, with an example, I guess. This will give a result that should prove familiar.

The *simplest* example of an orbital wave function (used by **the one for the** **orbital**, which happens to be:

#color(blue)(psi_(1s) = R_(10)(r)Y_(0)^(0)(theta,phi)#

#= stackrel(R_(10)(r))overbrace((2Z^"3/2")/(a_0^"3/2")e^(-"Zr/"a_0))*stackrel(Y_(0)^(0)(theta,phi))overbrace(1/sqrt(4pi))#

#= color(blue)(1/sqrt(pi) (Z/(a_0))^"3/2" e^(-"Zr/"a_0))# where:

#Z# is theatomic numberof the atom.#a_0# is the sameBohr radiuswe just defined as#5.29177xx10^(-11) "m"# .

In order to plot the "probability pattern", we have to grab what we need and turn it into

#4pir^2(R_(nl)(r))^2#

#= 4pir^2(stackrel(R_(nl)(r))overbrace((2Z^"3/2")/(a_0^"3/2")e^(-"Zr/"a_0)))^2#

#= 4pir^2(4Z^3)/(a_0^3)e^(-"2Zr/"a_0)#

*This is the function that corresponds to the red curve in the diagram above.*

Now, since we have the plot for **the region of highest electron density**, which is just the highest point in the

So...

- Take the derivative with respect to
#r# . - Set it equal to
#0# , since the peak has a slope of#0# .

#=> 16pi(Z/(a_0))^3d/(dr)[r^2e^(-"2Zr/"a_0)]#

Utilize the product rule:

#=> 16pi(Z/(a_0))^3[(-2Z)/(a_0)r^2e^(-"2Zr/"a_0) + 2re^(-"2Zr/"a_0)]#

Now setting it equal to

#0 = cancel(16pi(Z/(a_0))^3)[(-2Z)/(a_0)r^2e^(-"2Zr/"a_0) + 2re^(-"2Zr/"a_0)]#

#0 = (-2Z)/(a_0)r^2e^(-"2Zr/"a_0) + 2re^(-"2Zr/"a_0)#

Although

#0 = ((-2Z)/(a_0)r^2 + 2r)cancel(e^(-"2Zr/"a_0))#

#0 = (-2Z)/(a_0)r^2 + 2r#

#cancel(2)r = (cancel(2)Z)/(a_0)r^2#

#r = (Z)/(a_0)r^2 => color(green)(r = (a_0)/Z)# ...for a

Hydrogen-likeatom (e.g.#"He"^(+)# ,#"Li"^(2+)# , etc).

But for the Hydrogen atom,

#\mathbf(color(blue)(r = a_0 = 5.29177xx10^(-11) "m"))# .

*That tells us that the electron in the* *orbital of Hydrogen atom is most often one bohr radius distance away from the center of the nucleus, just as we would expect. Great!*