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Radial Distribution Function

  • Thread starter crador
  • Start date
94
11
1. Homework Statement
Hey Guys!

Here's my problem: ψ=2(Z/a)^3/2*e^ρ/2 for the 1s orbital of a hydrogen atom. Write down the radial distribution function expression (P) of a 1s electron and determine the most likely radius.

ρ=2Zr/a
Z nuclear charge
r radius
a Bohr's radius

2. Homework Equations

P=(ψ*)(ψ)∂τ= 4(Z/a)^3*e^ρ*4π*r^2*∂r

3. The Attempt at a Solution

We need to calculate ∂P/∂r

My professor solves r=a/Z which is all well and fine, but in an intermediate step he goes from P=(stuff)*e^(-2Zr/a)*r^2*∂r to ∂P/∂r=∂/∂r((stuff)*e^(-2Zr/a)*r^2)

To me it seems that he has neglected a ∂r and it should read ∂P/∂r=∂/∂r((stuff)*e^(-2Zr/a)*r^2*∂r), in which case I am not sure how to calculate something like this.

What would help me solve this is to know if I am making some stupid mistake, or if there is some rule in the form x=y∂r ---> ∂x=∂(y∂r)=? (my guess would be ∂x=∂y∂r+y(∂^2)r, but this doesn't agree with my professor's tricks, and I haven't taken a PDE class to know my way around)

Thanks guys!
 
94
11
Nevermind, realized my professor's mistake was P=4(Z/a)^3*e^ρ*4π*r^2 and not P=(ψ*)(ψ)∂τ (i.e. P=(ψ*)(ψ)∂τ/dr)
 

BvU

Science Advisor
Homework Helper
12,460
2,789
Note that P is a probability density function! i.e. ##P(r)\ dr## is the probability to find something between ##r## and ##r+dr##.

In other words, the ∂r should not be in your P expression. (there should also be a minus sign in the exponent).

And you don't want ∂P/∂r (it is negative definite).

If you want to calculate the most likely radius, that means integrating ψ*ψ dτ over a shell with radius r and thickness dr to get a radial probability density distribution. Differentiate that wrt r to find the most likely r.

(This works if your ψ is normalized and also if it is not)

Note that for the mean radius, you need the expectation value for ##|\vec r|##. that means integrating ψ* r ψ dτ and gives a different answer!
(This works if your ψ is normalized and if it is not, divide by ∫ψ*ψ dτ )

[edit] crossed your second post -- took me a while to type this together!
 

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