What is the Radial Probability Density for a Hydrogen Atom in the Ground State?

[w3390]

Homework Statement

Calculate the probability that the electron in the ground state of a hydrogen atom is in the region 0 < r < 3.75a0.

Homework Equations

a0=.0529 nm

P(r)=4(Z/a0)^3*r^2*e^(-2Zr/a0)

The Attempt at a Solution

I am confused because I am not sure if I am supposed to use 3.75a0 as my radius. I guess I am not supposed to since I did that and got it wrong. I cannot figure out what else to do. However, I have also seen the probability density written as:

P(r)[tex]\Delta[/tex]r=[4(Z/a0)^3*r^2*e^(-2Zr/a0)][tex]\Delta[/tex]r

If this is the correct formula, I am not sure what to use for delta r.

Any help would be much appreciated.

 

[kuruman]

You don't use dr. This is a volume integral in radial symmetry. You use dV = 4πr²dr. You carry the integral from lower limit r = 0 to upper limit r = 3.75a₀.

 

[w3390]

Okay I do that and I get a volume of 9.81e-29m^3. How do I get this into a probability function?

 

[kuruman]

You misunderstood what I said. The integral includes the probability density so that

[tex]P=\int \psi^*\psi\; 4\pi r^2 dr[/tex]

 

[w3390]

Okay. Gotcha. Thanks for the help.