# Radial probability density

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

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

Any help would be much appreciated.

## Answers and Replies

kuruman
Science Advisor
Homework Helper
Gold Member
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.

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

kuruman
Science Advisor
Homework Helper
Gold Member
You misunderstood what I said. The integral includes the probability density so that

$$P=\int \psi^*\psi\; 4\pi r^2 dr$$

Okay. Gotcha. Thanks for the help.