The mean value of (1/r^2) given a normalised wavefunction

[michael2k100]

The normalised wavefunction for the 1s electron in the hydrogen atom is

ψ=(1/((PI^1/2).a^3/2)).exp(-r/a)

where a is the bohr radius.

What is the mean value of (1/r2) in terms of the Bohr radius a0?



Answer: 2/(a^2)

 

[kuruman]

How do you think you should proceed? How does one calculate mean values?

 

[michael2k100]

How do you think you should proceed? How does one calculate mean values?

well basically i tried using the method of finding the expectation value of (1/r^2) although i was not ble to get the final answer, i thought about doing the integration using spherical polars, becuase there is a factor of pi that will need to be canceled out, but i still couldn't get the correct answer.

 

[kuruman]

Yes you need to find the expectation value of 1/r² and spherical coordinates is the way to go. Can you show exactly how you set up the integral?

 

[michael2k100]

i set it up as

integral of (1/(pi)a^3).(exp(-2r/a)/r^2)dr

but i do not no how to convert this into spherical polar i know that the volume element of spherical polar is (r^2)sin(theta)drd(theta)d(phi) but i don't no how to convert the dr line elemnt into spherical polars

 

[kuruman]

This is a triple integral. There is nothing to convert. You should just do the integral over r. Now the angular part of the integral is

\int_0^{\pi} sin\theta d\theta\int_{0}^{2\pi} d \phi

You need to do these two integrals separately and multiply the result with the radial integral.

 

[michael2k100]

i have done that and the integral of them is equal to 2pi

but once i do the integral of (exp(-2r/a)dr from 0<r<inifinity

i get a/2

this gives me a final answer of 1/(a^2) and the correct answer is 2/(a^2)

do u know where i am missing the extra factor of 2?

 

[michael2k100]

sorry i went over the angular part and got an answer of 4pi, so i have the right answer now

thanks for your help