Normalization constant for Hydrogen atom

[Reshma]

Help, I'm losing it .

Wavefunction of Hydrogen atom in the ground state is:
$$\Psi (r) = Ae^{-r/r_0}$$
Determine A.

I set about trying to obtain the Normalization factor.
$$\int \Psi^2 (r) dV = 1$$

$$\int \left(A^2 e^{-2r/r_0}\right)(4\pi r^2)dr = 1$$

What limits should I take for this integral?

 

[Tide]

What you have is the radial wave function so the limits of integration would be from 0 to infinity.

 

[Reshma]

What you have is the radial wave function so the limits of integration would be from 0 to infinity.

Thanks, Tide. It worked! I was trying to integrate between 0 to r₀ and ended up with a nutty answer !

So,
$$\int_0^{\infty} \left(A^2 e^{-2r/r_0}\right)(4\pi r^2)dr = 1$$

$$A^2 4\pi \int_0^{\infty} r^2 e^{-2r/r_0}dr = 1$$

$$A^2 4\pi \left[{-r_0r^2\over 2} - {r_0^{2}r\over 2} - {r_0^{3}\over 4}\right]e^{-2r/r_0} \vert_0^{\infty} = 1$$

The first two terms go to zero and only the last term survives.

$$A^2 4\pi \left({r_0^{3}\over 4}\right) = 1$$

$$A = \frac{1}{\sqrt{\pi r_0^{3}}}$$

So the normalized wavefunction will be:
$$\Psi = \frac{1}{\sqrt{\pi r_0^{3}}}e^{-r/r_0}$$

 

[the keck]

I was looking at your solution and a problem came into my head which relates to the ground state wave function. I noticed that the function is actually independent of theta and phi (Spherical coordinates). But what does that actually imply in terms of where the electrons are distributed in the atom?

Thanks

Regards,
The Keck

 

[malawi_glenn]

The ground state of Hydrogen atom does not contain angle dependance, it is spherical symmetric.

 

[anjor]

I was looking at your solution and a problem came into my head which relates to the ground state wave function. I noticed that the function is actually independent of theta and phi (Spherical coordinates). But what does that actually imply in terms of where the electrons are distributed in the atom?

Thanks

Regards,
The Keck

You are still trying to think with the particle picture... where electrons are points distributed somewhere in the space.

The ground state function you obtain by solving the schroedinger equation is the probability amplitude of the electron... Hence, in a sense... electron is like a cloud spread out around the nucleus...

 

[irenek]

Hi,
I don't understand why the first two terms go to zero and why exponential goes to 1. Is r always 0?
Irene