- #1
adh2
- 5
- 0
Hello!
I'm preparing for my quantum mechanics test. In the solutions of an old test I find this conversion, that I don't understand.
[tex] \Psi = Nze^{-r/2a_0} = Nre^{-r/2a_0}cos\Theta[/tex]
N is the normalization constant, which is to be calculated. I would have guessed that z is the atomic number ( =1), but it is apparently [tex] = r cos \Theta [/tex]. Can anyone please explain this to me?
Alfred
I'm preparing for my quantum mechanics test. In the solutions of an old test I find this conversion, that I don't understand.
[tex] \Psi = Nze^{-r/2a_0} = Nre^{-r/2a_0}cos\Theta[/tex]
N is the normalization constant, which is to be calculated. I would have guessed that z is the atomic number ( =1), but it is apparently [tex] = r cos \Theta [/tex]. Can anyone please explain this to me?
Alfred
, but it looks as if z is the component in the z-direction, and r and θ are the usual spherical coordinates, so that z = rcosθ.