Probability/Schrodinger's equation

  • Thread starter Thread starter NicolasM
  • Start date Start date
NicolasM
Messages
3
Reaction score
0
We have a hydrogen atom in its ground state. What is the probability of an electron being found within a sphere (with nucleus being the origin) of a radius of 2a_{o}\ =\ 0.5291772083(19)\ \times\ 10^{-10}α\ m?

Would 0 4πr3ψ2dr/3 x100% be correct?
 
Physics news on Phys.org
NicolasM said:
We have a hydrogen atom in its ground state. What is the probability of an electron being found within a sphere (with nucleus being the origin) of a radius of 2a_{o}\ =\ 0.5291772083(19)\ \times\ 10^{-10}α\ m?

Would 0 4πr3ψ2dr/3 x100% be correct?

No, it wouldn't even have correct dimensions. Why do you think that's right? Work it out from the three dimensional integral of |ψ|^2. Your differential factor is wrong.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top