Recent content by Cp.L

Hi, thanks alot. It really helped to use spherical coordinates :) got the correct answer now :)

It was listed in my book, for the groundstate of an hydrogen atom so it should actually be correct.

Oh yes that's logical since its the radius, thanks. Yes and the integration by parts should be u=r du=dr dv=e^{\frac{-2Zr}{a_{0}}} v=_{}\frac{a_{0}}{-2Z}e^{\frac{-2Zr}{a_{0}}} dr integrating \frac{z^{3}}{\pi a_{0}}∫e^{\frac{-2Zr}{a_{0}}} r dr from 0 to ∞ with these vaues i get <r>=...

Ok, i found a mistake, but still don't get it right. I write: <r>=\frac{1}{\pi}\frac{z_{0}^{3}}{a_{0}^{3}}∫e^{\frac{-2Zr}{a_{0}}}r dr Where the limits are from -∞ to ∞ Integrating i let U= r, du=1, dv= e^{\frac{-2Zr}{a_{0}}}, V=\frac{a_{0}}{-2Z}e^{\frac{-2Zr}{a_{0}}} this...

Hi, yes i used that C_{100} =(1/\sqrt{\pi}) (\frac{z}{a_{0}})^{3/2} Then i do the integral, but i must be doing it wrong cause i end up with e^{\frac{-2zr}{a_{0}}} in the answer and also a problem of e^{∞}

Homework Statement Show that the expectation value for r for an electron in the groundstate of a one-electron-atom is: <r>=(3/2)a_{0}/Z Homework Equations Expectationvalue: <f(x)>=∫\psi*f(x)\psidx, -∞<x>∞ \psi_{100}=C_{100} exp(-Zr/a_{0}), a_{o}\ =\ 0.5291\ \times\ 10^{-10}m , h\...