Recent content by Christofferk

Maybe i should also say that i am choosing between engineering degree in the industry, or taking the physics one and staying in academia, then the salary is lower.

It's so haaard :( I mean, with the engineering-masters i'll have a way higher salary and the stability a full time employment provides, but the physics one seems more fun :(

SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)

Maybe it's easier if you model the problem as one of "one" variable. If you write it as a function y of x, then it is y=x-2\sqrt{x}+1, do you know what the tangent if this is at an arbitrary point? Try to look at the equation for the tangentplane of a \textbf{R}^3-function. The tangent-plane for...

Hey there, i guess i just need to ventilate some and see what people think about my situation and about the choices i have. First of all i must say that i am from Sweden and therefore i have conditions that few other people have. I get my education for free and i am free to chose whatever field...

I see your point. I am taking this class in addition to the classes i currently have, i am majoring in an electrical engineering programme with emphasis on physics and i maybe want to take a masters in physics here that's given to the physics students. With that said i don't have the same...

Guys the mystery has just solved itself.. as i said in my OP, i was nudged to believe that the answer was 1/2... the source that told me this just clarified that it was 1/2 hartree units.. which translates to 13.6eV. The problem is solved, thank you guys for your help!

Oh i didn't understand your question earlier, let me dig into this for a while, i'll get back to you

The way i'd like to see it be is that the potential at the inner most radious for the electron (ground state) should be 0 because it can't jump back to a lower state but i guess this is not the case.. :/

a_0 is the bohr-radius

For this i get \frac{1}{a_0}=1.92*10^{10} which if i later also add in the rest of the equation for the potential energy gives me that the expected value for the potential energy should be -4.45*10^{-18} which is wrong i guess.. :/

Now i see it should ba 3 there... this vastly changes things.. ! I'll do the math once again for the expected potential and i'll get back!

The wavefunction i am using is the one for hydrigen in its m=0, l=0 and n=1-state, that is \frac{1}{\sqrt{\pi a_0}}e^{\frac{-r}{a_0}}

I still don't get it quite right it looks like. \langle\psi|V(r)|\psi\rangle=\langle\psi|-\frac{k}{r}|\psi\rangle=\frac{-k4\pi}{\pi a_0}\int_{0}^{\infty} re^{\frac{2r}{a_0}} dr which i get to become...

I'll get to that right now :) Get back to you in a second