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The wavefunction of the electron in the hydrogen atom is $ k* exp(-r/a)$ (k is the normalization constant), but it does not take n into account, whereas the solution of Schrödinger's equation ($H(wavefunction)=E*wavefunction$) says the energy of it is $E=E(0)/n^2$, but the variable in the wavefunction is $r$, and $n$ isn't there. If I apply the Hamiltonian operator on the wavefunction, no $n$ will appear, I don't understand!

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    $\begingroup$ The wavefunction you gave is for the ground state of the hydrogen atom (n=1). Incidentally, "I need an answer as fast as possible" is usually not how you get fast answers around here.... that's just how people who volunteer their expertise work. $\endgroup$ – Floris Jan 20 '16 at 22:23
  • $\begingroup$ here is a list of other sites that might answer a student question meta.physics.stackexchange.com/questions/391/… $\endgroup$ – anna v Jan 21 '16 at 4:48
  • $\begingroup$ I was simply very anxious when when I wrote the post, and worried about this point of the course I hadn't understood. Everyone is not a machine who writes perfect posts that could be published in scientifical reviews, Thank you for your answer, although it is shorter than your blame. $\endgroup$ – Chewie Jan 21 '16 at 5:06
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Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2.

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