Total energy of Coulomb model of Hydrogen atom

  • #1
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Homework Statement


Hi, my question is regardng a Coulomb model of an H atom specified with Hamiltonian operator, Hhat, by spherical coordinates of energy eigenfunction
ψ2,1,-1 (r,θ, ∅) =(1/ 64∏a02)1/2 r/a0 e-r/2a0 sinθ e-iθ

Principal quantum numer n = 2
orbital an mom l = 1
magnetic quantum number m = -1

I must specify magnitude of orbital ang momentum, Lhat2, and z-component of orbital ang momentum as well as total energy.


Homework Equations


so;
Lhat2 = l (l + 1) hbar2
Lz = m hbar (1/ √2∏) eim∅



The Attempt at a Solution


Lhat2 = 1 (1 + 1)hbar2
Lhat2 = 2 hbar2

Lz = m hbar (1/ √2∏) eim∅
Lz = -1 hbar / √2∏ ei-1∅
= -1 hbar

but what about total energy?
I'm lost in the forest again and can't see the wood for the trees.
 
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  • #2
There is a relation between principal quantum number and energy that you should know.
 
  • #3
Hmmm!

Well, En = (n + 1/2) hbar ω0

and

En = n22 hbar2 / 2mL2
where principal quantum number n = 0, 1, 2 . . .

but I can't evaluate ω0 and I don't have the mass for the other one . . do I?
 
  • #4
Ah!
n = principal quantum number
m = mass of electron
L2 = square of magnitude of orbital ang momentum.

so . . . .

E = 22 ∏2 hbar2 / 2(9.1x10-31) 2 hbar2
E = -2∏hbar/m

er! surely too messy and probably wrong.
 
  • #5
Roodles01 said:
but I can't evaluate ω0 and I don't have the mass for the other one . . do I?
Why? Your formula (edit: wait, those are the wrong formulas) assumes an infinite central mass, which is usually a reasonable approximation in a hydrogen atom.

The total energy of the system does not matter unless you want to consider special relativity.
 
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  • #6
Roodles01 said:
Hmmm!

Well, En = (n + 1/2) hbar ω0

and

En = n22 hbar2 / 2mL2
where principal quantum number n = 0, 1, 2 . . .

Does either of these formulas apply to the hydrogen atom?
 

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