- #1
Christofferk
- 18
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I saw another post about this but i didn't quite find what i was looking for there so i thought i'd give it a go instead with a thread.
Calculate the exact value of the kinetic energy of the hydrogen atom in its ground state. No more information is given, we are referred to litterature and stuff.
I am thinking that my approach is wrong in the sense that i am trying to do something in a way that it's not allowed to. We have the hamiltonian [tex]\hat{H}\psi=E\psi[/tex] which in this case is [tex](-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{4\pi\epsilon_0r})\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)=-13.6eV\psi(r,\theta,\phi)[/tex] and the last term is because we know that it is in its groundstate.
What i thought that i'd do that i now suspect is not "allowed" mathematically was to let the [itex]\psi(r,\theta,\phi)[/itex]s cancel themselves out and then since the first term on the left side represents the kinetic energy, just move the right term over to the right side and simply calculate the value. The value i get is incredibly small, since it is the hydrogenatom in its groundstate we can interpret the function with the values [tex]Z=1[/tex][tex]e=1.6*10^{-19}[/tex][tex]\epsilon_0=8.8*10^{-12}[/tex] and [tex]r=0.52Å=0.52*10^{-10}[/tex] and if i proceed to do as i intended the value for the kinetic energy i get is 13.6eV or so which is wrong, I've been nudge to believe that the kinetic energy should be [itex]\frac{1}{2}[/itex][tex]kinetic energy=-13.6+\frac{(1.6*10^{-19})^2}{(4\pi8.8*10^{-12}*0.52*10^{-10})}=-13.599999eV[/tex]
Should my approach instead be to try to calculate [tex](-\frac{\hbar^2}{2\mu}\nabla^2)\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] given that [tex]\psi(r,\theta,\phi)_{100}=\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] How would i go about and do this?
Homework Statement
Calculate the exact value of the kinetic energy of the hydrogen atom in its ground state. No more information is given, we are referred to litterature and stuff.
Homework Equations
I am thinking that my approach is wrong in the sense that i am trying to do something in a way that it's not allowed to. We have the hamiltonian [tex]\hat{H}\psi=E\psi[/tex] which in this case is [tex](-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{4\pi\epsilon_0r})\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)=-13.6eV\psi(r,\theta,\phi)[/tex] and the last term is because we know that it is in its groundstate.
The Attempt at a Solution
What i thought that i'd do that i now suspect is not "allowed" mathematically was to let the [itex]\psi(r,\theta,\phi)[/itex]s cancel themselves out and then since the first term on the left side represents the kinetic energy, just move the right term over to the right side and simply calculate the value. The value i get is incredibly small, since it is the hydrogenatom in its groundstate we can interpret the function with the values [tex]Z=1[/tex][tex]e=1.6*10^{-19}[/tex][tex]\epsilon_0=8.8*10^{-12}[/tex] and [tex]r=0.52Å=0.52*10^{-10}[/tex] and if i proceed to do as i intended the value for the kinetic energy i get is 13.6eV or so which is wrong, I've been nudge to believe that the kinetic energy should be [itex]\frac{1}{2}[/itex][tex]kinetic energy=-13.6+\frac{(1.6*10^{-19})^2}{(4\pi8.8*10^{-12}*0.52*10^{-10})}=-13.599999eV[/tex]
Should my approach instead be to try to calculate [tex](-\frac{\hbar^2}{2\mu}\nabla^2)\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] given that [tex]\psi(r,\theta,\phi)_{100}=\frac{1}{\sqrt{\pi a_0}}e^{-\frac{r}{a_0}}[/tex] How would i go about and do this?