Case when the potential energy of the 1st excited state is zero

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SUMMARY

The discussion revolves around calculating the potential energy of the first excited state of a hydrogen atom when its potential energy is set to zero. The ground state energy is defined as E=-13.6 eV, and the potential energy for the first excited state is -6.8 eV in the old reference frame. By changing the reference point to the first excited state, the potential energy at infinity becomes zero, and all energy levels are adjusted accordingly. The final energy for the third excited state is calculated to be 5.1 eV after adding 6.8 eV to the original potential energy.

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  • #31
PSN03 said:
Yes -6.8eV ...but after this.
The new energy is 0 therefore +6.8eV has been supplied and the change in energy is of 6.8-(-6.8)=13.6eV.

This calculation is wrong. You need some notation. Let's have: ##E_n, V_n## for the original energy and potential energy of the hydrogen energy states; and. ##E'_n, V'_n## for the energies with the changed zero potential to be at the first excited state.

I suggest you should write them all down so you can see the pattern.
 
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  • #32
PeroK said:
This calculation is wrong. You need some notation. Let's have: ##E_n, V_n## for the original energy and potential energy of the hydrogen energy states; and. ##E'_n, V'_n## for the energies with the changed zero potential to be at the first excited state.

I suggest you should write them all down so you can see the pattern.
Ohk...
So
##E_1##=-13.6
##E_2##=-3.4
##E_3##=-1.51
##E_4##=-0.85
##V_2##=-6.8
##V_3##=-3.02
##V_4##=-1.7

##V'_2##=0
 
  • #33
It's ##V'_4## you want, isn't it?
 
  • #34
PeroK said:
It's ##V'_4## you want, isn't it?
Yes but how to go about it?
 
  • #35
PSN03 said:
Yes but how to go about it?
You're so close!

Try this question: what is the new potential ##V'## at infinity now?
 
  • #36
PeroK said:
You're so close!

Try this question: what is the new potential ##V'## at infinity now?
I think it should be +6.8eV.
 
  • #37
PSN03 said:
According to my knowledge it should be infinity. Is it right or wrong?

No. It was ##V(\infty) = 0##. That's the "standard".

Another question: for ##V'_2## how did you get from ##V_2 = -6.8eV## to ##V'_2 = 0 eV##? What mathemtical process did that require?

Think simple!
 
  • #38
PeroK said:
No. It was ##V(\infty) = 0##. That's the "standard".

Another question: for ##V'_2## how did you get from ##V_2 = -6.8eV## to ##V'_2 = 0 eV##? What mathemtical process did that require?

Think simple!
Ohk so according to me we initially had ##V_2##=-6.8eV. Now we add 6.8eV to make it zero. So consequently at infinity we will get V"=0+6.8eV
 
  • #39
PSN03 said:
Ohk so according to me we initially had ##V_2##=-6.8eV. Now we add 6.8eV to make it zero. So consequently at infinity we will get V"=0+6.8eV
Yes, that's all the question is asking you to do. Add ##6.8eV## to all the energies. Note that the difference between any two energy levels remains the same, as it must.
 
  • #40
PeroK said:
Yes, that's all the question is asking you to do. Add ##6.8eV## to all the energies. Note that the difference between any two energy levels remains the same, as it must.
This means for the 3rd excited state we will get V'4=-1.7+6.8=5.1eV
 
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  • #41
PSN03 said:
This means for the 3rd excited state we will get V'4=-1.7+6.8=5.1eV
:partytime:
 
  • #42
PeroK said:
:partytime:
Ohh it was sooooo easy. Thanks a lottttt!
 

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