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

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Homework Help Overview

The discussion revolves around the potential energy of the first excited state of a hydrogen atom, particularly when this energy is considered to be zero. Participants explore the implications of changing the reference point for potential energy and how this affects the kinetic energy and overall energy states of the atom.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy, questioning how to handle the first excited state when its potential energy is set to zero. There are inquiries about the standard reference point for potential energy and how to calculate energies in a new frame of reference.

Discussion Status

There is an ongoing exploration of the implications of setting the potential energy of the first excited state to zero. Some participants have suggested calculating the potential energy for various states and discussing the necessary adjustments in energy calculations. Multiple interpretations of the problem are being considered, and guidance has been offered regarding the need to define potential energy for different states.

Contextual Notes

The problem involves understanding the energy states of a hydrogen atom, specifically how the potential energy is defined in relation to the first excited state. Participants are navigating the implications of using different reference points for potential energy and the associated calculations.

  • #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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