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

AI Thread 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. Initially, the user struggles with the concept of changing the reference point for potential energy, mistakenly thinking it complicates the problem. After clarifying that the potential energy at infinity is zero, participants guide the user to add 6.8 eV to all energy levels to adjust for the new reference point. Ultimately, they confirm that the energy difference between states remains unchanged, leading to the correct calculation for the third excited state. The user expresses relief and gratitude after resolving the problem.
PSN03
Messages
100
Reaction score
9
Homework Statement
For a neutral hydrogen atom where the ground state energy of the electron is taken as E=-13.6eV (with standard reference). Now if the potential energy of the first excited state is taken as zero reference potential energy then the potential energy of the third excited state is.
Relevant Equations
E=-13.6/n² where n is the orbit number.
Standard potential energy at infinity is zero.
I know how to solve this problem when the energy at ground state is zero but I don't know how to deal with 1st excited state energy as zero.
According to me since the potential energy is zero therefore the kinetic energy must be 13.6eV according to conservation of energy.
I also know that the energy is inversely proportional to the square of orbit number.
I don't know how to proceed further after this. Any kind of help or even a hint would be appreciated.
 
Physics news on Phys.org
This is just a change of reference point for potential energy. You're overthinking things.
 
Can u please give me some hints cause I can't think in other direction
 
PSN03 said:
Can u please give me some hints cause I can't think in other direction
It says in the question:

PSN03 said:
Homework Statement:: For a neutral hydrogen atom where the ground state energy of the electron is taken as E=-13.6eV (with standard reference).
What's the "standard reference"?
 
PeroK said:
It says in the question:What's the "standard reference"?
The standard reference is infinity I guess. Cause at infinity the potential energy becomes zero so any potential energy below that is negative
 
  • Like
Likes PeroK
PSN03 said:
The standard reference is infinity I guess. Cause at infinity the potential energy becomes zero so any potential energy below that is negative

Yes.

I suggest you first write down the energies of the relevant states in the standrad reference.
 
  • Like
Likes PSN03
PeroK said:
Yes.

I suggest you first write down the energies of the relevant states in the standrad reference.
Yes I have done that.
At standard state the energy of 2nd excited state is -13.6/3²=-1.51 and 3rd excited state is -0.85eV
 
PSN03 said:
Yes I have done that.
At standard state the energy of 2nd excited state is -13.6/3²=-1.51 and 3rd excited state is -0.85eV
I thought you wanted the first excited state and to take that as you new zero potential?
 
PeroK said:
I thought you wanted the first excited state and to take that as you new zero potential?
Yes that's exactly what has to be done with the new frame of reference. But I was answering it with respect to the old frame of potential energy. I don't know how to deal it with new frame of reference.
 
  • #10
PSN03 said:
Yes that's exactly what has to be done with the new frame of reference. But I was answering it with respect to the old frame of potential energy. I don't know how to deal it with new frame of reference.
Why didn't you calculate the energy of the first excited state?
 
  • #11
PeroK said:
Why didn't you calculate the energy of the first excited state?
I don't know how to find that in the new frame of reference. In the old frame it's -3.4eV
 
  • #12
PSN03 said:
I don't know how to find that in the new frame of reference. In the old frame it's -3.4eV

Have you learned about things like the Bohr Radius?
 
  • #13
PeroK said:
Have you learned about things like the Bohr Radius?
Yes I have learned about it.
 
  • #14
PSN03 said:
Yes I have learned about it.
Do you know the radius applicable to the energy states?
 
  • #15
PeroK said:
Do you know the radius applicable to the energy states?
According to my knowledge energy is inversely proportional to the square of radius.
E=k/n²
 
  • #16
PSN03 said:
According to my knowledge energy is inversely proportional to the square of radius.
E=k/n²
Okay, but you need to find the potential energy for each of these states. To do that you need to define the potential energy for each of these states. I assume you are supposed to calculate the potential for the 3rd Bohr radius and use that as the new zero. That being the second excited state.
 
  • #17
PeroK said:
Okay, but you need to find the potential energy for each of these states. To do that you need to define the potential energy for each of these states. I assume you are supposed to calculate the potential for the 3rd Bohr radius and use that as the new zero. That being the second excited state.
According to me me we are given that the energy of the 2nd excited state is zero and we are supposed to find the subsequent energy of the third excited state.
 
  • #18
PSN03 said:
According to me me we are given that the energy of the 2nd excited state is zero and we are supposed to find the subsequent energy of the third excited state.
That's one interpretation. Then the problem is very easy.

But, it does say "potential" energy. Which suggests that they want you to calculate the potential energy for the 2nd excited state. That would involve putting an appropriate radius into the Coulomb potential.

Perhaps you might want to confirm which it is?
 
  • #19
PeroK said:
That's one interpretation. Then the problem is very easy.

But, it does say "potential" energy. Which suggests that they want you to calculate the potential energy for the 2nd excited state. That would involve putting an appropriate radius into the Coulomb potential.

Perhaps you might want to confirm which it is?
Actually you are right. The problem says potential energy is zero and not the total energy.
But why should I calculate the potential energy for 2nd excited state?
 
  • #20
PSN03 said:
Actually you are right. The problem says potential energy is zero and not the total energy.
But why should I calculate the potential energy for 2nd excited state?
I believe the expected value of the Coulomb potential involves ##r_n = an^2##, where ##a## is the Bohr radius. I would use that.
 
  • #21
PeroK said:
I believe the expected value of the Coulomb potential involves ##r_n = an^2##, where ##a## is the Bohr radius. I would use that.
Yes that's absolutely right but it only gives us the radius of nth orbit. How to deal it with the new frame of reference?
 
  • #22
PSN03 said:
Yes that's absolutely right but it only gives us the radius of nth orbit. How to deal it with the new frame of reference?
There no frame of reference involved. There is only a zero point for potential energy.
 
  • #23
PeroK said:
There no frame of reference involved. There is only a zero point for potential energy.
Oh actually I meant that 2nd orbital should be new reference point for the potential energy...right?
 
  • #24
PSN03 said:
Oh actually I meant that 2nd orbital should be new reference point for the potential energy...right?
Yes.
 
  • #25
PeroK said:
Yes.
What to do after this. I mean how will this new frame affect the energy of the electron?
 
  • #26
PSN03 said:
What to do after this. I mean how will this new frame affect the energy of the electron?
Try to work out the potential energy of the first excited state. Until you have done that you can't go any further.
 
  • #27
PeroK said:
Try to work out the potential energy of the first excited state. Until you have done that you can't go any further.
The potential energy of first excited state would be -13.6*2/4=6.8eV in old reference frame. In new reference frame it's given to be zero.
 
  • #28
PSN03 said:
The potential energy of first excited state would be -13.6*2/4=6.8eV in old reference frame. In new reference frame it's given to be zero.
If you mean ##-6.8 eV## that's right.
 
  • #29
PeroK said:
If you mean ##-6.8 eV## that's right.
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.
 
  • #30
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.
It's the (potential) energy of the third excited state you want.
 
  • #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.
 
  • #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
 
  • Like
Likes PeroK
  • #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!
 

Similar threads

The energy of the first excited state of the system
Replies
17
Views
2K
Conservation of Momentum Problem - Mechanical and Kinetic Energies
Replies
2
Views
889
Is potential energy defined only for internal conservative forces?
Replies
15
Views
1K
Why is the elastic potential energy in position 2 zero?
Replies
6
Views
2K
How to calculate gamma ray energies following beta decay?
Replies
12
Views
1K
Challenging problem about an impact with a smooth frictionless surface
2
Replies
55
Views
5K
Minimizing the Energy of Two Conductors Very Far Apart
Replies
23
Views
1K
Minimum Kinetic energy of the hydrogen atom
Replies
22
Views
3K
What Does 'Mutual Potential Energy' Refer to in a Hydrogen Atom?
Replies
6
Views
4K
Statistics physics problem -- atom is in the ground state or excited state?
Replies
7
Views
3K

Hot Threads

Recent Insights

Back
Top