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

[PeroK]

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.

 

[PSN03]

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

 

[PeroK]

It's $V'_4$ you want, isn't it?

 

[PSN03]

It's $V'_4$ you want, isn't it?

Yes but how to go about it?

 

[PeroK]

Yes but how to go about it?

You're so close!

Try this question: what is the new potential $V'$ at infinity now?

 

[PSN03]

You're so close!

Try this question: what is the new potential $V'$ at infinity now?

I think it should be +6.8eV.

 

[PeroK]

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!

 

[PSN03]

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

 

[PeroK]

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.

 

[PSN03]

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

 

[PeroK]

This means for the 3rd excited state we will get V'4=-1.7+6.8=5.1eV

 

[PSN03]

Ohh it was sooooo easy. Thanks a lottttt!