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

[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!