What happens to electric PE as n increases in a hydrogen atom?

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In a hydrogen atom, as the principal quantum number n increases, the electric potential energy becomes less negative, indicating an increase in potential energy. The energy of the atom at the nth energy level is given by the formula -13.6z^2/n^2, which shows that while the energy value decreases in magnitude, it effectively increases when considering the negative sign. This reflects that higher n values correspond to excited states of the atom. The discussion clarifies that potential energy can be viewed from different reference points, affecting how it is interpreted. Ultimately, as n increases, the electric potential energy of the hydrogen atom increases.
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Homework Statement


As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom
a)increases
b)decreases
c)remains same
d)does not increase

Homework Equations



The Attempt at a Solution


Energy of the atom in nth energy level is -13.6z^2/n^2. As n increases E should decrease.
 
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utkarshakash said:
Energy of the atom in nth energy level is -13.6z^2/n^2. As n increases E should decrease.

Do you mean -13.6z^2/n^2 becomes more negative if n increases?


ehild
 
ehild said:
Do you mean -13.6z^2/n^2 becomes more negative if n increases?


ehild

I'm assuming the question asks about magnitude. The magnitude will decrease but if we consider -ve sign it will increase. I'm really confused!
 
No, it is not about the magnitude. You know that the potential energy is defined by an arbitrary constant. You could count the potential energy from the ground level of the hydrogen. Would the energy decrease in the excited states? Higher values of n mean excited states of the atom.

ehild
 
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