Quantum Mechanics Problem: ionization potential

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The discussion centers on the ionization potential of alkali atoms, specifically how the effective charge (Zeff) for the outer electron can be approximated using quantum numbers for Li, Na, K, Rb, and Cs. The formula for the potential energy of the valence electron is provided, but clarification is sought on the relationship between ionization potential and ionization energy. It is confirmed that ionization energy and ionization potential are essentially the same, with the distinction that ionization potential may be expressed in volts, requiring conversion to energy units. The differences in Zeff from unity are attributed to the shielding effect of inner electrons. Understanding these concepts is crucial for accurately calculating ionization potentials in quantum mechanics.
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The outer electron (valence electron) of an alkali atom may be treated in an approximate way, as if it were in a hydrogenic orbital. Suppose that one takes the quantum number for the valence electron to be 2, 3, 4, 5, and 6, respectively, for Li, Na, K, Rb, and Cs. What values of the corresponding effective charge, Zeff, must be assumed to account for the observed first ionization potentials of these atoms? Explain why they differ from unity.

So for the formula, I'm pretty sure it is

Vn eff = -(Zeff(n)e^2)/4pi(episilon0)

But I'm lost on the definition of ionization potential. Is it just ionization energy? How does this relate to the formula.. please help me! I'm lost.
 
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Ionization energy and ionization potential are the same thing. But there is a caveat. Ionization energy is measured in eV (or other energy units). Ionization potential MAY be measured in just volts, in which case it must be multiplied by the elementary charge e to get true ionization energy.
 
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