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Quantum Mechanics Problem: ionization potential

  • Thread starter wee00x
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  • #1
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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.
 

Answers and Replies

  • #2
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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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