Ionization Energy of Lithium problem

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Homework Help Overview

The discussion revolves around calculating the second ionization energy of lithium, specifically focusing on the energy required to remove the second electron from a Li+ ion. Participants are exploring the relationship between ionization energy and electron transitions, referencing specific equations related to energy levels.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand how to calculate the second ionization energy using given equations and questioning the applicability of these equations to multi-electron systems. There is also a discussion about the implications of ionization energy in relation to electron transitions.

Discussion Status

The discussion is ongoing, with participants raising questions about the calculations and the conditions under which certain equations apply. Some have provided insights into the limitations of the equations for multi-electron atoms, while others are seeking clarification on the calculations involved.

Contextual Notes

Participants are considering the implications of ionization energy definitions and the specific conditions under which the provided equations can be utilized. There is an acknowledgment of the need for additional information or measurements to accurately determine the second ionization energy.

wee00x
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The energy needed to strip all three electrons from a Li(g) atom was found to be 1.96*10^4 kJ/mol. The first ionization energy of Li is 520 kJ/mol. Calculate the second ionization energy of Lithium atoms (the energy required for the process)

Li+(g) ---> Li+2 + e-

Equation: frequency = (3.29x10^15s-1)Z^2(1/n^2i - 1/n^2f)

Ei + hv = Ef

My question: These equations can give me the energy of each energy level but how do I find the energy of required for ionization??
 
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Ionization means energy is higher than the one needed for the electron to jump from n=1 to n=∞.
 
But then how would that be calculated?
 
For very large n 1/n2 is for all practical purposes equal to zero.
 
wee00x said:
Equation: frequency = (3.29x10^15s-1)Z^2(1/n^2i - 1/n^2f)
That equation only works when there is one electron. If there are two or more, you pretty much have to measure (or be told) what the ionization energy is.

So, in what situation does the above equation apply to lithium?
 
Last edited:

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