Ionization Energy of Lithium problem

AI Thread Summary
The discussion focuses on calculating the second ionization energy of lithium, given that the total energy to remove all three electrons is 1.96*10^4 kJ/mol and the first ionization energy is 520 kJ/mol. The second ionization energy can be derived by subtracting the first ionization energy from the total energy. The equations provided relate to energy levels and ionization but are noted to be applicable primarily for single-electron systems. The complexity of multiple electrons in lithium makes it necessary to rely on known ionization energies rather than theoretical calculations alone. Understanding these principles is crucial for accurately determining ionization energies in multi-electron atoms like lithium.
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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?
 
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