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Ionization Energy

Submitted by wiggleurtoes on Tue, 08/12/2014 - 00:55

The ionization energy of vanadium from the ground state is 650.2 kJ/mol. Assume that the transition of 376 kJ/mol was from the ground state to an excited state. If that is the case, calculate the ionization energy from the excited state. 

The answer in my textbook is 274 kJ/mol, I just don't know how to get that answer, so any help would be greatly appreciated. 


To remove an unexcited electron requires 650.2 kJ/mol. To promote the electron to the excited state (ready to be lost) requires 376 kJ/mol so the difference must be needed to remove the electron from the excited state =  650.2 - 376 = 274.2 KJ