Ionization Energy of Helium Atom | 24.6 eV

AI Thread Summary
The ionization energy of a helium atom is 24.6 eV, but the energy required to remove both electrons is questioned. The relationship between the first ionization energy (IP1) and the second (IP2) is discussed, with a formula suggesting IP2 = IP1*Z^2. However, concerns arise regarding the applicability of Bohr's model to helium, as it is not a hydrogen-like atom. The discussion concludes that interpreting IP2 as the energy to remove the second electron from He+ and IP1 as the ionization energy of hydrogen provides clarity.
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Homework Statement


The energy required to ionize a helium atom is 24.6 eV. The energy required to remove both the electrons from He atom would be?

The Attempt at a Solution


My textbook says-
IP1= 24.6 eV
IP 2= IP1*Z2
How can they relate the first ionization energy to the second one? The Bohr's atom can be used only for Hydrogen like species and He atom is not Hydrogen like, only it's ion is. How did they relate it?
 
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He+ is a hydrogen like species as it has one electron.So Bohr's model is applicable on it.
 
harsh_sinha said:
He+ is a hydrogen like species as it has one electron.So Bohr's model is applicable on it.
But what about intial He atom with 2 electrons?
 
physicsmaths1613 said:
But what about intial He atom with 2 electrons?
Bohr's model is not applicable for He atom.Besides if we use this equation to calculate IP2 then the result would be twice of the actual value.
It would be great if you share the source of this equation.
 
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physicsmaths1613 said:
IP 2= IP1*Z2
I can only get the right answer from that by interpreting the IP2 as the energy to remove the second electron from He+, and the IP1 as the energy required to ionise H (which makes sense).
 
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