Ionization Energy of Helium Atom | 24.6 eV

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SUMMARY

The ionization energy of a helium atom is established at 24.6 eV. The second ionization potential (IP2) can be calculated using the formula IP2 = IP1 * Z², where Z is the atomic number. However, this approach is only valid for hydrogen-like species, such as He+, which has one electron. The Bohr model is not applicable to the neutral helium atom due to its two electrons, leading to potential inaccuracies if applied directly to He.

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physicsmaths1613
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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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