Question #d1d52

1 Answer
Jul 19, 2017

Answer:

#"54.4 eV"#

Explanation:

For starters, the ionization energy of a hydrogen atom is equal to #"13.6 eV"# because the ground state has an energy of #"-13.6 eV"# in a hydrogen atom.

http://dev.physicslab.org/Document.aspx?doctype=3&filename=AtomicNuclear_BohrModelDerivation.xml

So for the hydrogen atom, the energy of a given excited state, which is characterized by #n > 1#, is given by

#E_n = -"13.6 eV"/n^2#

For the ground state, you have #n=1# and

#E_ "ground state" = -"13.6 eV"#

Now, for a hydrogen-like atom like the helium cation, #"He"^(+)#, the energy of a given excited state is given by

#E_n = -"13.6 eV"/n^2 * Z^2#

Here #Z# represents the atomic number of the element. Consequently, the energy of the ground state will be given by

#E_ "ground state" = -"13.6 eV" * Z^2#

Since a helium atom has

#Z = 2#

you can say that its ground state will be at

#E_ "ground state He" = -"13.6 eV" * 2^2#

#E_ "ground state He" = -"54.4 eV"#

This implies that in order to remove the electron from a #"He"^(+)# ion to form a #"He"^(2+)# ion, you have to supply #"54.4 eV"# of energy.

#"He"^(+) + "54.4 eV" -> "He"^(2+) + "e"^(-)#

http://www.jyi.org/issue/bohr-revisited-model-and-spectral-lines-of-helium/