Question

Doubly-ionized lithium is hydrogen-like atom. Atomic ionization energy can be predicted by an analysis using electrostatic potential and the Bohr model of the atom. For this model ionization energy will be:
E= −(Z2 13.6[eV])/n2
where Z - nuclei charge. n - principal quantum number.
So, for Li2+ (Z=3) energy diagram will be something like this:
n=1. E1=-122.4 eV
n=2. E2=-30.6 eV
n=3. E3=-13.6 eV
….....................................
n = infinity E = 0 (vacuum level)
When n=1, corresponded E = 122.4 eV – this is ionization energy for ground state.

why the answer is not E= -122.4eV instead it is E= 122.4eV?

Answer 1

Answer on Question #59962- Engineering-Other

Doubly-ionized lithium is hydrogen-like atom. Atomic ionization energy can be predicted by an analysis using electrostatic potential and the Bohr model of the atom. For this model ionization energy will be:

$$E = \& \text{minus}; (Z2\ 13.6\,[\mathrm{eV}]) / n2$$

where Z - nuclei charge. n - principal quantum number.

So, for Li2+ (Z=3) energy diagram will be something like this:

$$n = 1. \ E1 = -122.4\, \mathrm{eV}$$ $$n = 2. \ E2 = -30.6\, \mathrm{eV}$$ $$n = 3. \ E3 = -13.6\, \mathrm{eV}$$ \text{…} $$n = \text{infinity } E = 0 \text{ (vacuum level)}$$

When $n=1$, corresponded $E = 122.4\, \mathrm{eV}$ — this is ionization energy for ground state.

Why the answer is not $E = -122.4\, \mathrm{eV}$ instead it is $E = 122.4\, \mathrm{eV}$?

Answer

The term ionization energy (Ei) (of an atom or molecule) is most commonly used to refer to the energy required to remove (to infinity) the outermost electron in the atom or molecule when the gas atom or molecule is isolated in free space and is in its ground electronic state. Thus,

$$E_i = E(\infty) - E_1 = 0 - (-122.4\, \mathrm{eV}) = 122.4\, \mathrm{eV}.$$

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