Question

The energy levels of one‑electron ions are given by the equation

=(−2.18 aJ)22
where is atomic number, is the energy level, and aJ is the abbreviation for the unit attojoules.
Ionization energy is the energy needed to eject an electron from an atom or ion.
Calculate the ionization energy, , of the one‑electron ion Li2+. The electron starts in the lowest energy level, =1.

Answer 1

The ionization energy for the one-electron ion Li2+ is calculated using the given energy level equation. It is found to be +19.62 x 10^−18 J by substituting the atomic number for lithium and the principal quantum number for the lowest energy level.

Step-by-step explanation:

The question asks to calculate the ionization energy for the one-electron ion Li2+, given the energy level equation for one-electron ions. The ionization energy (IE) is the energy required to remove an electron from an atom or ion. According to the provided equation, the energy levels (En) of one-electron ions can be expressed as En = (−2.18 x 10−18 J)(Z2/n2), where Z is the atomic number, and n is the principal quantum number corresponding to the energy level of the electron.

For Li2+, Z is 3 since lithium has an atomic number of 3, and the electron in question starts at the lowest energy level, n = 1. Plugging these values into the equation, we get E1 = (−2.18 x 10−18 J)(32)/12) = −19.62 x 10−18 J.

To calculate the ionization energy, we consider the energy required to remove the electron completely, which is the opposite in sign to E1, yielding an ionization energy of +19.62 x 10−18 J, since ionization energy is always positive.