Answer :

Answer: E = - 19.611×[tex]10^{-18}[/tex] J

Explanation: The lowest possible energy can be calculated using the formula:

[tex]E_{n} = - Z^{2}.\frac{k}{n^{2}}[/tex]

where:

Z is atomic number of the atom;

k is a constant which contains other constants and is 2.179×[tex]10^{-18}[/tex] J

n is a layer;

For the lowest possible, n=1.

Atom of Lithium has atomic number of Z=3

Substituing:

[tex]E_{1} = - 3^{2}.\frac{2.179.10^{-18}}{1}[/tex]

[tex]E_{1} =[/tex] [tex]-19.611.10^{-18}[/tex] J

The energy for the electron in the [tex]Li^{+2}[/tex] ion is - 19.611 joules

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The lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion is equal to [tex]1.96\times 10^{-17}\; Joules[/tex]

To determine the lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion, we would use the Bohr model:

Mathematically, Bohr's model is given by the equation:

[tex]Energy = -Z^2 \frac{k}{n^2}[/tex]

Where:

  • Z is the atomic number of an atom.
  • n is the number of energy level.
  • k is Rydberg constant.

We know that the atomic number of lithium (Li) is equal to 3.

Also, at the lowest possible energy, n = 1.

Rydberg constant = [tex]2.179 \times 10^{-18}[/tex]

Substituting the parameters into the equation, we have;

[tex]E_1 = -3^2 \times \frac{2.179 \times 10^{-18}}{1^2} \\\\E_1 =9 \times 2.179 \times 10^{-18}\\\\E_1 =1.96\times 10^{-17}\; Joules[/tex]

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