ANSWERS TO TOPIC E ADDITIONAL PROBLEM ASSIGNMENT:

Part a: The ground state of Li is [He]2s. We have the ionization energy of Li, which is the E value for taking the valence electron from the 2s orbital completely out of the atom. When the electron is completely removed, its energy is zero (by definition), so the energy of the 2s orbital must be -520 kJ/mol.

Now we can calculate the DE values that correspond to the five transitions listed in the problem. Use E_photon = hc/l to calculate the photon energy. E_electron will be the same number, but it will have the opposite sign.

2p 2s: l = 671 nm, so E_electron = -2.96 x 10^-19 J = -178 kJ/mol

E_electron = E_final - E_initial

-178 kJ/mol = -520 kJ/mol - E_2p

E_2p = -342 kJ/mol

3p 2s: l = 323 nm, so E_electron = -6.15 x 10^-19 J = -370 kJ/mol

E_electron = E_final - E_initial

-370 kJ/mol = -520 kJ/mol - E_3p

E_3p = -150 kJ/mol

4p 2s: l = 274 nm, so E_electron = -7.25 x 10^-19 J = -437 kJ/mol

E_electron = E_final - E_initial

-437 kJ/mol = -520 kJ/mol - E_4p

E_4p = -83 kJ/mol

5p 2s: l = 256 nm, so E_electron = -7.76 x 10^-19 J = -467 kJ/mol

E_electron = E_final - E_initial

-467 kJ/mol = -520 kJ/mol - E_5p

E_5p = -53 kJ/mol

6p 2s: l = 248 nm, so E_electron = -8.01 x 10^-19 J = -482 kJ/mol

E_electron = E_final - E_initial

-482 kJ/mol = -520 kJ/mol - E_6p

E_6p = -38 kJ/mol

Now we can draw the energy diagram!

Part b: The longest wavelength corresponds to light having energy of 520 kJ/mol (the ionization energy from the ground state). 520 kJ/mol corresponds to 8.635 x 10^-19 J; using E = hc/l gives l = 2.30 x 10^-7 m = 230. nm.

Part c: The ionization energy of Li(g) from its first excited state is DE for moving the electron from the 2p orbital to completely out of the atom (E = 0). Since the 2p state has E = -342 kJ/mol, the ionization energy is +342 kJ/mol.

Part d: This part is actually quite unrelated to the rest of the problem. The third ionization energy of an atom is the energy required to remove the third electron (having already removed the first two). In other words, it is DE for the reaction Li²⁺(g) Li³⁺(g) + e^-. The trick here is to recognize that since Li has only three electrons (check the periodic table), Li²⁺ is a one-electron species. For any one-electron atom or ion, the Bohr formula applies: E = -1312 kJ/mol x (Z²/n²). To ionize Li²⁺, we must move the one remaining electron from the 1s orbital (the ground state for any one-electron species) to infinity. For the 1s orbital, n = 1 and E = -1312 kJ/mol x (3²/1²) = -11808 kJ/mol. For the "infinite" energy level, E = 0. Therefore, the ionization energy must be 11808 kJ/mol.

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