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1. The orbital quantum number for the electron in a hydrogen atom is l = 3. What is
the smallest possible value (algebraically) for the total energy of this electron? Give
your answer in electron volts.
E = -13.6eV (Z2/n2) l = n -1 → n = l + 1
E = -13.6eV (Z2/(l +1)2)
E = (-13.6/16)eV
E = -0.85 eV
2. Suppose the value of the principal quantum number is n = 4. What are the
possible values for the magnetic quantum number ml ?
-3, -2, -1, 0, 1, 2, 3
3.
Write down every possible set of quantum numbers when the principal
quantum number is n = 3.
[3,0, 0,+½][3,0, 0,-½][3,1,-1,+½][3,1,-1,-½][3,1, 0,+½][3,1, 0,-½]
[3,1, 1,+½][3,1, 1,-½][3,2,-2,+½][3,2,-2,-½][3,2,-1,+½][3,2,-1,-½]
[3,2, 0,+½][3,2, 0,-½][3,2, 1,+½][3,2, 1,-½][3,2, 2,+½][3,2, 2,-½]
4. In the style indicated in Table 30.3 (1sx, 2sy, 2pz…), write down the ground state
electron configuration of copper (Z = 29).
Copper (Cu) 1s2, 2s2, 2p6, 3s2, 3p6, 3d9, 4s2
5. For an electron in a hydrogen atom, the z component of the angular momentum
has a maximum value of Lz = 4.22x10-34 J·s. Find the three smallest possible values
(algebraically) for the total energy (in eV) that this electron could have.
Electrons in higher orbits have more energy, thus we’re looking for the lowest
possible orbits—in other words the lowest values for the principal quantum number n.
Lz = ml (h/2π)
Lz = ml ħ
ml = Lz/ħ
ml = (4.22x10-34 J·s/1.055x10-34J·s)
ml = 4
n ≥ l + 1 and l ≥ ml thus the lowest value of n is 5, with the next lowest values being
6 and 7.
E = -13.6eV (Z2/n2)
E = -13.6eV (Z2/n2)
E = -13.6eV (Z2/n2)
E = -13.6eV (12/72)
E = -13.6eV (12/62)
E = -13.6eV (12/52)
E = -0.28eV
E = -0.38eV
E = -0.54eV
5.