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8 Atomic structure and
spectra
8A Hydrogenic Atoms
Answers to discussion questions
D8A.2 (i) �e principal quantum number n determines the energy of a hydrogenic
atomic orbital through [8A.8–306].
(ii) �e azimuthal quantum number l determines the magnitude of the or-
bital angular momentum, given by [l(l + 1)]1/2ħ.
(iii) �e magnetic quantum number m l determines the z-component of the
orbital angular momentum, given by m lħ.
(iv) �e spin quantum number s determines the magnitude of the spin angu-
lar momentum, given by [s(s + 1)]1/2ħ; for hydrogenic atomic orbitals s
can only be 12 .
(v) �e quantum numberms determines the z-component of the spin angu-
lar momentum, given bymsħ; for hydrogenic atomic orbitalsms can only
be ± 12 .
Solutions to exercises
E8A.1(b) �e energy of the level of a hydrogenic atom with quantum number n is given
by [8A.14–308], En = −hcR̃HZ2/n2, where Z is the atomic number of the atom.
As described in Section 8A.2(d) on page 309, the degeneracy of a state with
quantum number n is n2.
• With Z = 2, En = −4hcR̃H/n2; the state with E = −4hcRH has n = 1, and
hence degeneracy (1)2 = 1 .
• With Z = 4, En = −16hcR̃H/n2; the state with energy E = −hcRH/4 =
−16hcR̃H/(8)2 has n = 8, and hence degeneracy (8)2 = 64 .
• With Z = 5, En = −25hcR̃H/n2; the state with energy E = −hcRH =
−25hcR̃H/(5)2 has n = 5, and hence degeneracy (5)2 = 25 .
E8A.2(b) �e task is to �nd the value of N such that the integral ∫ ψ∗ψ dτ = 1, where
ψ = N(2−r/a0)e−r/2a0 .�e integration is over the range r = 0 to∞, θ = 0 to π,