Key 11
1. Key to Prob. 8.3
(a) For n = 1, l = 0, ml = 0, ms = ±1/2 (2 sets). From general formula:
2n2 = 2 × 12 = 2.
(b) For n = 2, l = 0, ml = 0, ±1, ms = ±1/2 (2 sets),l = 1, ml = −1, 0, +1, and
for each ms = ±1/2 . So total 8 sets. From general formula: 2n2 = 2 × 22 = 8.
(c) n = 3: l = 0, 1, 2 and for every l there are (2l + 1) values of ml and for every
value of n, l, ml there are two values of ms = ±1/2. So for n = 3,
2(1) + 2(3) + 2(5) = 18 = 2n2 = 2(3)2 = 18.
(d) n = 4:
2(1) + 2(3) + 2(5) + 2(7) = 32 = 2n2 = 2(4)2 = 32.
(e) n = 5:
2(1) + 2(3) + 2(5) + 2(7) + 2(9) = 50 = 2n2 = 2(5)2 = 50.
2. Key to Prob. 8.4
3p subshell. For a p state, l = 1. Thus ml = 0, ±1. For each ml , ms can be
±1/2.
3. Key to Prob. 8.10
n = 2; l = 0, 1; s = 1/2. So, j = 1 + 1/2, 1 − 1/2, 0 + 1/2, |0 − 1/2| = 3/2, 1/2.
For j = 3/2; mj = −3/2, −1/2, 1/2, 3/2. For j = 1/2; mj = −1/2, 1/2.
4. Key to Prob. 8.17
We know,
E=
h̄2 π 2 2
2
2
n
+
n
+
n
y
z
2mL2 x
1
Putting the values for the constants,
E=
(1.054 × 10−34 )2 (π 2 )(n2x + n2y + n2z )
= (1.5×10−18 J(n2x +n2y +n2z ) = (9.4 eV)(n2x +n2y +n2z ).
2(9.11 × 10−31 )(2 × 10−10 )2
(a) 2 electrons per state. The lowest states have,
(nx , ny , nz ) = (1, 1, 1) ⇒ E111 = (9.4 eV)(12 + 12 + 12 ) = 28.2 eV.
For, (nx , ny , nz ) = (1, 1, 2) or (1, 2, 1) or (2, 1, 1),
E112 = E121 = E211 = (9.4 eV)(12 + 12 + 22 ) = 56.4 eV.
Emin = 2 × (E111 + E112 + E121 + E211 ) = 2(28.2 + 3 × 56.5) = 394.8 eV.
(b) All 8 particles go into the (nx , ny , nz ) = (1, 1, 1) state, so
Emin = 8 × E111 = 225.6 eV.
5. Key to Prob. 8.21
(a) 1s2 2s2 2p4 .
(b) For the two 1s electrons, n = 1, l = 0, ml = 0, ms = ±1/2.
For the two 2s electrons, n = 2, l = 0, ml = 0, ms = ±1/2.
For the two 2p electrons, n = 2, l = 1, ml = 1, 0, −1; ms = ±1/2.
6. Key to Prob. 8.22
[Ar]3d4 4s2 is more energetic because it has 4 unpaired spins compared to 6 for
[Ar]3d5 4s1 . According to Hund’s rule, the latter, with 2 more unpaired spins,
should be more stable (and it is). The element is chromium (Cr).
2