12.24 The boundary surface represents the 90% probability value.12.25Angular momentum must be conserved and photons carry angular
momentum. Thus the change in lmust be 1.12.26 (a)(b) The wavefunctions with different angular dependences will no
longer be degenerate, and the dependence of the energy on
distance is now weaker, so the wavefunctions will be further
from the nucleus.12.27
12.28n^2 = 52 = 2512.29 An electron experiences a shielded nuclear charge due to the pres-
ence of the other electrons. This can be expressed by assigning an
effective nuclear charge that is lower than the atomic number.12.30
12.31Eatom=(Zeffatom)^2 EH=(2.4)^2 1312 kJ mol−^1 =7557 kJ mol−^112.32
12.33The outer electrons are both in 2s orbitals and beryllium has a larger
nuclear charge.−∇+∇ − +−
Z^2
12
22
122(^24012)
22
mrre
rr()()ψ
πε() ()
1
12
r rr^12 E rr^12⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
ψψ=Z
I
eff Iatom
atom
==H−
−2370
1312
1
1kJ mol
kJ mol= 134.
a
xxxex(^02654) x a
0
0
2
8
44
8
()(−+ = − 64 !(
∞
−
∫ d5544 345 0
!)+=( !)). a^2=−+
∞
−
∫()a(^0) xxxexx
2
654
(^80)
44 d
1
8
2
03 04
00
02a ax^000ax
a() eaxaxa/∞
−
∫ −⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ d ==−∞
−
∫ ()a(^0) xxexx
2
42
(^80)
2d
=−
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
∞
−
∫(^1) /
8
2
034(^00)
2
0
a
r
r
a
errad
ψψτ
π
*drr
a
r
a
(^22) era
0
3
0
2
1
16
1
2
=− 2 /^0
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ −
0 004 2
∞ ∞
∫∫ πrrd−∇ +
Z^22 −^2
0
24 m rr^3e
r()() rcos
,,θφψ θφ,, (,θ
πεψθ,,,,φψθφ)()=ErVre
r()
cos
,,θφ
πεθ
=−^2
04 3
470 ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 470