12.24 The boundary surface represents the 90% probability value.
12.25 Angular momentum must be conserved and photons carry angular
momentum. Thus the change in l must 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
12.28 n
2
= 5
2
= 25
12.29 An electron experiences a shielded nuclear charge due to the presence of the other electrons. This can be expressed by assigning an
effective nuclear charge that is lower than the atomic number.
12.30
12.31 E
atom
= (Z eff
atom )
2
E
H
= (2.4)
2
1312 kJ mol
−1
= 7557 kJ mol
−1
12.32
12.33 The outer electrons are both in 2s orbitals and beryllium has a larger
nuclear charge.
−
∇ + ∇
−
+ −
Z
2
1
2
2
2
1 2
2
0 1
2
2
4
2 2
m
r r
e
r
r
(
) ( )
ψ
πε
( )
( )
1
12
1 2
1 2
r
r r
E r r
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
ψ
ψ
Z
I
I
eff
atom
atom
H
=
=
−
−
2370
1312
1
1
kJ mol
kJ mol
.
= 1 34
a
x
x
x e
x
a
x
0
2
6
5
4
0
0
2
8
4
4
8
6 4
(
)
(!
(
−
+
=
−
∞
−
∫
d
5
5
44
345 0
2
!)
( !))
.
+
=
a
=
−
+
∞
−
∫ (
)
a
x
x
x e
x
x
0
2
6
5
4
0
8
4
4
d
1
8
2
0
3
0
4
0
0
0
2
0
0
0
a
a x
a x
a
e
a x
a x a
( )
/
∞
−
∫
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
d = =
−
∞
−
∫ ( )
a x
x e
x
x
0
2
4
2
0
8
2
d
=
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
∞
−
∫
/
1
8
2
0
3
4
0
0
2
0
a
r
r
a
e
r
r a d
ψ ψ τ
π
*
d
r
r
a
r
a
e
r a
2
2
0
3
0
2
1
16
1
2
2
0
/
=
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
−
0 0
0
2
4
∞
∞
∫
∫
πr r
d
−
∇
+
−
Z
2
2
2
0
3
2
4
m
r
r
e
r
r
(
) (
)
cos
(
, ,
, ,
,
θ ϕ ψ θ ϕ
θ
πε
ψ θ, ,
, ,
ϕ
ψ θ ϕ
)
(
)
= E r
V r
e
r
(
)
cos
, ,
θ ϕ
πε
θ
=
−
2
0
3
4
470
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 470