12.14
12.15
The limits are r=1.058 or x=2 and infinity so the probability is
equal to:
0 −(−e−^4 (2 × 42 + 2 × 4 +1)) =0.7512.16
The limits are r=0.5 Å or x=1 and r=1.058 Å or x=2, so the
probability is:− 4 e−^4 (2 × 4 + 4 +1) + 4 e−^2 (2 + 2 +1) =−0.44 +0.68 =0.2412.17 The most probable radial position of the electron is simply the peak
position of this term. We can find the peak by setting the derivative
equal to zero:12.18
=
3
2
a 04
44
03
02
032
00
0 2
arera r a xex x ae∞
− −∞
−
∫∫==/ ddxx⎛−− − −xxx
⎝⎜⎜
⎞
⎠
⎟⎟
32 ∞(^20)
3
4
3
4
3
8
ψψτ
π*drr π d
a= errra/⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
− =
∞ ∞
∫∫1
4
02
0 00 2 4 /
03
0(^20)
a
rera r
∞
−
∫ d
0
8
1
0
32
00=−/^0
⎛
⎝
⎜⎜
⎞
⎠
− ⎟⎟ =
aerr
ara or ra04
2 14
032
03= /^0 (
⎛
⎝
⎜⎜
⎞
⎠
− ⎟⎟=
d
dd
r dr
ae
arπ ra r
π(^22) e−^2 ra/ (^0) )
04 = ()^2
d
d*
rπψψrletxra/ then d d
a== 0 ∫err ex−ra/ −x
03(^4202242) xxex x=− x()+ +
∫
− (^22221)
1
4
4
0
32
2
0
3002 2
ππ
aerr
a−−ra erra⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
∫ //dd= ∫ dr
letxra/ then / d
a== 0 ∫err ex−ra −x
0
3(^4202242) ddxex x=− x()+ +
∫
− (^22221)
1
4
4
032
2
03
0 2 02
ππ
aerr
a−ra er−ra⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
∫ / dd= ∫ / dr
ψψψpppx=− ()()+−− =xf r1
2
468 ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 468