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 × 4
2
+ 2 × 4 + 1)) = 0.75
12.16
The limits are r = 0.5 Å or x = 1 and r = 1.058 Å or x = 2, so the
probability is:
−4e
−4
(2 × 4 + 4 + 1) + 4e
−2
(2 + 2 + 1) = −0.44 + 0.68 = 0.24
12.17 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
0
a
4
4
4
0
3
0
2
0
3 2
0
0
2
0
a
r e
r
a x e
x
a e
r a
x
∞
−
−
∞
−
∫
∫
=
=
/ d
d
x x
x
x
x
−
−
−
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
∞
3
2
0
2
3
4
3
4
3
8
ψ ψ τ
π
π
* d
d
r
r
a
e
r r
r a
/
=
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
=
−
∞
∞
∫
∫
1
4
0
2
0
0
2
0
/
4
0
3
0
2 0
a
r e
r
r a
∞
−
∫
d
0
8
1
0
3
2
0
0
0
/
=
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
=
−
a
e
r
r
a
r a
r a
or
0
4
1
4
2
0
3
2
0
3
0
(
/
=
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ =
−
d
d
d
d
r
r
a
e
a r
r
r a
π π
2 2 2 0
e
r a
− / )
0
4
2
(
)
=
d
d
*
r
r
π ψ ψ
let
/ then
d
d
x r a
a
e
r r
e x
r a
x
/
=
=
−
−
∫
0
0
3
2
2
2
2
4
4
0
x x
e
x
x
x (
)
= −
+
+
∫
−2
2
2
2
1
1
4
4
0
3
2
2
0
3
2
2
0
0
π
π
a
e
r r
a
e
r
r a
r a
−
−
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
=
∫
/
/
d
d dr
∫
let
/ then
d
/
x r a
a
e
r r
e x
r a
x
=
=
−
−
∫
0
0
3
2
2
2
2
4
4
0
d dx
e
x
x
x (
)
= −
+
+
∫
−2
2
2
2
1
1
4
4
0
3
2
2
0
3
2
2
0
0
π
π
a
e
r r
a
e
r
r a
r a
−
−
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
=
∫
/
/
d
d dr
∫
ψ
ψ
ψ
p
p
p
x
xf r
(
)
()
= −
−
=
+
−
1
2
468
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 468