212 3 Quantum Mechanics – II
Writing sin^2 θ= 1 −cos^2 θand simplifying we getu∗u=^29 A^23 e−^2 xx^4 which
is independent of bothθandφ. Therefore the 3dfunctions are spherically
symmetrical or isotropic.
3.71ψ 100 =
(
πa^30
)−^12
exp
(
−ar 0
)
The probabilitypof finding the electron within a sphere of radiusRis
P=
∫R
0
|ψ 100 |^2. 4 πr^2 dr=
(
4 π
πa^30
)∫R
0
r^2 exp
(
−
2 r
a 0
)
dr
Set^2 ar 0 =x;dr=
(a 0
2
)
dx
P=
(
4
a 03
)
·
(
a^20
4
)
·
(a 0
2
)∫
x^2 e−xdx=^1 / 2
∫
x^2 e−xdx
Integrating by parts
P=^1 / 2 [−x^2 e−x+ 2
∫
xe−xdx]
=^1 / 2
[
−x^2 e−x+ 2
{
−xe−x+
∫
e−xdx
}]
=^1 / 2
[
−x^2 e−x− 2 xe−x− 2 e−x
] 2 R/a 0
0
=^1 / 2
[
−
(
2 R
a 0
) 2
exp
(
−
2 R
a 0
)
− 2
(
2 R
a 0
)
exp
(
−
2 R
a 0
)
−2exp
(
−
2 R
a 0
)
+ 2
]
P= 1 −e−
(^2) aR
0
(
1 +
2 R
a 0
+
2 R^2
a^20
)
3.72 The hydrogen wave function forn=2 orbit is
ψ 200 =
(
1
4
)
(
2 πa 03
)−^12
(
2 −
r
a 0
)
e−r/^2 a^0
The probability of finding the electron at a distancerfrom the nucleus
P=|ψ 200 |^2 · 4 πr^2 =
1
8
r^2
a^30
(
2 −
r
a 0
) 2
exp
(
−
r
a 0
)
Maxima are obtained from the condition dp/dr=0.
The maxima occur atr=(3−
√
3)a 0 andr=(3+
√
3)a 0 while minimum
occur atr= 0 , 2 a 0 and∞(Fig. 3.25) (the minima are found from the condi-
tionp=0).
3.73 (a)hν= 13. 6 Z^2
(
mμ
me
)(
1
22 −
1
32
)
Putmμ=106 MeV (instead of 106.7 MeV for muon)
Z=15 andme= 0 .511 MeV