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 radiusRisP=∫ R
0|ψ 100 |^2. 4 πr^2 dr=(
4 π
πa^30)∫R
0r^2 exp(
−
2 r
a 0)
drSet^2 ar 0 =x;dr=(a 0
2)
dxP=
(
4
a 03)
·
(
a^20
4)
·
(a 0
2)∫
x^2 e−xdx=^1 / 2∫
x^2 e−xdxIntegrating by partsP=^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^0The probability of finding the electron at a distancerfrom the nucleusP=|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