206 3 Quantum Mechanics – II
=
Nn^2
α^3∫∞
−∞Hn^2 (ξ)ξ^2 e−ξ2
dξ=
(
α
√
π 2 nn!)(
1
α^3)
(2n+1)
22 nn!√
π=
1
α^2(
n+1
2
)
=
√
km(
n+1
2
)
∴Δx.ΔP=√
mEn√
km(
n+1
2
)
Now,ω=√
k
m∴Δx.ΔPx=√
ω(
n+1
2
)
√
m
k(
n+1
2
)
=
(
n+1
2
)
Thus,Δx.ΔP≥ 2 is in agreement with uncertainty principle.3.59 The vibrational levels are equally spaced and so with the ruleΔn=1, the lines
should coincide. The rotational levels are progressively further spaced such
that the difference in the wave number of consecutive lines must be constant.
This can be seen as follows:
EJ=
J(J+1)^2
2 I 0
c
λJ=ΔE=EJ+ 1 −EJ=
[(J+ 1 )(J+2)−J(J+1)]^2
2 I 0
=
(J+1)^2
2 I 0
ThereforeλJ^1 + 1 −λ^1 J=Δ(
1
λJ)
α[(J+2)−(J+1)]=constant103. 73 − 83. 03 = 20 .70 cm−^1 ; 124. 30 − 103. 73 = 20 .57 cm−^1 ;
145. 03 − 124. 30 = 20 .73 cm−^1 ; 165. 51 − 145. 03
= 20 .48 cm−^1 ; 185. 86 − 165. 51 = 20 .35 cm−^1The data are consistent with a constant difference, the mean value being,
20 .556 cm−^1. Thus the transitions are rotational.
EJ+ 1 EJ
EJ+ 2 −EJ+ 1=
2(J+1)
2(J+2)
=
83. 03
103. 73
= 0. 8
ThereforeJ= 3
The levels are characterized byJ= 3 , 4 , 5 , 6 , 7 , 8 , 9