Key 12
1. Key to Prob. 10.2
For the l = 1 to l = 2 transition,
4E = hf =
[2(2 + 1) − 1(1 + 1)]h̄2
,
2I
or, hf = 2h̄2 /I. Solving for I gives
I=
2h̄2
h
6.626 × 10−34 J.s
= 2 =
) = 1.46 × 10−46 kg.m2 .
hf
2π f
(2π 2 )(2/3 × 1011 Hz
The effective mass is,
µ=
m1 m2
= 1.14 × 10−26 kg,
m1 + m2
and the bond length is
s
R0 =
I
= 1.13AA.
µ
2. Key to Prob. 10.4
(a) The separation between two adjacent rotational levels is given by, 4E =
(h̄2 /I)l, where l is the quantum number of the higher level. Therefore,
4E1,0 = 4E6,5 /6.
So, λ1,0 = 6λ6,5 = 6(1.35 cm) = 8.10 cm. The frequency is then,
f1,0 =
c
3 × 1010 cm/s
=
= 3.70 GHz.
λ1,0
8.10 cm
(b) 4E1,0 = hf1,0 = h̄2 /I. SOlving for I gives,
h̄
1.055 × 10−34 J.s
I=
=
= 4.53 × 10−45 kg.m2 .
9
2πf1,0
(2π)(3.7 × 10 Hz)
1
3. Key to Prob. 10.6
HCl molecule in the l = 1 rotational energy level: R0 = 1.275 Å. Erot =
√
(h̄2 /2I)l(l + 1). For l = 1 , Erot = h̄2 /I = Iω 2 /2. So, ω = 2(h̄/I). Now,
I=
m1 m2 2 (1 u)(35; u) 2
R =
R ,
m1 + m2 0
1 u + 35; u 0
or,
I=
√
2[0.9722 u×1.66×10−27 kg/u]×(2.62×10−47 kg.m2 ) = 2.62×10−47 kg.m2 .
Therefore,
ω=
√
√
2h̄/I = (1.055 × 10−34 J.s)/(2.62 × 10−47 kg.m2 )) 2 = 5.69 × 1012 rad/s.
4. Key to Prob. 10.7
(a) HI, 4E0→1 = hf = (h/2π)(k/µ)1/2 or 2πf =
q
K/µ or K = 4π 2 f 2 µ, where
µHI = m1 m2 /(m1 + m2 ). Putting all the values we get, µHI = 1.65 × 10−27 kg
and µN O = 1.24 × 10−26 kg. So, KHI = 292 N/m, and KN O = 1550 N/m.
q
q
2
(b) KA2 /2 = (h̄/2) K/µ or A2 = h̄ 1/Kµ. Then, KHI
= 1.519 × 10−22 m2 .
So, AHI = 0.123 Å= 0.0123 nm. Likewise, AN O = 0.0492 Åor 0.00492 nm.
(c) N and O are “cemented” by more electrons.
5. Key to Prob. 10.9
We take Evib = 4.5 eV in the formula
Evib = 4.5 eV = (ν + 1/2)h̄ω,
= (ν + 1/2)(6.582 × 10−16 eV.s)(8.277 × 1014 rad/s),
= (ν + 1/2)(0.5448 eV).
2
Solving for ν we get, ν = 7.760. Ofcourse, ν must be an integer, so that ν = 7
represents the highest vibrational level that can be excited without the molecule
coming apart.
3