# ATKINS` PROBLEMS

```ATKINS’ PROBLEMS
Q. 13.5(a) Calculate the frequency of the J = 4
Theequilibriu bond length is 115 pm.
3 transition in the pure roatational spectru of 14N16O.
Q. 13.6(a) If the wavenumber of the J = 3
2 rotational transition of 1H35Cl considered as a rigid
rotor is 63.56 cm-1, what is (a) the moment of inertia of the molecule, (b) the bond length ?
Q.13.7(a) Given that the spacing of the lines in the microwave spectrum of 27Al1H is constant at 12.604
cm-1, calculate the monment of inertia and bond length of the molecule. (m(27Al) = 26.9815mu)
Q.13.8(a) The rotational constant of 127I35Cl is 0.1142 cm-1. Calculate the ICl bond length. (m(35Cl) =
34.9688mu, m(127I) = 126.905mu)
MOLECULAR
E13,a)
1.0 x 1013S = 0.10 ps, implying that
(b)
T ~
(100) x (1.0 x 1O-13s) = l Ops.Implying that Sf ~10.S3cm-ll.
Ii
= --,
4ncl
meff
265
IS3 cm-I I.
+ I) [13.31]
=
!
= meffR2
mNmO
[nuclide masses from
(14.003 u)
Then,!
and B
=
=
[Table 13.1], and
Data Section]
mN+mO
= ((14.003U)
S
SPECTRA
NO is a linear rotor and we assume there is little centrifugal distortion; hence
.
transitions
ov ~
AND VIBRATIONAL
T ~
WithB
-A
8p
1: ROTATIONAL
(a)
F(l) = Rl(l
mtaneous
SPECTROSCOPY
x (1.660S x 1O-27k u-I) = l.240 x 1O-26k .
g
g
x (1S.99SU))
+ (lS.99S
u)
(1.240 x 10-26 kg) x (I.IS x 1O-IOm)2
=
l.640 x 1O-46kgm2
1.0S46 x 10-34 J s
(4n) x (2.998 x 108 m S-I) x (l.640 x 10-46 kg m2)
=
-170.7 m-I
=
13.6cm-l.
=
l.707 cm-I.
The wavenumber of the 1 = 4 +- 3 transition is
.
by
Wit'h P given
v
ditional conclusions
= 2B(l + 1)[13.37] = 8B[l = 3] =
(8) x (1.707cm-l)
The frequency is
v
=
vc
=
(l3.65cm-l)
x (\O~:~I)
x (2.998 x 108m -I)
=14.09
x lOll Hz
I.
Question. What is the percentage change in these calculated values if centrifugal distortion is included?
E13.6(a)
(a)
The wavenumber of the transition is related to the rotational constant by
hcv
=
6.£
= hcB[l(J + 1) -
(l - I)l]
where 1 refers to the upper state (l
structure by
=
= 2hcBl
[13.2S, 13.27]
3). The rotational constant is related to molecular
B = _li_ [13.24]
4ncl
where I is moment of inertia. Putting these expressions together yields
_
v =2Bl=--
IiJ
2ncl
so
hi
(l.OS46 X 10-34 J s) x (3)
I - - - -:------:--::--::--:------:-:-:;;----;-:----,-:-:----:--:---.,-
- cv - 2n(2.998 x 1010cm s=') x (63.S6 cm-I)
= 12.642
X
10-47 kg m21.
266
STUDENT'S
SOLUTIONS
MANUAL
(b) The moment of inertia is related to the bond length by
I = meffR
m
-I
eff -
m
so
2
-I
H
R =
-I
+ m CI
if
-.
meff
(1.0078u)-1 + (34.9688u)-1
1.66054 x 10-27 kg u"!
-
-
26
= 6.1477 x 10 kg-
I
x 1026kg-I) x (2.642 x 1O-47kgm2) = 1.274 x lO-lOm =1 127.4pml·
andR = )(6.1477
7
3.7(a) If the spacing of lines is constant, the effects of centrifugal distortion are negligible. Hence we may use
for the wavenumbers of the transitions
F(]) - F(] -
1) =
2B] [13.27].
Since] = 1,2,3, ... , the spacing of the lines is 2B.
12.604cm-1 = 2B,
B = 6.302cm-1
I = --
Ii.
= 6.302 x 102 m-I.
[Section 13.4] =
4ncB
meffR
2
.
1.0546 x 10-34 J S
-----,--------::------:,.,...
(4n) x (2.9979 x 108ms-l)
Ii
4nc
= 2.7993 x 10
_I 4.442 xl
44
_ 2.7993 x 10- kg m
1- 6.302 x 102m-1
-
0-47 k
-44
kg m.
21
gm.
mAlmH
+mH
meff =
mAl
= (26.98) x (1.008))
(26.98) + (1.008)
u x (1.6605 x 1O-27k u-I)
g
I )1/2
(4.442 x 1O-47kgm2)1/2
R= ( =
-27
meff
1.6136 x 10
kg
Ii
,.8(a)
B = [13.24],
4ncl .
Then, with I =
meffR2,
We use meff =
_m_Im_2_
ml
+ mz
implying that
Ii
=1.659xlO
= 1.6136 x 1O-27k .
g
-10
m=1165.9pml·
.
Ii
1=--.
4ncB
)1/2
R = ( 4nmeffCB
= (126.904) x (34.9688) u = 27.4146u
(126.904) + (34.9688)
and hence obtain
II
1.05457 x 1O-34J s
) .
R = ( (4n) x (27.4146) x (1.66054 x 10-27 kg) x (2.99792 x IOlOcm S-I) x (0.1142 cm")
= 1232.1 pm
I.
```