J 5 →J6: line position 12 B124.3 cm^1 ,
therefore B10.36 cm^1
J 6 →J7: line position 14 B145.0 cm^1 ,
therefore B10.36 cm^1
J 7 →J8: line position 16 B165.5 cm^1 ,
therefore B10.34 cm^1
The average value for Bfrom this data is 10.37 cm^1. The reciprocal of
this gives an appropriate equivalent wavelength in centimeters, which is
0.09643 cm, or 9.643
10 ^4 m. Using c , this converts into a frequency
of3.109
1011 s^1 , and using the equation Eh, this converts to an
energy of 2.060
10 ^22 J. (You should satisfy yourself on the reasons for all
these steps in the conversion to joules.) Since the energy unit is joules, which
is defined in terms of kg, the reduced mass of the^1 H^35 Cl atom must be ex-
pressed in units of kg. The H atom has a mass of 1.66
10 ^27 kg, and the
Cl atom has a mass of 5.81
10 ^26 kg. The reduced mass is therefore
1.61
10 ^27 kg. The only unknown in the expression
B
22
r^2is the distance r. Substituting for the known values (and not forgetting the 2
part of):
2.060
10 ^22 J Rearranging and solving for the radius r:
r1.29
10 ^10 m 1.29 ÅThis is how rotational spectroscopy gives us an idea about the actual sizes of
molecules.
Example 14.9
NO has a bond length of 1.151 Å. Predict where the first four lines of the pure
rotational spectrum of NO will appear, in units of GHz. Refer to Example
14.8 for calculational details.
Solution
The reduced mass of NO is calculated to be 1.24
10 ^26 kg. We can calcu-
late a value ofB:
Bwhere we have used the definition 1 Å 10 ^10 m to get all the units consis-
tent with each other. We get
B6.77
10 ^23 JIn units of frequency, we need to use Ehto get 1.02
1011 s^1. Since
this can be written as 102
109 s^1 , and since a GHz is 10^9 Hz, this corre-
sponds to a value of 102 GHz. The first four lines, appearing at 2B,4B,6B,
and 8B, are therefore found at 204, 408, 612, and 816 GHz.
(6.626
10 ^34 J s)^2
(2)^2 2(1.24
10 ^26 kg)(1.151
10 ^10 m)^2(6.626
10 ^34 J s)^2
(2)^2 2(1.61
10 ^27 kg)r^214.6 Rotational Spectroscopy 475