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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 369
From eqn 11.1 it is seen that for each transition (J + 1 ← J) there are
separate lines for each value of K2.�e K quantum number takes values
from −J to +J in integer steps, and so has (2J + 1) values. However, the
energy goes as K2, of which there are only (J + 1) distinct values. �e
transition 1 ← 0 has J = 0 and hence K = 0 only: there is one line. �e
transition 2 ← 1 has J = 1 and hence K = 0, 1 : there are two lines, and
because the term in K2 is negative, the line for K = 1 is at lower frequency
that that for K = 0.�e transition 3 ← 2 has J = 2 and hence K = 0, 1, 2
: there are three lines, running to lower frequency as K increases.
�e following table indicates the quantumnumbers for each line and their
frequencies expressed in terms of the various constants.
line observed freq./GHz J K ν(J ,K)
1 51.0718 0 0 2B − 4DJ
2 102.1408 1 1 4B − 32DJ − 4DJK
3 102.1426 1 0 4B − 32DJ
not given 2 2 6B − 108DJ − 24DJK
4 153.2076 2 1 6B − 108DJ − 6DJK
5 153.2103 2 0 6B − 108DJ
�e assignment of lines 4 and 5 needs some comment. With the assign-
ment given the separation of lines 4 and 5 is 6DJK , and the separation of
lines 2 and 3 is 4DJK ; these separations are therefore expected to be in the
ratio 6/4 = 1.5. From the data the ratio is (153.2103−153.2076)/(102.1426−
102.1408) = 1.5, which is consistent with the assignment. Such consis-
tency is not achieved with any other values of K for lines 4 and 5.
With this assignment it is possible to �nd the constants but taking dif-
ferences between multiples of the frequencies of di�erent lines from the
table:
line 3 − line 2 = 4DJK = 102.1426 − 102.1408 = 0.0018 GHz
hence DJK = 1
4 × 0.0018 GHz = 0.45 MHz .
line 3 − 2 × line 1 = −24DJ = 102.1426 − 2 × 51.0718 = −0.0010 GHz
hence DJ = 1
24 × 0.0010 GHz = 0.042 MHz . A di�erent choice gives a
di�erent value, indicating that the accuracy of the data is perhaps lower
than it appears
2 × line 5 − 3 × line 3 = −120DJ = 2 × 153.2103 − 3 × 102.1426
= −0.0072 GHz
hence DJ = 1
120 × 0.0072 GHz = 0.060 MHz .
line 3 − 8 × line 1 = −12B = 102.1426 − 8 × 51.0718 = −306.43 GHz
hence B = 1
12
× 306.4 GHz = 25.536 GHz