CHEM 155 Instrumental Analysis Prof. Terrill DH-007 924-4970 [email protected] Page 1 of 3
Homework 8 Solutions
8-1 The flame absorption peak for calcium in a low-temperature flame is broad because the
calcium is largely present as the calcium molecule CaOH, which has many vibrational and
rotational states and thus many excited energy levels. Thus a broad molecular absorption
peak is observed. In contrast, barium is apparently present only as atoms that only absorb at
a few discrete wavelengths.
8-2 Resonance fluorescence is a type of fluorescence in which the emitted radiation has a
wavelength that is identical to the wavelength of the radiation used to excite the
fluorescence.
8-4 Natural line widths in atomic spectroscopy are the widths of lines when only the uncertainty
principle, and not Doppler and pressure broadening, contribute to the broadening.
8-5 In the presence of KCI ionization of sodium is avoided because of the high concentration of
electrons from ionization of potassium. In the absence of KCI some of the sodium is
ionized, which leads to a lower intensity of the emission line for atomic sodium.
8-6 The energy necessary to promote a ground state s electron to the next p level is so high for
Cs that only a small fraction of the Cs atoms are excited at the temperature of the natural gas
flame. At the higher temperature of the hydrogen/oxygen flame a much larger fraction of
the atoms is excited and thus a more intense emission line for Cs is observed.
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CHEM 155 Instrumental Analysis Prof. Terrill DH-007 924-4970 [email protected] Page 2 of 3
8-10. The key to solving this is simply in seeing the ordering of the energy levels.
3s Æ 3pÆ 4s – the energy level diagram for Na Figure 8-1 is essential. The
energy difference between the 3p and 3s state for sodium was 3.37x10-19 J. The
energy difference between the 4s and 3p state can be calculated from the
emission wavelength, 1139 nm. The energy difference from 4s to 3s is simply
the sum of these two steps.
8-11. This was a tricky one – but not that complex once you understand the
dynamic. The low absorbance at low concentration implies the loss of Uatoms at low concentration. One loss mechanism is ionization. Ionization,
just like for a weak acid for example, is more extensive at low
concentration because of the equilibrium expression:
KION = [Un+] [e-] / [U]
Total U atom species entering the plasma - [UO]
U atoms left after ionization
- [U] = [UO] - x (this is what we want)
U ions formed
- [x]
Electrons released
- [e-] = [x]
2
KION = x / [UO] – x
Now drop the ‘ion’ subscript and the parentheses
KUo – Kx = x2
x2 + Kx – KUO = 0 Æ Quadratic… a = 1, b = K, c = -KUO
x = -b±(b2-4ac)1/2 / 2a
x = -K±(K2+4KUO)1/2 / 2
Æ [U] = [UO] – x = nonlinear at low [UO] concentrations… see graphs on following
page:
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CHEM 155 Instrumental Analysis Prof. Terrill DH-007 924-4970 [email protected] Page 3 of 3
Assumed ionization constant:
Concentration of ions:
Uo
8 .10
4
6 .10
4
x K , U o 4 .10
4
2 .10
4
0
2 .10
0
K
x( K , Uo )
4
4 .10
10
4
2
K
K
4 . K . Uo
2
4
Uo
6 .10
4
8 .10
4
0.001
Fraction ionized:
1
x K, U o
Uo
z
0.1
z
0.01
7
1 .10
1 .10
6
1 .10
5
4
1 .10
Uo
z
C:\SJSU\Teaching\155\2004 ASpring\Homework\Hw8 solutions.doc
1 .10
3
0.01
0.1
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