Chemistry 660 Problem Set Keys--Fall 2003

Dr. James Ingle

answers to problem set 13 (practice) - posted 12/06/03

13-11. A = ebc = 1.00 x 105 L mol-1 cm-1 x 1.00 cm x 2.00 x 10-5 M

A = 2.00, T = 0.01; stray radiation readout fraction r = 0.02

A' = - log T' = - log (0.01 + 0.02) = 1.52

13-22. A = - log T = - log { [(is)1 + (is)2]/ [(ir)1 + (ir)2]}

A = - log {[(is )1 + (is )2] /2ir }

But, - log [(is )1/ir] = e1bc, and - log [(is)2/ir] = e2bc

Use e1 = 1.0 x 104 L mol-1 cm-1 and e2 = 1.0 x 105 L mol-1 cm-1

For 1 mM, A = 0.053; for 10, 20 and 30 mM, A = 0.349, 0.494 and 0.600

 answers to problem set 6 (practice) - posted 10/19/03

4-7. a) 0.25, b) rap = 6.1 x 104 s-1 , iap = 9.8 x 10-15 A

4-8. icp = 3.9 x 10-16 A, F = 5.5 x 10-14 W or 1 x 105 photons/s, rap = 2.4 x 103 s-1

4-16. a) t = q/(ER(l)) = 0.3 s, b) # e-hole pairs = q/e = 2.25 x 107

note that ER(l) has units of pA or pC/s and is equivalent to ΦpK(l)e

practice problem set 3

7.  NA  = 0.244, theta zero = 14.1 degrees, F/n = 1.99

13. f = 6 cm, omega = 0.022 sr, E = 0.028 W cm-2

practice problem set 2

1.

400 nm

7.5 x 1014 Hz

4.97 x 10-19 J

4.97 x 10-12 ergs

3.10 eV

2.5 x 104 cm-1

11. 0.010 mM

 

NOT revised below this point for F03

Practice problem set 11.

 

12-1. fF = kF/(kF + knr) = 1 x 109 s-1/(1 x 109 s-1 + 9 x 109 s-1) = 0.1

tF = 1/(kF + knr) = 1/(1 x 109 s-1 + 9 x 109 s-1) = 10-10 s = 0.1 ns

(fF)r = 1/kF = 1/1 x 109 s-1 = 1 ns

12-7. fFo/fF = 1 + Kq [Q]

fF  proportional to tF

6/2 = 1 + Kq x 10-2 M

Kq = 2 x 102 L mol-1

tFo = 1/(kF + kisc)

tFo kF + tFo kisc = 1

kF = (1 - tFo kisc)/tFo

kF = (1 - 6 x 10-9 s-1 x 1 x 108 s-1)/6 x 10-9 s = 6.7 x 107 s-1

Kq = kq tFo

kq = Kq/tFo = 2 x 102 L mol-1 /6 x 10-9 s = 3.3 x 1010 s-1 M-1

Practice problem set 1.

1.

400 nm

7.5 x 1014 Hz

4.97 x 10-19 J

4.97 x 10-12 ergs

3.10 eV

2.5 x 104 cm-1

11. 0.010 mM.

Practice problem set 3.

2.

S2 is at infinity because collimated, the magnification is undefined

F/n of system is 2

the solid angle is 0.196 sr

6. 4.88 mm

10. 19.2 degrees

31.  3.16 x 10 13 mol cm-3

 

Key for Problem Set 14 (Practice) - posted 11/30/01

10-1. a. When the monochromator is set to 213.9 nm to measure the absorption of Zn, some of the Cu emission lines will also be passed by a monochromator if the spectral bandpass is 5 nm. This means that part of the reference or blank signal will be due to Cu emission radiation that cannot be absorbed by Zn atoms in the flame. (i) If there is no Cu in the standards, the unabsorbed radiation represents a stray radiation component and will cause a lower initial calibration slope and a negative deviation. (ii) If there is Cu in the standards, some of the Cu radiation will be absorbed causing a non-zero intercept. There will also be a lower initial slope and negative deviation as above, but to a lesser extent because some of the of the Cu emission radiation is absorbed.

b. The separation between the Zn line and the closest Cu line is 2.6 nm. As the wavelength is fixed (no scanning), the spectral bandpass must be less or equal to the wavelength difference to resolve the lines. Hence W = 2.6 nm/(2 nm/mm) = 1.3 mm.

10-18. For absorption measurements in the UV-visible region, the limiting noise sources at or near the detection limit (where the lamp photon signal is greatest), are usually signal shot noise or signal flicker noise. When the spectral bandpass is increased, the lamp photon flux reaching the detector increases. If signal shot noise was limiting, the relative shot noise would decrease in proportion to the square root of the increase in the lamp signal and the DL would improve. As the DL did not change, the limiting noise source must be signal flicker noise which could be lamp flicker noise or flame transmission noise in AAS.

Note that analyte absorption flicker noise or analyte emission noise is never limiting at the detection limit as they are proportional to the analyte concentration which is very low at the detection limit. Also note that dark current noise or amplifier noise cannot be limiting as they are independent of the spectral bandpass. Thus the ratio of the lamp signal to noise from these sources would increase with increasing s and the detection limit would improve if they were limiting.

The independence of the DL on spectral bandpass would be observed if the background emission was a continuum and background emission shot noise was limiting. This is very unlikely as the lamp signal is normally much greater than the background signal.

kq = Kq/tFo = 2 x 102 L mol-1 /6 x 10-9 s = 3.3 x 1010 s-1 M-1

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Last updated on December 06, 2003