A maior rede de estudos do Brasil

Grátis
235 pág.
principios de analise instrumental 6ed skoog resolucoes

Pré-visualização | Página 33 de 33

recorded by detector 2, integrated, and displayed as shown in the accompanying diagram. 
 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 33
 
 5
 
 
 
 
 
Skoog/Holler/Crouch Chapter 34
Principles of Instrumental Analysis, 6th ed. Instructor’s Manual
 
 
 1 
CHAPTER 34 
34-1. The particle size distribution is a plot of the number of particles having a particular size 
(diameter, volume, surface area) versus that quantity. The cumulative distribution 
represents the fraction of particles bigger than or smaller than a particular size plotted 
against the size. 
34-2. The problem is that a single quantity, such as diameter, radius, volume, etc., cannot 
adequately describe a three dimensional particle unless it is spherical. 
34-3. Usually diameter (or radius), volume, or surface area. 
34-4. Generally, we assume the particles are spherical and proceed as we would with spherical 
particles, or the values are converted to those of an equivalent sphere. 
34-5. With particles that are not perfect spheres, even the same technique can give different 
results depending on what quantity is used to describe the size. Different instruments 
often measure different quantities or make assumptions related to the theoretical models. 
Also, other quantities may be needed like refractive index, viscosity, density, etc. These 
quantities may not be known accurately for the system under study. 
34-6. Mie scattering is scattering from finite objects with diameters larger than the wavelength 
of the incident light. Mie scattering shows an intensity distribution that varies with 
scattering angle because of constructive and destructive interference between the rays 
emitted from one part of the particle and those emitted from another. Mie scattering 
shows predominately forward scattering. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 34
 
 2
34-7. An Airy pattern is a diffraction pattern that is a function of x =2πrs/λf. It shows a 
maximum at x = 0 and small maxima to either side. The Airy pattern is broader for 
particles of smaller size and narrower for larger particles. The extrema shift to higher 
values of x as the particle size decreases. 
34-8. The cumulative undersize distribution shows the fraction of particles having diameters 
less than or equal to a particular diameter as a function of diameter. 
34-9. In laser diffraction, the pattern resulting from scattering of the incident laser beam is 
analyzed using Mie theory or Franhofer diffraction theory. The intensity distribution of 
the scattered light is used to obtain the particle size from the theoretical model. In 
dynamic light scattering, the time fluctuations of the Doppler broadened Rayleigh 
scattered light are analyzed by correlation techniques to obtain translational diffusion 
coefficients. These are then converted to particle diameters via a model such as the 
relationship. 
34-10. We use the Stokes-Einstein relationship to obtain the hydrodynamic diameter, dh. Note 
that 1 cp = 10–3 Pa s 
23
h
T
1.38066 10 J = = 3
kTd Dπη
−× K 1 293.15 K− × 31 m Pa× / J
33 1.002 10 Paπ −× × × s 7 23.75 10 cm−× × s 4 2 210 m / cm−×
8 = 1.14 10 m = 11.4 nm−×
 
34-11. 
23
T
h
1.38066 10 J = 3
kTD dπη
−×= K
1 293.15 K− × 31 m Pa× / J
33 1.002 10 Paπ −× × × 6
14 2 1 4 2 2 10 2 1
 s 35 10 m
 = 1.22 10 m s 10 cm / m = 1.22 10 cm s
−
− − −
× ×
× × × −
 
 
34-12. Here we note 1 G = 6.6720 × 10–11 N m2/kg2, and 1 cp = 10–3 N s m–2 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 34
 
 3
( )F 2
3 3 3 11 2
6 2
3 2
12 1 10 1
 = 18
(1.05 0.998) g cm 10 kg/g (100 cm/m) 6.6720 10 N m /kg = (10.0 10 m)
18 1.002 10 N s m
 = 1.92 10 m s 100 cm/m 1.92 10 cm s
d d g
u Dη
− − − −
− −
− − − −
−
− × × × × × ×× ×
× × = ×
 
To fall 10 mm would require t = h/u = 10 mm/(1.92 × 10–10 cm/s × 10 mm/cm) = 5.2 × 
109 s 
34-13.
 
( )
3 4 1 1 2 4
3 1 1 1
2 2
F
(1.05 0.998) g cm (1.0 10 rev/min 2 rad rev /60 s min ) 8.0 cm (10.0 10 cm)
18 1.002 10 kg m s 1000 g/kg 0.01 m cm
 = 
18
=
ard d Du
π
ω
η
− − −
− − − −
− × × × × × × ×
× × × ×
2−
−
= 2.53× 10–4 cm s–1
t = ln(ra/r0)ra/u = ln(80/70) × 8.0 cm/ 2.53 × 10–2 cm s–1 = 4.22 × 103 s or 70.4 min 
 The centrifugal acceleration is 
ω2ra = (10,000 rev/min× 2 × π rev–1/60 s/min)2 × 8.0 cm = 8.77 × 106 cm s–2 = 8.77 × 104 m s–2 
Near the earth’s surface, g is approximately 9.8 m s–2. Hence a 10000 rpm centrifuge produces 
an acceleration that is 8.77 × 104/9.8 = 8952 G or approximately 9000 G. 
34-14. u = ln(ra/r0)ra/t = ln(80/70) × 8.0 cm/2.9 s = 0.368 cm s–1
 
( )
( ) ( )
0
Stokes 2 2
18 ln / 18 = a a
a aF F
rr r u
r t d d r d d
D η ηω ω=− − 
 
2 1 1 1
2 3
18 1.002 10 g cm s 0.368 cm s = 145 nm or 0.145 μm
9000 9.8 m s 100 cm/m (1.05 0.998) g cm
− − − −
− −
× × ×= × × − 
 
	Ch01
	Ch02
	Ch03
	Ch04
	Ch05
	Ch06
	Ch07
	Ch08
	Ch09
	Ch10
	Ch11
	Ch12
	Ch13
	Ch14
	Ch15
	Ch16
	Ch17
	Ch18
	Ch19
	Ch20
	Ch21
	Ch22
	Ch23
	Ch24
	Ch25
	Ch26
	Ch27
	Ch28
	Ch29
	Ch30
	Ch31
	Ch32
	Ch33
	Ch34