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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