When solving this problem none of the fruits or vegetables was destroyed! In 1611 German mathematici — Analytical Chemistry Chemistry Question
Fruits, vegetables, atoms
When solving this problem none of the fruits or vegetables was destroyed!
In 1611 German mathematician and astronomer Johannes Kepler observed the stacking of cannonballs in a pyramid. He asserted there is the only way to fill the space the tightest possible with equal hard spheres, "…so that in no other arrangement could more pellets be stuffed into the same container". He was the first to formulate such a problem termed later as Kepler Conjecture. In 1998 Professor Thomas Hales announced a solution to the Kepler Conjecture, which was published in a series of papers in "Discrete and Computational Geometry" starting from 1997. He considered 150 more variants of space filling besides that asserted by Kepler. Hales’ solution required about 250 pages in a printed version and a size of 3 Gb in computer files. Thus, the term of close-packing of spheres (c.p.s.) widely accepted in the field of solid state chemistry passed through the rigorous mathematical proof and remained valid.
We do not request that you provide an alternative solution to this problem. However, you can check with our help how the basic law of space filling is applicable to our everyday life.
In order to avoid smashing tomatoes during their transportation, it is useful to arrange them on a shelf in a uniform single layer. Let us consider two types of packing (Fig. 16.1).
a) Calculate the density of tomatoes packing (φ) for the case A and B as
φ = Stomato / (Svoid + Stomato).
b) Which type of the packing requires less shelf area?
[VISUAL]
Model Answer
a) Since tomatoes touch each other in layers A and B, regular n-polygons (where n is the number of the nearest neighbors) with touch points located in the middle of their sides define the square relevant to one tomato. Among n-polygons only squares and hexagons fill space without voids. Therefore, φ = Stomat/Spolygon.
R is the radius of vegetable or fruit.
Ssquare = 4R^2, Shexagon = 2 * sqrt(3) * R^2. Stomat = π R^2.
Case A: φ = π / 4 ≈ 0.7854
Case B: φ = π / (2 * sqrt(3)) ≈ 0.9069
b) Type B requires less shelf area (higher packing density).
Hard vegetables such as potatoes or cabbage heads can be packed in containers. Consider several types of packing:
(1) The first layer is of the type A (see Fig. 16.1). The second layer is an exact copy of the first, a vegetable in the second layer is above another one in the first layer (such a packing is termed usually as simple cubic packing , or s.c.).
(2) The first layer is of the type A. In the second layer each vegetable is above a void space in the first layer (body centered cubic packing, or b.c.c.).
(3) The first layer is of the type B. The second layer is an exact copy of the first, a vegetable in the second layer is above another one in the first layer (hexagonal packing, or h.p.).
(4) The first layer is of the type B. In the second layer each vegetable is above a void space in the first layer (hexagonal close packing, or h.c.p.).
a) Calculate the densities of packing for the cases (1) – (4).
b) Which type of packing is more efficient in the sense of van filling?
c) There are two alternatives to arrange the third layer in the case B: i) by placing vegetables right above the vegetables of the first layer (that is to place them into the voids of the second layer) or ii) by arranging vegetables right above the voids of the first layer (see the case B in Fig. 2). Calculate the density of packing φ for the second alternative which is called the face centered cubic packing – f.c.c.
d) A farmer filled the third layer in the way of f.c.c. and now can not figure out where the voids and vegetables of the first layer are. How does the value of φ vary due to the faults in regular sequence of closed packed layers?
Model Answer
a) The density of a packing can be estimated as the ratio of the volume of all tomatoes (Z) with the radius R filling the space inside of an arbitrarily chosen bulk polyhedron (P) of a certain volume VP.
φ = 4/3 * π * Z * R^3 / VP
- s.c. (1):
P: Cube, a = 2R
VP = 8R^3
Z = 8 * (1/8) = 1
φ = 0.5236
- b.c.c. (2):
P: Cube, a = 4 * sqrt(3) / 3 * R
VP = 64 * sqrt(3) / 9 * R^3
Z = 1 + 8 * (1/8) = 2
φ = 0.6802
- h.p. (3):
P: Rhombic prism, h = 2R, L = 2R
VP = 4 * sqrt(3) * R^3
Z = 4 * (1/12) + 4 * (1/6) = 1
φ = 0.6046
- h.c.p. (4):
P: Rhombic prism, h = 4 * sqrt(6) / 3 * R, L = 2R
VP = 8 * sqrt(2) * R^3
Z = 1 + 4 * (1/12) + 4 * (1/6) = 2
φ = 0.7405
b) The case (4) (h.c.p.) corresponds to the most efficient way to fill space.
c) Calculation for f.c.c.: P is a cube a = 2 * sqrt(2) * R, Z = 6 * (1/2) + 8 * (1/8) = 4, VP = 16 * sqrt(2) * R^3.
φ = 4 * (4/3 * π * R^3) / (16 * sqrt(2) * R^3) = π / (3 * sqrt(2)) ≈ 0.7405.
d) For close-packed spheres (c.p.s.), φ does not depend on the type of the layer sequence (so it remains constant at ≈ 0.7405 despite sequence faults).
Assume now that the enterprising farmer decided to place peaches into the van with watermelons. His bright idea was to place peaches into the voids of watermelon packing.
a) Estimate the maximal value of the Rpeach / Rwatermelon radii ratio that allows to avoid peach smashing in cases of:
(1) cubic void within s.c.
(2) octahedral void within b.c.c.
(3) octahedral void within f.c.c.
b) How many peaches (maximum) per one watermelon can the farmer place using s.c. (referred to as c.s.), h.c.p., b.c.c. and f.c.c. types of packing?
c) What is the maximal φ value for s.c. (referred to as c.s.), b.c.c. and f.c.c. packings containing peaches in voids?
Model Answer
a) In order to avoid peaches smashing, the radius of a void should be less than the radius of a peach (r – radius of a peach, R – radius of a watermelon).
(1) s.c.: 2r < (a_s.c. * sqrt(3) - 2R) ⇒ r(max)/R = sqrt(3) - 1 ≈ 0.7321
(2) b.c.c.: 2r < (a_b.c.c. - 2R) ⇒ r(max)/R = 2/sqrt(3) - 1 ≈ 0.1547
(3) f.c.c.: 2r < (a_f.c.c. - 2R) ⇒ r(max)/R = sqrt(2) - 1 ≈ 0.4142
b) The number of peaches cannot exceed that of corresponding voids:
- s.c.: Zpeach / Zwatermelon = 1 / 1 = 1
- b.c.c.: Zpeach / Zwatermelon = 6 / 2 = 3
- h.c.p.: Zpeach / Zwatermelon = 2 / 1 = 2
- f.c.c.: Zpeach / Zwatermelon = 4 / 4 = 1
c) Maximal density is calculated by: φ = 4/3 * π * (Z_watermelon * R^3 + Z_peach * r^max^3) / VP
- s.c.: P: Cube, a = 2R, VP = 8R^3, Zpeach/Zwatermelon = 1, (1 + Zpeach/Zwatermelon * (r/R)^3) = 1.3924, φ = 0.721
- b.c.c.: P: Cube, a = 4 * sqrt(3) / 3 * R, VP = 64 * sqrt(3) / 9 * R^3, Zpeach/Zwatermelon = 3, (1 + Zpeach/Zwatermelon * (r/R)^3) = 1.0111, φ = 0.6878
- f.c.c.: P: Cube, a = 2 * sqrt(2) * R, VP = 16 * sqrt(2) * R^3, Zpeach/Zwatermelon = 1, (1 + Zpeach/Zwatermelon * (r/R)^3) = 1.0711, φ = 0.7931
The fruits can go bad due to insufficient ventilation in the van.
a) In order to keep the voids in b.c.c. and f.c.p. packings the go-ahead farmer decided to put peaches only in the octahedral voids which are not linked by edges and faces. How many peaches per one watermelon can be packed in this case?
b) The enterprising farmer has got another idea: to feel all the octahedral voids in f.c.c. with peaches (you know about it), whereas (it’s brilliant!) the tetrahedral voids with apples. How many apples per one watermelon can he arranged in this way?
Model Answer
a) In the case of b.c.c., ventilation of voids can be achieved by filling 1/4 of the voids, so the optimal ratio is 1/4 peach per watermelon.
Similarly, in the case of f.c.c., the optimal ratio is also 1/4 peach per watermelon.
b) For f.c.c., Zapple = 8 accounting for 4 watermelons (forming the f.c.c. unit cell), giving Zapple / Zwatermelon = 2 apples per watermelon.
a) Find the minimal values of Miller indices – (h k l) related to the first “permitted” reflection in f.c.c.
b) Calculate the wavelength of light if the first reflection is observed for 2 θ = 60°. The radius of SiO2 microspheres is equal to 450 nm. The dispersion of SiO2 refraction index (that is, its dependence on wavelength) can be neglected.
Model Answer
a) The condition of diffraction maximum for f.c.c. is: h + k = 2n, k + l = 2m, h + l = 2q, where m, n, q are integers. Hence, the first permitted reflection with the minimal Miller indices is (1 1 1).
b) For f.c.c. with sphere radius r = 450 nm:
a = 2 * sqrt(2) * r
d = a / sqrt(h^2 + k^2 + l^2) = 2 * sqrt(2) * r / sqrt(3) = 2 * sqrt(2/3) * 450 ≈ 734.85 nm
Using Bragg's law as solved in the text:
λ = d * sin(30°) = 734.85 * 1/2 ≈ 367.42 nm.