If equal, rigid spheres are arranged in a simple cubic lattice in the usual way (that is, in such a way that they touch each other), what fraction of the corresponding solid will be empty space? [The volume of a sphere is (4/3) $\pi$r³, with $\pi$ = 3.14.] A) 0.52 B) 0.32 C) 0.68 D) 0.48 E) none of these

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Quizplus · Accepted Answer

Consider a single sphere in the simple cubic lattice. Its volume is (4/3)F1F1F1S1F1F1F10rS1U1P13S1S1P0, and since the spheres are touching each other, all of this volume is occupied by the sphere itself. However, there are 6 nearest neighbor spheres surrounding each sphere, and their volumes are overlapping with that of the original sphere. Each neighboring sphere contributes a "wedge" to the space around the original sphere. The wedge can be thought of as forming when the neighboring sphere is cut into two halves, and one half is removed. The resulting shape is a frustrum of a sphere (i.e. a slice of the sphere that is bounded by two parallel planes). The volume of this frustrum can be calculated by subtracting the volume of a conical frustum (formed by cutting a cone into two halves and removing one half) from the volume of a cylinder. The volume of a frustrum of a sphere is (1/3)F1F1F1S1F1F1F1S1S1P02(h3 + r3 - R3), where h is the height of the frustrum, r is the radius of the small base, and R is the radius of the large base (i.e. the radius of the original sphere). The height of the frustrum is just the radius of the neighboring sphere, minus the distance between the centers of the two spheres (which is also the radius of the original sphere). Using the Pythagorean theorem, we can calculate this distance to be S1S1P02. Putting everything together, the volume of the empty space around each sphere is (1/3)F1F1F1S1F1F1F1S1S1P02(3rS1S1P02 - S1S1P01)S1S1P03, and the total volume of empty space is 6 times this. Hence, the fraction of empty space is 6(1/3)F1F1F1S1F1F1F1S1S1P02(3rS1S1P02 - S1S1P01)S1S1P03/(4F1F1F1S1F1F1F10rS1U1P13S1S1P0) = 3 - 8/3F1F1F1S1S1P00. Solving for this, we get 0.48, which is answer choice D.

If Equal, Rigid Spheres Are Arranged in a Simple Cubic π\piπ

If Equal, Rigid Spheres Are Arranged in a Simple Cubic $\pi$