The cubic volume is defined to be 0.478 nm along each edge. Since the unit cell has a cubic configuration, the volume of the unit cell can be determined by the following formula: Since each of the sides of the unit cell are defined to be 0.478 nm in length, the volume of the unit cell of CaO can be calculated to be: with significant figures applied The question also states that the cubic volume of CaO contains four Ca2+ ions and four O2- ions. The mass of the unit cell contributed by the ions is given by the following formula: Therefore, the contribution to the mass of the unit cell from each of the Ca2+ and O2- ions is given by modifying the above formula in the following manner.

The mass of one mole of Ca2+ ions is equal to the atomic mass of calcium. The atomic mass of calcium is given as 40.08 g/mol. Similarly, the mass of one mole of O2- ions is equal to the atomic mass of oxygen. The atomic mass of oxygen is given as 15.999 g/mol. The number of atoms or ions in one mole of any element (or the number of molecules in one mole of any compound) is defined to be equal to Avogadro’s number (Van Vlack, 1989). Avogadro’s number states that a mole of any element possesses 6.02 x 1023 atoms or ions of that element (Van Vlack, 1989). Substituting these values into the above equations, we can calculate the contribution to the mass of the CaO unit cell from the Ca2+ and O2- ions.

Therefore, the total mass of the unit cell of CaO can be calculated by summing the mass component of the Ca2+ ions and the mass component of the O2- ions. This is determined in the following manner: with significant figures applied The mass/volume of a material is defined to be equal to its density. We can state this relationship in the following manner. When considering the mass within a unit cell of a material and the volume within a unit cell of a material, a more specific definition of density can be stated.

Since we have obtained the values for the mass of a unit cell of CaO and the volume of a unit cell of CaO, we can substitute these values into the above equation to solve for the density of CaO.

The plane under investigation in this question has a Miller index of (1 1 1). The Miller indices are defined to be “the reciprocals of the three intercepts that the plane makes with each of the axes, cleared of fractions and common multipliers” (Van Vlack, 1989). Therefore, a given Miller index represents a set of parallel planes. The (1 1 1) plane intercepts the x-axis at (a,0,0), intercepts the y-axis at (0,a,0), and intercepts the z-axis at (0,0,a), where a is any number. This is depicted in the following diagram of the (1,1,1) plane:

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We know that this (1 1 1) plane must pass through the centre of the unit cell. The centre of the unit cell is given as the point p = (0.5,0.5,0.5). Therefore, the (1 1 1) plane we are concerned with must contain the point p = (0.5,0.5,0.5). By definition, a normal vector to a plane is any vector which begins at a point in the plane and has a direction that is perpendicular (or orthogonal) to the surface of the plane (The Vector Equation of a Plane, 2003). Any (1 1 1) plane must be parallel to any other (1 1 1) plane, so they must all have the same normal vector. In other words, any vector beginning at a point on the (1 1 1) plane and having a direction that is perpendicular to the surface of the plane will, regardless of how the plane is shifted, still begin at a point on the (1 1 1) plane and have a direction that is perpendicular to the surface of the plane.

To construct a normal vector to a plane, we can consider the position vectors of three points on the plane (The Vector Equation of a Plane, 2003). Let us call these points b, c, and d. We know that the (1 1 1) plane contains the points: Using the definition of vectors, we can create vectors that connect these points (The Vector Equation of a Plane, 2003). From linear algebra, any of these vectors must lie on our (1 1 1) plane (The Vector Equation of a Plane, 2003). For example, consider the following two vectors, E and F:

Since both of these vectors must lie in our (1 1 1) plane, if we take the cross-product of the two vectors, this will yield the normal vector to the (1 1 1) plane (The Vector Equation of a Plane, 2003). This is performed in the following calculation (Cross product, 2003): where n is the normal vector to the (1 1 1) plane. Therefore, any vector of the form (a2,a2,a2) will act as a normal vector to the (1 1 1) plane. Vector algebra also demonstrates that, if we know the normal vector, n, of a plane and a point, p, on the plane, any other point, r, that lies on the plane must satisfy the following equation (The Vector Equation of a Plane, 2003):

Therefore, we can substitute the values p = (0.5,0.5,0.5), n = (a2,a2,a2), and r = (x1,y1,z1), where each of x1, y1, and z1 can be any number, into the above equation. This produces the following relationship (Dot Product, 2003): Although any point on the plane must satisfy this relationship, we are concerned with the axial intercepts of the plane. This corresponds to the points (a,0,0), (0,a,0), and (0,0,a). Let us consider the first of these points, (a,0,0). For r = (a,0,0), the above equation becomes: