43. A topaz crystal has an interplanar spacing (d) of 1.36stackrel (circ )(A) (1stackrel (circ )(A)= 1times 10^-10m ). Calculate the wavelength of the X ray that should be used if Theta =15.0^circ (assume n=1 44. X rays of wavelength 2.63stackrel (circ )(A) were used to analyze a crystal. The angle of first -order diffraction (n=1 in the Bragg equa- tion) was 1555 degrees. What is the spacing between crystal planes, and what would be the angle for second-order diffrac- tion (n=2) 45. Calcium has a cubic closest packed structure as a solid . Assum- ing that calcium has an atomic radius of 197 pm, calculate the density of solid calcium. 46. Nickel has a face-centered cubic unit cell. The density of nickel is 6.84g/cm^3 . Calculate a value for the atomic radius of nickel.

43. To calculate the wavelength of the X-ray that should be used, we can use the Bragg equation:<br /><br />nλ = d sin(θ)<br /><br />where n is the order of diffraction, λ is the wavelength of the X-ray, d is the interplanar spacing, and θ is the angle of incidence.<br /><br />Given:<br />n = 1 (first-order diffraction)<br />d = 1.36 Å = 1.36 × 10^-10 m<br />θ = 15.0°<br /><br />Substituting the values into the Bragg equation:<br /><br />1λ = (1.36 × 10^-10 m) sin(15.0°)<br /><br />Solving for λ:<br /><br />λ = (1.36 × 10^-10 m) sin(15.0°) ≈ 3.64 × 10^-11 m<br /><br />Therefore, the wavelength of the X-ray that should be used is approximately 3.64 × 10^-11 m.<br /><br />44. To find the spacing between crystal planes, we can use the Bragg equation:<br /><br />nλ = d sin(θ)<br /><br />Given:<br />n = 1 (first-order diffraction)<br />λ = 2.63 Å = 2.63 × 10^-10 m<br />θ = 15.55°<br /><br />Substituting the values into the Bragg equation:<br /><br />1(2.63 × 10^-10 m) = d sin(15.55°)<br /><br />Solving for d:<br /><br />d = (2.63 × 10^-10 m) / sin(15.55°) ≈ 7.23 × 10^-10 m<br /><br />Therefore, the spacing between crystal planes is approximately 7.23 × 10^-10 m.<br /><br />To find the angle for second-order diffraction (n = 2), we can use the Bragg equation again:<br /><br />2λ = d sin(θ)<br /><br />Substituting the values:<br /><br />2(2.63 × 10^-10 m) = (7.23 × 10^-10 m) sin(θ)<br /><br />Solving for θ:<br /><br />sin(θ) = (2 × 2.63 × 10^-10 m) / (7.23 × 10^-10 m) ≈ 0.727<br /><br />θ ≈ sin^-1(0.727) ≈ 46.57°<br /><br />Therefore, the angle for second-order diffraction is approximately 46.57°.<br /><br />45. To calculate the density of solid calcium, we need to know the number of atoms per unit cell and the mass of each atom.<br /><br />Calcium has a cubic closest packed structure, which means that each unit cell contains 8 atoms.<br /><br />The atomic radius of calcium is given as 197 pm (picometers), which is equal to 197 × 10^-12 m.<br /><br />The edge length of the unit cell can be calculated using the formula:<br /><br />a = 2r√2<br /><br />where a is the edge length of the unit cell and r is the atomic radius.<br /><br />Substituting the value of r:<br /><br />a = 2(197 × 10^-12 m)√2 ≈ 5.59 × 10^-10 m<br /><br />The volume of the unit cell is:<br /><br />V = a^3 = (5.59 × 10^-10 m)^3 ≈ 1.66 × 10^-28 m^3<br /><br />The mass of each atom can be calculated using the atomic mass of calcium (40.08 g/mol) and Avogadro's number (6.022 × 10^23 atoms/mol):<br /><br />m = (40.08 g/mol) / (6.022 × 10^23 atoms/mol) ≈ 6.65 × 10^-23 g<br /><br />The total mass of the unit cell is:<br /><br />m_total = 8 × m = 8 × (6.65 × 10^-23 g) ≈ 5.32 × 10^-22 g<br /><br />Finally, the density of solid calcium can be calculated as:<br /><br />ρ = m_total / V = (5.32 × 10^-22 g) / (1.66 × 10^-28 m^3) ≈ 3.21 g/cm^3<br /><br />Therefore, the density of solid calcium is approximately 3.21 g/cm^3.<br /><br />46. To calculate the atomic radius of nickel, we can use the given density and the fact that nickel has a face-centered cubic unit cell.<br /><br />The number of atoms per unit cell in a face-centered cubic unit cell is 4.<br /><br />The mass of each atom can be calculated using the atomic mass of nickel (58.69 g/mol) and Avogadro's number (6.022 × 10^23 atoms/mol):<br /><br />m = (58.69 g/mol) / (6.022 × 10^23 atoms/mol) ≈ 9.74 × 10^-23 g<br /><br />The volume of the unit cell can be calculated using the density and the number of atoms per unit cell:<br /><br />V = m_total / ρ = (4