Вопрос
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.
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мастер · Репетитор 5 лет
Ответ
43. To calculate the wavelength of the X-ray that should be used, we can use the Bragg equation:nλ = d sin(θ)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.Given:n = 1 (first-order diffraction)d = 1.36 Å = 1.36 × 10^-10 mθ = 15.0°Substituting the values into the Bragg equation:1λ = (1.36 × 10^-10 m) sin(15.0°)Solving for λ:λ = (1.36 × 10^-10 m) sin(15.0°) ≈ 3.64 × 10^-11 mTherefore, the wavelength of the X-ray that should be used is approximately 3.64 × 10^-11 m.44. To find the spacing between crystal planes, we can use the Bragg equation:nλ = d sin(θ)Given:n = 1 (first-order diffraction)λ = 2.63 Å = 2.63 × 10^-10 mθ = 15.55°Substituting the values into the Bragg equation:1(2.63 × 10^-10 m) = d sin(15.55°)Solving for d:d = (2.63 × 10^-10 m) / sin(15.55°) ≈ 7.23 × 10^-10 mTherefore, the spacing between crystal planes is approximately 7.23 × 10^-10 m.To find the angle for second-order diffraction (n = 2), we can use the Bragg equation again:2λ = d sin(θ)Substituting the values:2(2.63 × 10^-10 m) = (7.23 × 10^-10 m) sin(θ)Solving for θ:sin(θ) = (2 × 2.63 × 10^-10 m) / (7.23 × 10^-10 m) ≈ 0.727θ ≈ sin^-1(0.727) ≈ 46.57°Therefore, the angle for second-order diffraction is approximately 46.57°.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.Calcium has a cubic closest packed structure, which means that each unit cell contains 8 atoms.The atomic radius of calcium is given as 197 pm (picometers), which is equal to 197 × 10^-12 m.The edge length of the unit cell can be calculated using the formula:a = 2r√2where a is the edge length of the unit cell and r is the atomic radius.Substituting the value of r:a = 2(197 × 10^-12 m)√2 ≈ 5.59 × 10^-10 mThe volume of the unit cell is:V = a^3 = (5.59 × 10^-10 m)^3 ≈ 1.66 × 10^-28 m^3The 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):m = (40.08 g/mol) / (6.022 × 10^23 atoms/mol) ≈ 6.65 × 10^-23 gThe total mass of the unit cell is:m_total = 8 × m = 8 × (6.65 × 10^-23 g) ≈ 5.32 × 10^-22 gFinally, the density of solid calcium can be calculated as:ρ = m_total / V = (5.32 × 10^-22 g) / (1.66 × 10^-28 m^3) ≈ 3.21 g/cm^3Therefore, the density of solid calcium is approximately 3.21 g/cm^3.46. 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.The number of atoms per unit cell in a face-centered cubic unit cell is 4.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):m = (58.69 g/mol) / (6.022 × 10^23 atoms/mol) ≈ 9.74 × 10^-23 gThe volume of the unit cell can be calculated using the density and the number of atoms per unit cell:V = m_total / ρ = (4