CHEM 471 - Solid State Chemistry Problems in Inorganic
Chemistry
Problems and Solutions
1. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
2. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
3. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
4. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
5. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
6. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
7. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
8. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
9. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
10. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 Ev/K)(300 K)ln(9)=0.0663 Ev
11. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
12. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
13. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
14. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
15. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
16. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
17. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
18. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
19. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
20. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
21. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
22. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
23. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
24. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
25. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
26. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
27. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
28. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
29. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
30. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
31. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
32. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
33. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
34. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
35. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
36. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
37. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
38. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
39. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
40. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
41. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
42. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
43. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
44. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
45. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
46. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
47. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
48. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
49. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
50. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
51. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
52. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
53. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
54. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
55. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
56. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
57. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
58. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
59. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
60. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
61. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
62. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
63. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
64. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
65. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
66. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
67. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
68. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
69. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
70. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
71. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
72. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
73. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
74. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
75. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
76. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
77. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
78. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
79. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
80. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
81. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
82. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
83. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
84. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
85. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
86. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
87. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
88. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
89. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
90. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
91. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
92. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
93. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
94. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
95. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
96. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
97. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐 =3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
98. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎 =𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
99. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
100. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
101. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
102. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
103. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
104. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
105. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
106. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
107. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
108. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
109. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
110. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
111. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
112. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
113. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
114. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
115. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
116. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
117. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
118. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
119. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
120. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
121. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
122. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
123. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
124. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
125. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
126. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
127. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
128. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
129. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
130. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
131. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
132. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
133. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
134. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
135. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
136. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
137. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
138. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
139. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
140. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
141. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
142. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
143. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
144. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
145. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
146. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
147. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
148. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
149. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
150. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
151. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
152. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
153. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
154. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
155. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
156. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
157. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
158. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
159. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
160. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
161. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
162. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
163. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
164. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
165. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
166. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
167. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
168. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
169. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
170. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
171. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
172. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
173. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
174. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
175. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
176. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
177. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
178. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
179. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
180. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
181. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
182. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
183. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
184. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
185. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
186. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
187. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
188. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
189. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
190. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
191. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
192. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
193. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
194. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
195. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
196. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
197. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
198. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
199. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
200. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
201. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
202. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
203. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
204. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
205. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
206. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
207. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
208. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
209. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
210. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
211. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
212. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
213. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
214. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
215. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
216. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
217. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
218. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
219. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
220. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
221. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
222. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
223. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
224. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
225. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
226. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
227. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
228. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
229. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
230. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
231. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
232. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
233. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
234. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
235. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
236. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
237. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
238. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
239. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
240. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
241. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
242. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
243. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
244. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
245. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
246. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
247. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
248. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
249. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
250. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
251. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
252. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
253. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
254. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
255. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
256. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
257. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
258. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
259. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
260. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
261. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
262. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
263. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
264. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
265. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
266. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
267. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
268. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
269. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
270. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
271. Density Calculation: Copper crystallizes in a face-centered cubic (FCC) structure
with an edge length of 3.61 Å. Calculate the density of copper. (Atomic mass of Cu =
63.55 g/mol)
Solution:
– Number of atoms per unit cell in FCC = 4
– Volume of unit cell = (3.61×10−10 m)3=4.70×10−29 m3
– Mass of atoms in unit cell = 4⋅ 63.55
6.022×1023 =4.22×10−22 g
– Density = 4.22×10−22 g
4.70×10−29 m3=8.98 g/cm3
272. Miller Indices: A plane in a cubic crystal intersects the a, b, and c axes at 2, 3, and 6
unit cell lengths respectively. Determine the Miller indices of this plane.
Solution:
– Take reciprocals: 1
2,1
3,1
6
– Find least common denominator: 6
– Multiply all terms by 6: 3, 2, 1
– Miller indices: (321)
273. Bragg’s Law: X-rays with a wavelength of 1.54 Å are diffracted from a crystal plane
with a d-spacing of 2.1 Å. Calculate the angle of diffraction for the first-order
reflection.
Solution:
– Bragg’s Law: 𝑛𝜆=2𝑑sin𝜃
– For first-order reflection, 𝑛=1
– 1⋅1.54=2⋅2.1⋅sin𝜃
– sin𝜃=1.54
4.2 =0.3667
– 𝜃=arcsin(0.3667)=21.5°
274. Coordination Number: In a body-centered cubic (BCC) structure, calculate the
coordination number and the atomic packing factor (APF).
Solution:
– Coordination number for BCC = 8
– Volume of unit cell = 𝑎3, where 𝑎 is edge length
– Volume of atoms = 2⋅4
3𝜋(√3𝑎
4)3
– APF = Volume of atoms
Volume of unit cell =2⋅4
3𝜋(√3𝑎
4)3
𝑎3=𝜋√3
8≈0.68
275. Defect Concentration: The energy required to form a Schottky defect in NaCl is 2.3
eV. Calculate the fraction of lattice sites that are vacant at 800°C.
Solution:
– Fraction of defects = exp(− 𝐸
2𝑘𝑇)
– 𝐸=2.3 eV =3.68×10−19 J
– 𝑇=800+273=1073 K
– 𝑘=1.38×10−23 J/K
– Fraction = exp(− 3.68×10−19
2⋅1.38×10−23⋅1073)=1.1×10−5
276. Ionic Radii Ratio: The ionic radius of Na+ is 1.02 Å and that of Cl- is 1.81 Å. Predict
the coordination number of Na+ in NaCl based on the radius ratio rule.
Solution:
– Radius ratio = 𝑟cation
𝑟anion =1.02
1.81 =0.56
– 0.414 < 0.56 < 0.732
– This range corresponds to a coordination number of 6 (octahedral)
277. Unit Cell Volume: A tetragonal unit cell has dimensions 𝑎=𝑏=4.5 Å and 𝑐=3.0 Å.
Calculate the volume of the unit cell in nm^3.
Solution:
– Volume = 𝑎2𝑐
– 𝑉=(4.5×10−10 m)2⋅(3.0×10−10 m)
– 𝑉=60.75×10−30 m3=0.06075 nm3
278. Electrical Conductivity: The electrical conductivity of a material is given by 𝜎=𝑛𝑒𝜇,
where 𝑛 is the number of charge carriers per unit volume, 𝑒 is the elementary
charge, and 𝜇 is the mobility. If 𝑛=8.5×1028 m^-3 and 𝜇 =0.0044 m^2/Vs,
calculate the conductivity.
Solution:
– 𝑒=1.602×10−19 C
– 𝜎=(8.5×1028 m−3)(1.602×10−19 C)(0.0044 m2/Vs)
– 𝜎=5.98×107 S/m
279. Band Gap Energy: The conductivity of a semiconductor varies with temperature
according to the equation 𝜎=𝜎0exp(−𝐸𝑔/2𝑘𝑇), where 𝐸𝑔 is the band gap energy. If
the conductivity doubles when the temperature is raised from 300 K to 350 K,
calculate 𝐸𝑔.
Solution:
– 𝜎2
𝜎1=exp(−𝐸𝑔
2𝑘(1
𝑇2−1
𝑇1))
– ln(2)=−𝐸𝑔
2𝑘(1
350−1
300)
– 𝐸𝑔=−2𝑘ln(2)
1
350−1
300 =0.54 eV
280. Fermi-Dirac Distribution: At what energy above the Fermi level is the probability of
finding an electron equal to 0.1 at room temperature (300 K)?
Solution:
– Fermi-Dirac distribution: 𝑓(𝐸)=1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Given 𝑓(𝐸)=0.1 and 𝑇=300 K
– 0.1= 1
1+exp((𝐸−𝐸𝐹)/𝑘𝑇)
– Solving: 𝐸−𝐸𝐹=𝑘𝑇ln(9)=(8.617×10−5 eV/K)(300 K)ln(9)=0.0663 eV
