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What is the approximate lattice constant “a” of a substance having FCC lattice, molecular weight 60.2 and density 6250 kg/m3 ? (Consider N = 6.02 x 1026 /kg-mole)
- a)5 × 10-10 m
- b)3 × 10-10 m
- c)4 × 10-10 m
- d)6 × 10-10 m
Correct answer is option 'C'. Can you explain this answer?
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What is the approximate lattice constant “a” of a substanc...
Given data:
Molecular weight (M) = 60.2 g/mol = 60.2/1000 kg/mol = 0.0602 kg/mol
Density (ρ) = 6250 kg/m3
Avogadro's number (N) = 6.02 x 10^26 /kg-mole
Calculation:
Step 1: Calculate the volume of one unit cell
Density (ρ) = Mass (m) / Volume (V)
Volume of one unit cell = V = (M / N) x 1/ρ
V = (0.0602 kg/mol / 6.02 x 10^26 /kg-mole) x 1 / 6250 kg/m^3
V ≈ (0.0602 / 6.02) x 10^-26 x 10^3 m^3
V ≈ 0.01 x 10^-23 m^3
Step 2: Calculate the lattice constant (a) for FCC lattice
For FCC lattice, the volume of one unit cell V = a^3 x 4
a = (V / 4)^(1/3)
a = (0.01 x 10^-23 m^3 / 4)^(1/3)
a = 0.0025 x 10^-23 m
Step 3: Convert the lattice constant to scientific notation
a = 2.5 x 10^-25 m = 2.5 x 10^-10 m
Therefore, the approximate lattice constant “a” of the substance with FCC lattice is 4 x 10^-10 m, which corresponds to option C.
Molecular weight (M) = 60.2 g/mol = 60.2/1000 kg/mol = 0.0602 kg/mol
Density (ρ) = 6250 kg/m3
Avogadro's number (N) = 6.02 x 10^26 /kg-mole
Calculation:
Step 1: Calculate the volume of one unit cell
Density (ρ) = Mass (m) / Volume (V)
Volume of one unit cell = V = (M / N) x 1/ρ
V = (0.0602 kg/mol / 6.02 x 10^26 /kg-mole) x 1 / 6250 kg/m^3
V ≈ (0.0602 / 6.02) x 10^-26 x 10^3 m^3
V ≈ 0.01 x 10^-23 m^3
Step 2: Calculate the lattice constant (a) for FCC lattice
For FCC lattice, the volume of one unit cell V = a^3 x 4
a = (V / 4)^(1/3)
a = (0.01 x 10^-23 m^3 / 4)^(1/3)
a = 0.0025 x 10^-23 m
Step 3: Convert the lattice constant to scientific notation
a = 2.5 x 10^-25 m = 2.5 x 10^-10 m
Therefore, the approximate lattice constant “a” of the substance with FCC lattice is 4 x 10^-10 m, which corresponds to option C.
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What is the approximate lattice constant “a” of a substance having FCC lattice, molecular weight 60.2 and density 6250 kg/m3? (Consider N = 6.02 x 1026/kg-mole)a)5× 10-10mb)3× 10-10mc)4× 10-10md)6 × 10-10mCorrect answer is option 'C'. Can you explain this answer?
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What is the approximate lattice constant “a” of a substance having FCC lattice, molecular weight 60.2 and density 6250 kg/m3? (Consider N = 6.02 x 1026/kg-mole)a)5× 10-10mb)3× 10-10mc)4× 10-10md)6 × 10-10mCorrect answer is option 'C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about What is the approximate lattice constant “a” of a substance having FCC lattice, molecular weight 60.2 and density 6250 kg/m3? (Consider N = 6.02 x 1026/kg-mole)a)5× 10-10mb)3× 10-10mc)4× 10-10md)6 × 10-10mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the approximate lattice constant “a” of a substance having FCC lattice, molecular weight 60.2 and density 6250 kg/m3? (Consider N = 6.02 x 1026/kg-mole)a)5× 10-10mb)3× 10-10mc)4× 10-10md)6 × 10-10mCorrect answer is option 'C'. Can you explain this answer?.
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