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Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?
- a)0.115 nm
- b)0.144 nm
- c)0.235 nm
- d)0.156 nm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Gold (atomic mass 197 u) crystallises in a face-centred unit cell. Wha...
Given,
Edge length of the Gold unit cell (a) = 0.407 x 10-7m
For FCC unit cell, the atomic radius (r) = a/(2√2)
= 0.407 x 10-9/(2√2)
= 0.144 nm.
Edge length of the Gold unit cell (a) = 0.407 x 10-7m
For FCC unit cell, the atomic radius (r) = a/(2√2)
= 0.407 x 10-9/(2√2)
= 0.144 nm.
Most Upvoted Answer
Gold (atomic mass 197 u) crystallises in a face-centred unit cell. Wha...
Given:
- Atomic mass of gold = 197 u
- Edge length of the gold unit cell = 0.407 x 10^(-9) m
To find:
The atomic radius of gold.
Solution:
The face-centered unit cell of gold consists of atoms located at the corners and the face centers of the unit cell.
Step 1: Calculate the number of atoms in the unit cell:
In a face-centered unit cell, there are:
- 8 corner atoms, where each atom contributes 1/8th to the unit cell
- 6 face-centered atoms, where each atom contributes 1/2 to the unit cell
Total number of atoms in the unit cell = (8 x 1/8) + (6 x 1/2) = 4
Step 2: Calculate the volume of the unit cell:
The volume of a unit cell can be calculated using the formula:
Volume = (edge length)^3
Given:
Edge length of the unit cell = 0.407 x 10^(-9) m
Volume of the unit cell = (0.407 x 10^(-9))^3 = 6.835 x 10^(-29) m^3
Step 3: Calculate the volume occupied by each atom:
Since there are 4 atoms in the unit cell, the volume occupied by each atom can be calculated by dividing the total volume of the unit cell by the number of atoms:
Volume occupied by each atom = Volume of unit cell / Number of atoms = 6.835 x 10^(-29) / 4 = 1.70875 x 10^(-29) m^3
Step 4: Calculate the radius of each atom:
The volume occupied by each atom can also be calculated using the formula for the volume of a sphere:
Volume of a sphere = (4/3) x π x (radius)^3
Given:
Mass of gold atom = 197 u
Avogadro's number = 6.022 x 10^23
The mass of one gold atom can be converted to grams:
Mass of one gold atom = (197 u) x (1.66 x 10^(-27) kg/u) = 3.26 x 10^(-25) kg
The volume occupied by one gold atom can be calculated using its mass and the density of gold:
Density of gold = Mass / Volume
Volume = Mass / Density = (3.26 x 10^(-25) kg) / (19.3 x 10^3 kg/m^3) = 1.69 x 10^(-29) m^3
Equating the two expressions for the volume occupied by each atom:
1.70875 x 10^(-29) m^3 = (4/3) x π x (radius)^3
Simplifying the equation:
(radius)^3 = (1.70875 x 10^(-29) m^3) x (3/4π)
(radius)^3 = 3.226 x 10^(-29) m^3
Taking the cube root of both sides:
radius = (3.226 x 10^(-29))^(1/3) m
radius ≈
- Atomic mass of gold = 197 u
- Edge length of the gold unit cell = 0.407 x 10^(-9) m
To find:
The atomic radius of gold.
Solution:
The face-centered unit cell of gold consists of atoms located at the corners and the face centers of the unit cell.
Step 1: Calculate the number of atoms in the unit cell:
In a face-centered unit cell, there are:
- 8 corner atoms, where each atom contributes 1/8th to the unit cell
- 6 face-centered atoms, where each atom contributes 1/2 to the unit cell
Total number of atoms in the unit cell = (8 x 1/8) + (6 x 1/2) = 4
Step 2: Calculate the volume of the unit cell:
The volume of a unit cell can be calculated using the formula:
Volume = (edge length)^3
Given:
Edge length of the unit cell = 0.407 x 10^(-9) m
Volume of the unit cell = (0.407 x 10^(-9))^3 = 6.835 x 10^(-29) m^3
Step 3: Calculate the volume occupied by each atom:
Since there are 4 atoms in the unit cell, the volume occupied by each atom can be calculated by dividing the total volume of the unit cell by the number of atoms:
Volume occupied by each atom = Volume of unit cell / Number of atoms = 6.835 x 10^(-29) / 4 = 1.70875 x 10^(-29) m^3
Step 4: Calculate the radius of each atom:
The volume occupied by each atom can also be calculated using the formula for the volume of a sphere:
Volume of a sphere = (4/3) x π x (radius)^3
Given:
Mass of gold atom = 197 u
Avogadro's number = 6.022 x 10^23
The mass of one gold atom can be converted to grams:
Mass of one gold atom = (197 u) x (1.66 x 10^(-27) kg/u) = 3.26 x 10^(-25) kg
The volume occupied by one gold atom can be calculated using its mass and the density of gold:
Density of gold = Mass / Volume
Volume = Mass / Density = (3.26 x 10^(-25) kg) / (19.3 x 10^3 kg/m^3) = 1.69 x 10^(-29) m^3
Equating the two expressions for the volume occupied by each atom:
1.70875 x 10^(-29) m^3 = (4/3) x π x (radius)^3
Simplifying the equation:
(radius)^3 = (1.70875 x 10^(-29) m^3) x (3/4π)
(radius)^3 = 3.226 x 10^(-29) m^3
Taking the cube root of both sides:
radius = (3.226 x 10^(-29))^(1/3) m
radius ≈
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Community Answer
Gold (atomic mass 197 u) crystallises in a face-centred unit cell. Wha...
Given,
Edge length of the Gold unit cell (a) = 0.407 x 10-7m
For FCC unit cell, the atomic radius (r) = a/(2√2)
= 0.407 x 10-9/(2√2)
= 0.144 nm.
Edge length of the Gold unit cell (a) = 0.407 x 10-7m
For FCC unit cell, the atomic radius (r) = a/(2√2)
= 0.407 x 10-9/(2√2)
= 0.144 nm.
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Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer?.
Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer?.
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Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?a)0.115 nmb)0.144 nmc)0.235 nmd)0.156 nmCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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