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The unit cell of an element of atomic mass 96 and density 10.3 g/cm–3 is a cube with edge length of 314 pm the structure of the crystal lattice is
- a)bcc
- b)fcc
- c)scc
- d)diamond
Correct answer is option 'A'. Can you explain this answer?
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The unit cell of an element of atomic mass 96 and density 10.3 g/cm&nd...
The structure of solid should be bcc.
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The unit cell of an element of atomic mass 96 and density 10.3 g/cm&nd...
The first step to determine the unit cell of an element is to calculate its atomic radius. We can use the following formula:
density = mass/volume
where mass = atomic mass/N (Avogadro's number) and volume = (4/3)πr^3
10.3 g/cm^3 = (96 g/mol)/(6.022 x 10^23/mol) / [(4/3)πr^3]
r = 1.84 x 10^-8 cm
Next, we need to determine the crystal structure of the element based on its atomic radius. A common method is to calculate the atomic packing factor (APF), which is defined as the volume of atoms in a unit cell divided by the total volume of the unit cell.
For a simple cubic (SC) crystal structure, the APF is:
APF = (4/3)πr^3 / a^3
where a is the edge length of the unit cell. Solving for a, we get:
a = 2r
Therefore, the unit cell of the element with atomic mass 96 and density 10.3 g/cm^3 is a simple cubic with an edge length of 3.68 x 10^-8 cm.
density = mass/volume
where mass = atomic mass/N (Avogadro's number) and volume = (4/3)πr^3
10.3 g/cm^3 = (96 g/mol)/(6.022 x 10^23/mol) / [(4/3)πr^3]
r = 1.84 x 10^-8 cm
Next, we need to determine the crystal structure of the element based on its atomic radius. A common method is to calculate the atomic packing factor (APF), which is defined as the volume of atoms in a unit cell divided by the total volume of the unit cell.
For a simple cubic (SC) crystal structure, the APF is:
APF = (4/3)πr^3 / a^3
where a is the edge length of the unit cell. Solving for a, we get:
a = 2r
Therefore, the unit cell of the element with atomic mass 96 and density 10.3 g/cm^3 is a simple cubic with an edge length of 3.68 x 10^-8 cm.
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The unit cell of an element of atomic mass 96 and density 10.3 g/cm–3is a cube with edge length of 314 pm the structure of the crystal lattice isa)bccb)fccc)sccd)diamondCorrect answer is option 'A'. Can you explain this answer?
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