Respuesta :
Answer:
Density of unit cell ( rhodium) = 12.279 g/cm³
Explanation:
Given that:
The radius (r) of a rhodium atom = 135 pm
The atomic mass of rhodium = 102.90 amu
For a face-centered cubic unit cell,
[tex]r = \dfrac{a}{2\sqrt{2}}[/tex]
where;
a = edge length.
Making "a" the subject of the formula:
[tex]a = 2 \sqrt{2} \times r[/tex]
[tex]a = 2 \times 1.414 \times 135 \ pm[/tex]
a = 381.8 pm
to cm, we get:
a = 381.8 × 10⁻¹⁰ cm
However, recall that:
[tex]density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}[/tex]
where;
mass of unit cell = mass of atom × numbers of atoms per unit cell
Also;
[tex]mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}[/tex]
[tex]mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}[/tex]
Recall also that number of atoms in a unit cell for a face-centered cubic = 4
So;
[tex]mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4[/tex]
mass of unit cell = 6.83380375 × 10⁻²² g
[tex]Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}[/tex]
Density of unit cell ( rhodium) = 12.279 g/cm³
The density of rhodium is equal to 12.4g/cm^3
Data;
- radius = 135pm
- atomic mass = 102.90 amu
The radius of a FCC is calculated as
[tex]r = \frac{a}{2\sqrt{2} }\\a = 2\sqrt{2} * r\\a = 2 * 1.414 * 135 = 381.8pm = 381.8 * 10^-^1^0cm\\[/tex]
The Density of Rhodium
The formula of density is given as
[tex]\rho = \frac{mass}{volume}\\[/tex]
The mass of a unit cell = mass of atom * number of atoms per unit cell.
[tex]mass of atom = \frac{atomic mass}{avogadro's number} = \frac{102.90}{6.023*10^2^3}\\[/tex]
For BCC number of atom in unit cell = 4
mass of unit cell = [tex]\frac{102.90}{6.02*10^2^3} * 4[/tex]
This makes the density of the atom equal to
[tex]\rho = \frac{\frac{102.90}{6.02*10^2^3} }{(381*10^-^1^0)^3}* 4\\ \rho = 12.4 g/cm^3[/tex]
The density of Rhodium is 12.4 g/cm^3
Learn more on radius of face centered cubic;
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