Respuesta :
Answer:
The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.
Explanation:
Given that,
Weight of metal A = 12.5%
Weight of metal B = 87.5%
Length of unit cell = 0.395 nm
Density of A = 4.27 g/cm³
Density of B= 6.35 g/cm³
Weight of A = 61.4 g/mol
Weight of B = 125.7 g/mol
We need to calculate the density of the alloy
Using formula of density
[tex]\rho=n\times\dfrac{m}{V_{c}\times N_{A}}[/tex]
[tex]n=\dfrac{\rho\timesV_{c}\times N}{m}[/tex]....(I)
Where, n = number of atoms per unit cells
m = Mass of the alloy
V=Volume of the unit cell
N = Avogadro number
We calculate the density of alloy
[tex]\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100[/tex]
[tex]\rho=5.98[/tex]
We calculate the mass of the alloy
[tex]m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100[/tex]
[tex]m=111.15[/tex]
Put the value into the equation (I)
[tex]n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}[/tex]
[tex]n=1.99\approx 2\ atoms/cell[/tex]
Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.
The number of atoms in the unit cell of hypothetical alloy composed of metal A and B is 2 so the crystal structure is body-centered cubic.
What is crystal structure?
The crystal structure is the order of the atom arrangement in a repeating way. types of crystal structure are-
- Simple cubic-In this arrangement, one atom is placed at each corner of the unit cell.
- Face-centered cubic-In this arrangement, one atom is placed at each corner of the unit cell and one atom is placed at each face of the unit cell.
- Body-centered cubic-In this arrangement, one atom is placed at each corner of the unit cell and one atom is placed at center of the unit cell.
Number of atom can be find out with the following formula,
[tex]n=\dfrac{N\rho}{m}[\tex]
Here, (m) is the mass of the object and (N) is Avogadro's number.
Given information-
Hypothetical alloy is composed with metal A of 12.5 wt%.
Hypothetical alloy is composed with metal B of 87.5 wt%.
Densities of metals A and B are 4.27 and 6.35 g/cm 3,
Atomic weights of A and B are 61.4 and 125.7 g/mol,
The density of the alloy is,
[tex]\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100=5.98[\tex]
Similarly the mass of the alloy is,
[tex]m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100=111.15[\tex]
As the mass and the density of the alloy has been obtained and the Avogadro's number is 6.023*10²³.
Put the value in the above equation to find the number of atom, we get,
[tex]n=2[/tex]
Thus the number of atoms in the unit cell is 2. The crystal structure is body-centered cubic.
Learn more about the crystal structure here;
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