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A hypothetical metal has the BCC crystal structure, a density of 7.24 g/cm3, and an atomic weight of 48.9 gmol. a) Sketch a BCC unit. b) cell determine the atomic radius of this metal. c) calculate the side length of the BCC unit cell

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Answer :

(a) The sketch of BCC unit cell is shown below.

(b) The atomic radius of this metal is, [tex]1.221\times 10^{-8}cm[/tex]

(c) The side length of the BCC unit cell is, [tex]2.82\times 10^{-8}cm[/tex]

Solution : Given,

Number of atom in unit cell of BCC (Z) = 2

Atomic mass (M) = 48.9 g/mole

Density = [tex]7.24g/cm^3[/tex]

Avogadro's number [tex](N_{A})=6.022\times 10^{23} mol^{-1}[/tex]

First we have to calculate the edge length of unit cell.

Formula used :  

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]      .............(1)

where,

[tex]\rho[/tex] = density  = [tex]7.24g/cm^3[/tex]

Z = number of atom in unit cell  = 2

M = atomic mass  = 48.9 g/mole

[tex](N_{A})[/tex] = Avogadro's number  = [tex]6.022\times 10^{23} mol^{-1}[/tex]

a = edge length of unit cell  or the side length of the BCC unit cell =?

Now put all the values in above formula (1), we get:

[tex]7.24g/cm^3=\frac{2\times (48.9g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (a)^3}[/tex]

[tex]a=2.82\times 10^{-8}cm[/tex]

The edge length of unit cell is, [tex]2.82\times 10^{-8}cm[/tex]

Now we have to determine the atomic radius of this metal.

Formula used :

[tex]a=\frac{4r}{\sqrt{3}}[/tex]

where,

a = edge length of unit cell

r = nearest neighbor distance  or atomic radius of metal

Now put all the given values in this formula, we get :

[tex]2.82\times 10^{-8}cm=\frac{4\times r}{\sqrt{3}}[/tex]

[tex]r=1.221\times 10^{-8}cm[/tex]

The atomic radius of this metal is, [tex]1.221\times 10^{-8}cm[/tex]

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Lanuel

The atomic radius of this hypothetical metal with a BCC crystal structure is equal to [tex]1.22 \times 10^{-8}\;cm[/tex]

Given the following data:

Density of metal = 7.24 g/cm³

Atomic weight of metal = 48.9 gmol.

Scientific data:

Avogadro's number = [tex]6.02 \times 10^{23}[/tex]

Number of atom in a unit cell of BCC (Z) = 2.

How to determine the atomic radius of this metal.

First of all, we would have to calculate the edge length (a) of this unit cell by using this formula:

[tex]\rho = \frac{ZM}{a^3A_N}[/tex]

Where:

  • [tex]\rho[/tex] is the density.
  • M is the atomic mass.
  • Z is the number of atom in a unit cell.
  • [tex]A_N[/tex] is Avogadro number.
  • a is the edge length.

Making a the subject of formula, we have:

[tex]a=\sqrt[3]{\frac{ZM}{\rho A_N} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]a=\sqrt[3]{\frac{2 \times 48.9}{7.24 \times 6.02 \times 10^{23}} }\\\\a= 2.82 \times 10^{-8}\;cm[/tex]

For the atomic radius, we have:

[tex]a=\frac{4r}{\sqrt{3} } \\\\r=\frac{a\sqrt{3}}{4} \\\\r=\frac{2.82 \times 10^{-8} \times \sqrt{3}}{4}\\\\r=1.22 \times 10^{-8}\;cm[/tex]

Read more on atomic radius here: https://brainly.com/question/14831455

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