Answer :
Answer:
[tex]5.796\times 10^{-29}m^3[/tex]
Step-by-step explanation:
Atomic radius of metal=0.137nm=[tex]0.137\times 10^{-9}[/tex]m
[tex]1nm=10^{-9}m[/tex]
Structure is FCC
We know that
The relation between edge length and radius in FCC structure
[tex]a=2\sqrt 2r[/tex]
Where a=Edge length=Side
r=Radius
Using the relation
[tex]a=2\sqrt 2\times 0.137\times 10^{-9}=0.387\times 10^{-9}m[/tex]
We know that
Volume of cube=[tex](side)^3[/tex]
Using the formula
Volume of unit cell=[tex](0.387\times 10^{-9})^3=5.796\times 10^{-29} m^3[/tex]
If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.137 nm, The volume of its unit cell is
[tex]\rm =5.796\times 10^{-29}m^3[/tex]
Given : Cubic crystal structure = 0.137 nm
To find : The volume of its unit cell.
According to the question,
→Atomic Radius of metal=0.137nmm
i.e. [tex]\rm =0.137 \times 10^{-9} m\\\\\rm 1 \;nm= 10^{-9} m[/tex]
We know that structure is in face-centered cubic crystal.
Now,
The relation between the Edge length and Radius in face-centered cubic crystal structure will be
→ [tex]\rm a=2\sqrt{2r}[/tex]
Where, a = Edge length
r = Radius
Now, we will use the relation here,
[tex]\rm a=2\sqrt{2} \times 0.137 \times 10^{-9}m \\\\\rm a=0.387 \times 10^{-9} m[/tex]
Hence, Volume of cube = [tex]\rm (Edge\; length^{3} )[/tex]
Now we will apply the formula and on further solving we get,
[tex]\rm Volume \;of\; unit\; cell = (0.387 \times 10^{-9})^3\\\\\rm Volume \;of\; unit\; cell = \rm =5.796\times 10^{-29}m^3[/tex]
Therefore, The volume of its unit cell [tex]\rm =5.796\times 10^{-29}m^3[/tex]
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