Q 24: Aluminium crystallises in a
cubic close–packed structure. Its metallic radius is 125 pm.
(i) What is
the length of the side of the unit cell?
(ii) How
many unit cells are there in 1.00 cm3 of aluminum?
Solution:
(i)
For the cubic close–packed
structure
Let a is the edge of the
cube and r is the radius of atom
Given that r = 125 pm
a = 2√2 r
plug the value of r we get
= 2 x 1.414 x125 pm
= 354 pm (approximately)
(ii) Volume of one unit cell = side 3
= (354 pm)3
1 pm = 10–10 cm
= (354 x 10–10
cm)3
= (3.54 x 10–8
cm)3
= 44.36 x 10–24
cm3
= 4.4 × 10−23 cm3
Total number of unit cells in 1.00 cm3
= total volume / size of each cell
= (1.00cm3)/( 4.4 × 10−23 cm3)
= 2.27 × 1022 unit cell
Awsm notes
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ReplyDelete1pm=10^-12 not 10^-10 Please correct it as 10^-10 is 1 angstrom
ReplyDeleteHe is coverting it into cm not M
DeletePlz help reply me fast tomorrow is my preboard exam how u get a=2√2r I don't know about this formula ???replay as soon as possible thank
ReplyDeletea=√8r=2√2r
ReplyDeleteHello
ReplyDeleteValue of pico metre is 10^-12 you are saying its 10^-10 which makes the solution wrong
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ReplyDeleteGood boy
ReplyDeleteTake a cube of side"a" having FCC .then the distance btw two atoms "d" is equal to the half of hypotenuse of any face. We can find hypotenuse with the help of Pythagoras theorem then d=√2a/2...as we know radius is equal to half of the distance btw two atoms..
ReplyDeleter=d/2
r=√2a/4
r=a/2√2
a=2√2r