Niobium crystallizes bcc structure. If the density of niobium is 8.55 g/cm and length of unit cell edge is 330.6 pm,
then find out:
i. Number of atoms in 0.2g of niobium and ,
ii. Number of unit cells present in o.2g of niobium ?
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Answer:
ANSWER
Density (D), number of atoms per unit cell (Z), atomic mass (M), Avogadro's number (N
0
) and edge length (a) are related as below:
D=
N
0
a
3
ZM
Given D=8.55 g/cm
3
and M=93 u.
Edge length =a=(
DN
0
ZM
)
1/3
=(
8.55×6.023×10
23
2×93
)
1/3
=3.306×10
−8
cm
Atomic radius =r=
4
3
a=
4
3
×3.306×10
−8
=1.432×10
−8
cm=14.32 nm
Explanation:
hpoe u got the answer
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