Niobium crystallises in body-centred cubic structure. If density is 8.55
g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.
Answers
Answered by
1
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
Answered by
1
Density (D), number of atoms per unit cell (Z), atomic mass (M), Avogadro's number (NO) and edge length (a) are related as below:
D= ZM ÷ NOa³
Given D=8.55 g/cm and M=93 u.
Edge length = a =( ZM÷DNO)¹/³ = 2×93 ÷. 8.55 × 6.023 ×. 10²³
= 3.306 × cm
Atomic radius = r = √3a÷4 = √3÷4 ×3.306 × 10-⁸= 1.432
×10-⁸cm = 14.32 nm
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