Q 13 Niobium crystallises in body–centred
cubic structure. If density is 8.55 g cm–3, alculate atomic radius
of niobium using its atomic mass 93 u.
Edge of length
of cell a = ?
Density p = 8.55
g /cm3
Number of atoms
in unit cell of BCC lattice Z = 2
Avogadro number
NA = 6.022x1023
Use formula
Density
Cross multiply
we get
pa3NA=
ZM
Divide by PNA
we get
Plug the values
we get
Plug the value
of a we get
Very good quality
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ReplyDeleteCan you plz explain how the cubic root of (36.1×10^-24) is calculated by logarithmic approach?
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