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 PNAwe get
Plug the values we get
Plug the value of a we get
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