Niobium crystallizes in body-centred cubic structure. If the density is `8.55 g cm^(-3)`, calculate the atomic radius of niobium using its atomic mass `93 u`.
Text Solution
Verified by Experts
Atomic mass of `Nb=(Density xx6.023xx10^(23)xxVolume)/(n)` `(`For `b.c.c.,z=2,` Atomic mass of `Nb=92.90)` `:. 92.90=(8.55xx6.023xx10^(23)xxV)/(2)` `:. V=3.6xx10^(-23)cm^(3)` Now, `V=a^(3)` `:. a=3sqrt((3.6xx10^(-23)))` `=3.3xx10^(-8)cm` For `b.c.c.` structure, `r=(sqrt(3))/(4)a` `=1.43xx10^(-8)cm`
Topper's Solved these Questions
SOLID STATE
P BAHADUR|Exercise Exercise 9|1 Videos
MOCK TEST PAPER
P BAHADUR|Exercise Exercise|92 Videos
SURFACE CHEMISTRY
P BAHADUR|Exercise Exercise 5|1 Videos
Similar Questions
Explore conceptually related problems
Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium. (Atomic mass = 93u).
Niobium crystallises in body centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of the element. (Atomic mass = 93 u)
A metal (atomic mass = 50 ) has a body centred cubic crystal structure. If the density of the metal is 5.96 g cm^(-3) , calculate the volume of the unit cell.
The metal nickel crystallizes in a face centered cubic structure. Its density is 8.90 gm//cm^(3) . Calculate (a) the length of the edge of a unit cell. (b) the radius of nickel atom. (atomic weight of Ni = 58.89).
