Chromium metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be `287`pm. Calculate the atomic radius. What woulds be the density of chromium in `g cm^(-3)`?
Text Solution
Verified by Experts
We know, For `b.c.c.` lattice `r=(sqrt(3)a)/(4)=(sqrt(3))/(4)xx287=124.27p m` Now Density `=(zxxat.wt)/(Vxx Av. no .)=(zxxat.wt.)/(a^(3)xxAv. no .)` `=z=2` for `b.c.c.,a=287xx10^(-10)cm` `:. rho=(2xx51.99)/((2.87xx10^(-10))^(3)xx6.023xx10^(23))` `=7.30g//cm^(3)`
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
A metal crystallises with a body-centred cubic lattice. The edge length of the unit cell is 360 pm. Radius of the metal atom is
a metal crystallizes with a face-centered cubic lattice.The edge of the unit cell is 408 pm. The diameter of the metal atom is :
A metal crystallizes with a face -centred cubic lattice.The edge of the unit cells is 408 pm.The dimeter of the metal atoms is
A metal has a body-centred cubic lattice and the length of the unit cell is \(3A^{0}\) .If the density is 10gm/cc .Calculate its atomic weight.
A metal crystallizes in a body-centred cubic lattice with the unit cell length 320 pm. The radius of the metal atom (in pm) will be
