Question

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)`?

Answer 1

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)`