Lithium has a bcc structure .Its density is `530 kg m^(-3)` and its atomic mass is `6.94 g mol^(-1)` .Calculate the edge length of a unit cell of lithium metal `(N_(A) = 6.02 xx 10^(23) mol^(-1))`

Options

A. 352 pm
B. 527 pm
C. 264 pm
D. 154 pm

Answer

A

Explanation

Given, Li has a bcc structure
Density `(rho)=530"kg-m"^(-3)`
Atomic mass (M) = `6.94"g mol"^(-1)`
Avogadro's number `(N_(A))=6.02xx10^(23)"mol"^(-1)`
We know that number of atoms per unit cell in bcc
(Z) = 2.
`therefore` We have the formula for density,
`rho=(ZM)/(N_(A)a^(3))`
where a = edge-length of a unit cell.
`"or a"=root(3)((2xx6.94" g mol"^(-1))/(0.53"g cm"^(-3)xx06.02xx10^(23)"mol"^(-1)))`
`=root(3)(4.35xx10^(-23)"cm"^(-3))`
`=3.52xx10^(-8)cm`
`a=352"pm"`