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Density of Li is 0.53"g cm"^(-3). The ed...

Density of Li is `0.53"g cm"^(-3)`. The edge length of Li is `3.5Å`. Find the number of Li atoms in a unit cell `(N_(0)=06.023xx10^(23),M=6.94)`.

Text Solution

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Edge length of unit cell `(a) = 3.5Å=3.5xx10^(-8)cm.`
Volume of unit cell `(a^(3))=(3.5xx10^(-8))^(3)cm^(3)`
Density of Li`(rho)=0.53"g cm"^(-3)`
Avogadro's no. `(N_(0))=6.023xx10^(23)mol^(-1)`
Atomic mass of Li (M) = `6.94"g mol"^(-1)`
We know that `rho=(ZxxM)/(a^(3)xxN_(0))" or "Z=(rhoxxa^(3)xxN_(0))/(M)`
`Z=((0.53gcm^(-3))xx(3.5xx10^(-8))^(3)cm^(3)xx(6.023xx10^(23)mol^(-1)))/((6.94gmol^(-1)))~=2`
`therefore` No. of atoms in a unit of Li = 2 (Body centred unit cell)
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