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InterviewSolution
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The density of lead is `11.35 "g cm"^(-3)` and the metal crystallizes with fcc unit cell. Estimate the radius of lead atom. (At. Mass of lead = `207 "g mol"^(-1)` and `N_A=6.02xx10^23 "mol"^(-1)`). |
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Answer» Correct Answer - 174.7 pm `rho=(ZxxM)/(a^3xxN_A)` or `a^3=(ZxxM)/(N_Axxrho)=(4xx207 g mol^(-1))/(6.02xx10xx23 mol^(-1)xx11.35 g cm^(-3))=12.118 xx10^(-23) =121.18xx10^(-24)` or `a=(121.18)^(1//3)xx10^(-8)` Let `x=(121.18)^(1//3) therefore log x =1/3 log 121.18=1/3(2.083)=0.694` or x =antilog 0.694=4.943 `therefore a=4.943xx10^(-8) cm=494.3xx10^(-10) `cm=494.3 pm For fcc, `r=a/(2sqrt2)`=0.3535a=0.3535 x 494.3 pm =174.7 pm |
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