InterviewSolution
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InterviewSolution
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The density of lead is `11.35 g cm^(-3)` and the metal crystallizes with fee unit cell. Estimate the radius of lead atom. (At. Mass of lead `= 207 g mol^(-1) and NA = 6.02xx10^23 mol^(-1))` |
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Answer» Density of Lead `=11.35g//cm^3` Atomic mass of lead `=270g//mol` `N_A+6.02xx10^23mol^(-1)," Edges"=a" PM"=axx10^(-10)` Valume `=a^3xx10^(-30)cm^3` `"Density of the unit cell"=("Mass of unit cell")/(a^3xxN_Axx10^(-30))g//cm^3` `11.35g//cm^3=(4xx207g//mol)/(a^3xx6.02xx10^23mol)g//cm^3` `a^3=(4xx207cm^3)/(11.35xx6.02xx10^(-3))` `a^3=121.18xx10^(-24)cm^3` `a=4.948xx10^(-8)cm=494.8p m`. `"In" Fcc,r=(sqrt2a)/4=(1.414xx494.8)/4=175 p m` |
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