InterviewSolution
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InterviewSolution
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Niobium crystallize in a body centred cubic structure. If density is ` 8.55 " g cm"^(-3)` , calculate atomic radius of niobium, given that its atomic mass is 93n. |
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Answer» ` a^(3) = ( M xx Z)/(p xx N_(0) xx 10^(-30)) = ( 93 " g mol"^(-1) xxx2)/( 8.55 " g cm"^(-3) xx 6.02 xx 10^(23) "mol"^(-1) xx 10^(-30))= 3.61 xx 10^(7) = 36.1 xx 10^(7) = 36.1 xx 10^(6)` ` a = (36.1)^(1//3) xx 10^(2) "pm" = 3.304 xx 10^(2) "pm" = 330.4 "pm"` ` [x = (36.1)^(1//3) , log x = 1/3 log 36.1 = 1/3 xx 1.5575 = 0.519 or x = " antilog" 0.519 = 3.304`) For body - centred cubic , ` r= sqrt3/4 or = 0.433 a,= 0.433 xx 330.3 "pm" = 143.1 "pm" ` |
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