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Niobium crystallises in body-centred cubic structure. If density is 8.55 "g cm"^(-3), calculate atomic radius of niobium using its atomic mass 93 u. |
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Answer» Solution :Density (d) = `8.55 "G cm"^(-3)` Atomic radius (r) = ? Number of atoms per unit cell (Z) = 2 By formula `d = (Z XX M)/(N_A xx a^3) ` we get `a^3 = (2 xx 93)/(8.55 xx 6.022 xx 10^(23))` `therefore a= 3.304 xx 10^(-8)` cm `therefore a = 330.4 "pm"` For BCC, `r = (SQRT(3))/(4) a` `r = (sqrt3)/(4) xx (330.4)` = 143.1 pm. |
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