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Niobium crystallizes in body-centred cubic structure. If the density is `8.55 g cm^(-3)`, calculate the atomic radius of niobium using its atomic mass `93 u`. |
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Answer» Density, `rho = 8.55 g cm^(-3)` Atomic mass of niobrium, `Aw = 94.0` `rho = (Z_(eff) xx Aw)/(a^(3) xx N_(A))` or `a^(3) = (Z_(eff) xx Aw)/(rho xx N_(A))` Or `a^(3) = (2 xx 94)/(8.55 xx 6.02 xx 10^(23)) = (108)/(51.47) xx 10^(-23)` `= 3.27 xx 10^(-23) cm^(3)` Or `a = (3.27 xx 10^(-23))^(1//3) cm` `= 14.29 cm` |
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