Niobium crystallizes in a body-centered cubic | vedclass

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(B) Given: Density $(d)$ = $8.55 \ g \ cm^{-3}$,$M_w = 93 \ g \ mol^{-1}$,$Z = 2$ (for $BCC$ structure),$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Using

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$1.43 	imes 10^{-8} \ cm$

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(B) Given: Density $(d)$ = $8.55 \ g \ cm^{-3}$,$M_w = 93 \ g \ mol^{-1}$,$Z = 2$ (for $BCC$ structure),$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Using the density formula: $d = \frac{Z 	imes M_w}{a^3 	imes N_A}$.$a^3 = \frac{Z 	imes M_w}{d 	imes N_A} = \frac{2 	imes 93}{8.55 	imes 6.022 	imes 10^{23}} \approx 3.61 	imes 10^{-23} \ cm^3$.$a = (36.1 	imes 10^{-24})^{1/3} \approx 3.305 	imes 10^{-8} \ cm$.For $BCC$ structure,the relation between radius $(r)$ and edge length $(a)$ is $4r = \sqrt{3}a$.$r = \frac{\sqrt{3} 	imes a}{4} = \frac{1.732 	imes 3.305 	imes 10^{-8}}{4} \approx 1.43 	imes 10^{-8} \ cm$.

Niobium crystallizes in a body-centered cubic $(BCC)$ structure. If the density is $8.55 \ g \ cm^{-3}$,calculate the atomic radius of niobium. (Atomic mass of niobium $M_w = 93 \ g \ mol^{-1}$)

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