The unit cell of an element has an edge lengt | vedclass

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(B) The density formula is given by $ho = \frac{n 	imes M}{a^3 	imes N_A}$.Rearranging for the number of atoms per unit cell $(n)$: $n = \frac{h

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Body-centred cubic

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(B) The density formula is given by $ho = \frac{n 	imes M}{a^3 	imes N_A}$.Rearranging for the number of atoms per unit cell $(n)$: $n = \frac{ho 	imes N_A 	imes a^3}{M}$.Given: $ho = 4 \ g \ cm^{-3}$,$M = 149 \ g \ mol^{-1}$,$a = 5 \mathring{A} = 5 	imes 10^{-8} \ cm$,and $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $n = \frac{4 	imes 6.022 	imes 10^{23} 	imes (5 	imes 10^{-8})^3}{149}$.$n = \frac{4 	imes 6.022 	imes 10^{23} 	imes 125 	imes 10^{-24}}{149} = \frac{301.1}{149} \approx 2.02$.Since $n \approx 2$,the crystal structure is Body-centred cubic $(BCC)$.

The unit cell of an element has an edge length of $5 \mathring{A}$ and a density of $4 \ g \ cm^{-3}$. If its atomic mass is $149 \ g \ mol^{-1}$,identify the crystal structure.

Similar Questions

An element $A$ has a face-centred cubic $(fcc)$ structure with an edge length equal to $361 \ pm$. The radius of atom $A$ is ............... $pm$.

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