The density of a body-centered cubic (BCC) cr | vedclass

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(B) For a $BCC$ unit cell,the number of atoms per unit cell $n = 2$. Given: Density $\rho = 10.3 \ g \ cm^{-3}$,Molar mass $M = 95.94 \ g \ mol^{-1}$,

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$313$

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(B) For a $BCC$ unit cell,the number of atoms per unit cell $n = 2$. Given: Density $ho = 10.3 \ g \ cm^{-3}$,Molar mass $M = 95.94 \ g \ mol^{-1}$,Avogadro's number $N_A = 6.022 	imes 10^{23} \ mol^{-1}$. Using the formula: $ho = \frac{n 	imes M}{a^3 	imes N_A}$. Rearranging for $a^3$: $a^3 = \frac{n 	imes M}{ho 	imes N_A} = \frac{2 	imes 95.94}{10.3 	imes 6.022 	imes 10^{23}}$. $a^3 = \frac{191.88}{62.0266 	imes 10^{23}} \approx 3.0935 	imes 10^{-23} \ cm^3$. $a = (30.935 	imes 10^{-24})^{1/3} \approx 3.139 	imes 10^{-8} \ cm$. Converting to $pm$: $a = 3.139 	imes 10^{-8} \ cm 	imes 10^{10} \ pm/cm = 313.9 \ pm$.

The density of a body-centered cubic $(BCC)$ crystal of Molybdenum is $10.3 \ g \ cm^{-3}$. Calculate the edge length of the unit cell in $pm$. (Atomic mass of $Mo = 95.94 \ g \ mol^{-1}$) (in $.9$)

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