Lithium has a bcc structure. Its density is 5 | vedclass

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(D) For a $bcc$ structure, the number of atoms per unit cell $z = 2$. Density $d = 530 \ kg \ m^{-3} = 0.530 \ g \ cm^{-3}$. Atomic mass $M = 6.94 \ g

Vedclass · Accepted Answer

$352$

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(D) For a $bcc$ structure, the number of atoms per unit cell $z = 2$. Density $d = 530 \ kg \ m^{-3} = 0.530 \ g \ cm^{-3}$. Atomic mass $M = 6.94 \ g \ mol^{-1}$. Using the formula $d = \frac{z M}{a^3 N_A}$, we have $a^3 = \frac{z M}{d N_A}$. $a^3 = \frac{2 	imes 6.94 \ g \ mol^{-1}}{0.530 \ g \ cm^{-3} 	imes 6.02 	imes 10^{23} \ mol^{-1}}$. $a^3 = 4.352 	imes 10^{-23} \ cm^3$. $a = (43.52 	imes 10^{-24})^{1/3} \ cm = 3.517 	imes 10^{-8} \ cm$. Since $1 \ cm = 10^{10} \ pm$, $a = 3.517 	imes 10^{-8} 	imes 10^{10} \ pm = 351.7 \ pm \approx 352 \ pm$.

Lithium has a $bcc$ structure. Its density is $530 \ kg \ m^{-3}$ and its atomic mass is $6.94 \ g \ mol^{-1}$. Calculate the edge length of a unit cell of lithium metal in $pm$ $(N_A = 6.02 \times 10^{23} \ mol^{-1})$.

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