An element is found to crystallize with BCC s | vedclass

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Question

(A) The density formula for a unit cell is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.For a $BCC$ structure,the number of atoms per unit cell is $Z = 2$.Gi

Vedclass · Accepted Answer

$(3.61 	imes 10^{-23})^{1/3} \ cm$

Vedclass · Answer

(A) The density formula for a unit cell is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.For a $BCC$ structure,the number of atoms per unit cell is $Z = 2$.Given: $d = 8.55 \ g \ cm^{-3}$,$M = 93 \ g \ mol^{-1}$,$N_A \approx 6.022 	imes 10^{23} \ mol^{-1}$.Rearranging for $a^3$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$.Substituting the values: $a^3 = \frac{2 	imes 93}{8.55 	imes 6.022 	imes 10^{23}} \approx 3.61 	imes 10^{-23} \ cm^3$.Therefore,the edge length $a = (3.61 	imes 10^{-23})^{1/3} \ cm$.

An element is found to crystallize with $BCC$ structure having density $8.55 \ g \ cm^{-3}$. What is the edge length of the unit cell? (Atomic mass of element $= 93$)

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