An element (atomic weight = 250 u) crystalli | vedclass

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(C) For a simple cubic lattice,the number of atoms per unit cell $(Z)$ is $1$. The formula for density $(d)$ is $d = \frac{Z 	imes M}{N_A 	imes a^3}

Vedclass · Accepted Answer

$1.93$

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(C) For a simple cubic lattice,the number of atoms per unit cell $(Z)$ is $1$. The formula for density $(d)$ is $d = \frac{Z 	imes M}{N_A 	imes a^3}$,where $M$ is the atomic weight,$N_A$ is Avogadro's number,and $a$ is the edge length. Substituting the values: $7.2 = \frac{1 	imes 250}{6.02 	imes 10^{23} 	imes a^3}$. $a^3 = \frac{250}{7.2 	imes 6.02 	imes 10^{23}} \approx 5.767 	imes 10^{-23} \ cm^3$. $a = (57.67 	imes 10^{-24})^{1/3} \approx 3.86 	imes 10^{-8} \ cm = 3.86 \ \mathring{A}$. For a simple cubic lattice,the radius $(r)$ is related to the edge length $(a)$ by $r = \frac{a}{2}$. $r = \frac{3.86}{2} = 1.93 \ \mathring{A}$.

An element (atomic weight $= 250 \ u$) crystallises in a simple cubic lattice. If the density of the unit cell is $7.2 \ g \ cm^{-3}$,what is the radius (in $\mathring{A}$) of the atom of the element? $(N_A = 6.02 \times 10^{23} \ mol^{-1})$

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