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The edge length of unit cell of metal having molecular weight 75g/mol is $5 A^{0}$ which crystallizes in simple cubic lattice. If the density is 2g/cc then the radius of metal atom in pm is $x \times 10^{2}$ , then x is $[N_{A} = 6 \times 10^{23}]$

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Detailed Solution

$d = Z M / a^{3} N; 2 = 1 \times 75 / a^{3} \times 6 \times 10^{23}$

In simple cube $2 r = a; r = a / 2$

Original density =1

$\frac{d_{1}}{d_{2}} = \frac{{(a_{2})}^{3}}{5 \times 10^{- 8} \times 5 \times 10^{- 8} \times 5 \times 10^{- 8}}$

$\frac{1}{2} = \frac{a_{2}^{3}}{5 \times 10^{- 8} \times 5 \times 10^{- 8} \times 5 \times 10^{- 8}}$

On simplifying we get r =2pm

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