Question.
The diameter of zinc atom is $2.6 \mathrm{~A}$. Calculate (a) radius of zinc atom in $\mathrm{pm}$ and (b) number of atoms present in a length of $1.6 \mathrm{~cm}$ if the zinc atoms are arranged side by side lengthwise.
The diameter of zinc atom is $2.6 \mathrm{~A}$. Calculate (a) radius of zinc atom in $\mathrm{pm}$ and (b) number of atoms present in a length of $1.6 \mathrm{~cm}$ if the zinc atoms are arranged side by side lengthwise.
Solution:
(a) Radius of zinc atom $=\frac{\text { Diameter }}{2}$
$=\frac{2.6 \mathrm{~A}}{2}$
$=1.3 \times 10^{-10} \mathrm{~m}$
$=130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm}$
(b) Length of the arrangement $=1.6 \mathrm{~cm}$
$=1.6 \times 10^{-2} \mathrm{~m}$
Diameter of zinc atom $=2.6 \times 10^{-10} \mathrm{~m}$
$\therefore$ Number of zinc atoms present in the arrangement
$=\frac{1.6 \times 10^{-2} \mathrm{~m}}{2.6 \times 10^{-10} \mathrm{~m}}$
$=0.6153 \times 10^{8} \mathrm{~m}$
$=6.153 \times 10^{7}$
(a) Radius of zinc atom $=\frac{\text { Diameter }}{2}$
$=\frac{2.6 \mathrm{~A}}{2}$
$=1.3 \times 10^{-10} \mathrm{~m}$
$=130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm}$
(b) Length of the arrangement $=1.6 \mathrm{~cm}$
$=1.6 \times 10^{-2} \mathrm{~m}$
Diameter of zinc atom $=2.6 \times 10^{-10} \mathrm{~m}$
$\therefore$ Number of zinc atoms present in the arrangement
$=\frac{1.6 \times 10^{-2} \mathrm{~m}}{2.6 \times 10^{-10} \mathrm{~m}}$
$=0.6153 \times 10^{8} \mathrm{~m}$
$=6.153 \times 10^{7}$
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