The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Solution and Explanation
(a) Radius of zinc atom = \(\frac{Diameter}{2}\) = \(\frac{26\,\AA}{2}\)
= 1.3 × 10-10 m = 130 × 10-12 m = 130 pm
(b) Length of the arrangement = 1.6 cm = 1.6 × 102 m
Diameter of zinc atom = 2.6 × 1010 m
∴ The number of zinc atoms present in the arrangement= \(\frac{1.6\times 10^{-2}}{2.6\times 10^{-10}}\) m
= 0.6153×108 m = 6.153×107
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