InterviewSolution
Saved Bookmarks
| 1. |
The diameter of zinc atom is `2.6 A`. 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. |
|
Answer» (a) Radius of zinc atom = `("Diameter")/(2)` = `(2.6 Å)/(2)` = `1.3 xx 10^(-10) m ` = `130 xx 10^(-12) m = 130 `pm (b) Length of the arrangement = `1.6` cm = `1.6 xx 10^(-2)` m Diameter of zinc atom = `2.6 xx 10^(-10) m ` `therefore` Number of zinc atoms present in the arrangement = `(1.6 xx 10^(-2) m)/(2.6 xx 10^(-10 m )` = `0.6153 xx 10^(8) m ` = `6.153 xx 10^(7) ` |
|